Complex Numbers

Prove that $\left(1 + i\right)^n + \left(1 - i\right)^n = 2 ^{\frac{n + 2}{2}} \cos \left(\dfrac{n \pi}{4}\right)$, $\;$ $n \in N$

Let $\;$ $\left(1 + i\right) = r \left(\cos \theta + i \sin \theta\right)$ $\;\;\; \cdots \; (1)$

Equating the real and imaginary parts separately, we have

$r \; \cos \theta = 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $r \; \sin \theta = 1$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $r = \sqrt{\left(1\right)^2 + \left(1\right)^2} = \sqrt{2}$

Substituting the value of $r$ in equations $(2a)$ and $(2b)$, we have

$\cos \theta = \dfrac{1}{\sqrt{2}}$, $\;\;$ $\sin \theta = \dfrac{1}{\sqrt{2}}$ $\implies$ $\theta = \dfrac{\pi}{4}$

Substituting the values of $r$ and $\theta$ in equation $(1)$ we have,

$\left(1 + i\right) = \sqrt{2} \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]$

$\begin{aligned} \therefore \; \left(1 + i\right)^n & = \left\{\sqrt{2} \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]\right\}^{n} \\\\ & = \left(\sqrt{2}\right)^n \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]^n \\\\ & = 2^{\frac{n}{2}} \left[\cos \left(\dfrac{n \pi}{4}\right) + i \sin \left(\dfrac{n \pi}{4}\right)\right] \;\;\; \cdots \; (3a) \end{aligned}$

Replacing $+i$ with $-i$ in equation $(3a)$, we have

$\left(1 - i\right)^n = 2^{\frac{n}{2}} \left[\cos \left(\dfrac{n \pi}{4}\right) - i \sin \left(\dfrac{n \pi}{4}\right)\right]$ $\;\;\; \cdots \; (3b)$

Adding equations $(3a)$ and $(3b)$, we have

$\left(1 + i\right)^n + \left(1 - i\right)^n = 2^{\frac{n}{2}} \left[2 \cos \left(\dfrac{n \pi}{4}\right)\right]$

i.e. $\;$ $\left(1 + i\right)^n + \left(1 - i\right)^n = 2 ^{\frac{n + 2}{2}} \cos \left(\dfrac{n \pi}{4}\right)$

Hence proved.

Complex Numbers

If $\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma$, prove that

1. $\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma = 3 \cos \left(\alpha + \beta + \gamma\right)$


2. $\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma = 3 \sin \left(\alpha + \beta + \gamma\right)$

Given $\;\;$ $\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma$ $\;\;\; \cdots \; (1)$

Let

$a = \cos \alpha + i \sin \alpha = e^{i \alpha}$ $\;\;\; \cdots \; (2a)$

$b = \cos \beta + i \sin \beta = e^{i \beta}$ $\;\;\; \cdots \; (2b)$

$c = \cos \gamma + i \sin \gamma = e^{i \gamma}$ $\;\;\; \cdots \; (2c)$

$\therefore \; a + b + c = \left(\cos \alpha + \cos \beta + \cos \gamma\right) + i \left(\sin \alpha + \sin \beta + \sin \gamma\right)$

i.e. $\;$ $a + b + c = 0$ $\;\;$ [from equation $(1)$] $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ We have from equation $(3)$, $\;$ $a^3 + b^3 + c^3 = 3 abc$ $\;\;\; \cdots \; (4)$

[Note:

$a + b + c = 0$ $\implies$ $a + b = -c$ $\;\;\; \cdots \; (A)$

$\therefore$ $\;$ $\left(a + b\right)^3 = \left(-c\right)^3$

i.e. $\;$ $a^3 + b^3 + 3ab \left(a + b\right) = - c^3$

i.e. $\;$ $a^3 + b^3 + c^3 = - 3 ab \left(a + b\right)$

i.e. $\;$ $a^3 + b^3 + c^3 = 3abc$ $\;$ (from equation $A$) ]

From equation $(2a)$,

$a^3 = \left(\cos \alpha + i \sin \alpha\right)^3 = \cos 3 \alpha + i \sin 3 \alpha$ $\;\;\; \cdots \; (5a)$

[By De Moivre's theorm: $\;$ $\left(\cos \theta + i \sin \theta \right)^n = \cos n \theta + i \sin n \theta$]

From equation $(2b)$,

$b^3 = \left(\cos \beta + i \sin \beta\right)^3 = \cos 3 \beta + i \sin 3 \beta$ $\;\;\; \cdots \; (5b)$

From equation $(2c)$,

$c^3 = \left(\cos \gamma + i \sin \gamma\right)^3 = \cos 3 \gamma + i \sin 3 \gamma$ $\;\;\; \cdots \; (5c)$

Adding equations $(5a)$, $(5b)$ and $(5c)$ we get,

$a^3 + b^3 + c^3 = \left(\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma\right)$

$\hspace{3cm}$ $+ i \left(\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma\right)$ $\;\;\; \cdots \; (6a)$

We have from equations $(2a)$, $(2b)$ and $(2c)$,

$\begin{aligned} 3abc & = 3 e^{i \alpha} \cdot e^{i \beta} \cdot e^{i \gamma} \\\\ & = 3 e^{i \left(\alpha + \beta + \gamma\right)} \\\\ & = 3 \left[\cos \left(\alpha + \beta + \gamma\right) + i \sin \left(\alpha + \beta + \gamma\right)\right] \\\\ & = 3 \cos \left(\alpha + \beta + \gamma\right) + 3 i \sin \left(\alpha + \beta + \gamma\right) \;\;\; \cdots \; (6b) \end{aligned}$

$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(4)$ becomes

$\left(\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma\right) + i \left(\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma\right)$

$\hspace{4cm}$ $= 3 \cos \left(\alpha + \beta + \gamma\right) + 3 i \sin \left(\alpha + \beta + \gamma\right)$ $\;\;\; \cdots \; (7)$

1. Equating the real parts on either side of equation $(7)$ we have,
   
   $\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma = 3 \cos \left(\alpha + \beta + \gamma\right)$



2. Equating the imaginary parts on either side of equation $(7)$ we have,
   
   $\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma = 3 \sin \left(\alpha + \beta + \gamma\right)$

Complex Numbers

Simplify: $\;\;\;$ $\dfrac{\left(\cos 2 \theta - i \sin 2 \theta\right)^7 \left(\cos 3 \theta + i \sin 3 \theta\right)^{-5}}{\left(\cos 4 \theta + i \sin 4 \theta\right)^{12} \left(\cos 5 \theta - i \sin 5 \theta\right)^{-6}}$

$\begin{aligned} \dfrac{\left(\cos 2 \theta - i \sin 2 \theta\right)^7 \left(\cos 3 \theta + i \sin 3 \theta\right)^{-5}}{\left(\cos 4 \theta + i \sin 4 \theta\right)^{12} \left(\cos 5 \theta - i \sin 5 \theta\right)^{-6}} & = \dfrac{\left(e^{-i2 \theta}\right)^7 \cdot \left(e^{i 3 \theta}\right)^{-5}}{\left(e^{i 4 \theta}\right)^{12} \cdot \left(e^{-i 5 \theta}\right)^{-6}} \\\\ & = \dfrac{\left(e^{- i 14 \theta}\right) \cdot \left(e^{- i 15 \theta}\right)}{\left(e^{i 48 \theta}\right) \cdot \left(e^{i 30 \theta}\right)} \\\\ & = \dfrac{e^{- i 29 \theta}}{e^{i 78 \theta}} \\\\ & = e^{- i 107 \theta} \\\\ & = \cos \left(107 \theta\right) - i \sin \left(107 \theta\right) \end{aligned}$

Complex Numbers

Solve the equation $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = 0$ $\;$ if $\;$ $3 + i$ $\;$ is a root.

Given: $\;\;$ One root is $\left(3 + i\right)$

$\implies$ $\left(3 - i\right)$ is also a root.

Sum of roots $= 6$

Product of roots $= \left(3 + i\right) \left(3 - i\right) = 9 - i^2 = 10$

$\therefore$ $\;$ The corresponding factor is $\;$ $x^2 - 6x + 10$

$\therefore$ $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = \left(x^2 - 6x + 10\right) \left(x^2 + \lambda x + 2\right)$ $\;\;\; \cdots \; (1)$

$\lambda$ $\;$ is a real number (constant).

Equating the coefficients of the $x$ term in equation $(1)$ we have,

$- 32 = -12 + 10 \lambda$

i.e. $\;$ $10 \lambda = -20$ $\implies$ $\lambda = -2$

Substituting the value of $\lambda$ in equation $(1)$ we have,

$x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = \left(x^2 - 6x + 10\right) \left(x^2 - 2x + 2\right)$

$\therefore$ $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = 0$ $\implies$ $x^2 - 6x + 10 = 0$ $\;$ or $\;$ $x^2 - 2x + 2 = 0$

Now,

$\begin{aligned} x^2 - 2x + 2 = 0 \implies x & = \dfrac{2 \pm \sqrt{4 - 8}}{2} \\\\ & = \dfrac{2 \pm 2i}{2} \\\\ & = 1 \pm i \end{aligned}$

$\therefore$ $\;$ The roots are $\left(3 \pm i\right)$ and $\left(1 \pm i\right)$.

Complex Numbers

$P$ represents the variable complex number $z$. Find the locus of $P$, if $\;$ $arg \left(\dfrac{z - 1}{z + 3}\right) = \dfrac{\pi}{4}$

Let $z = x + iy$

Given: $\;\;\;$ $arg \left(\dfrac{z - 1}{z + 3}\right) = \dfrac{\pi}{4}$

i.e. $\;$ $arg \left(z - 1\right) - arg \left(z + 3\right) = \dfrac{\pi}{4}$

i.e. $\;$ $arg \left[\left(x - 1\right) + i y\right] - arg \left[\left(x + 3\right) + i y\right] = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left(\dfrac{y}{x - 1}\right) - \tan^{-1} \left(\dfrac{y}{x + 3}\right) = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left[\dfrac{\dfrac{y}{x - 1} - \dfrac{y}{x + 3}}{1 + \left(\dfrac{y}{x - 1}\right) \left(\dfrac{y}{x + 3}\right)} \right] = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left[\dfrac{4y}{x^2 + 2x - 3 + y^2}\right] = \dfrac{\pi}{4}$

i.e. $\;$ $\dfrac{4y}{x^2 + y^2 + 2x - 3} = \tan \left(\dfrac{\pi}{4}\right) = 1$

i.e. $\;$ $x^2 + y^2 + 2x - 4y - 3 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required equation of locus of $P$.

Complex Numbers

$P$ represents the variable complex number $z$. Find the locus of $P$, if $\;$ $\left|z - 5i\right| = \left|z + 5i\right|$

Let $z = x + iy$

Then, $z - 5i = x + i \left(y - 5\right)$; $\;\;\;$ $z + 5i = x + i \left(y + 5\right)$

$\left|z - 5i\right| = \sqrt{\left(x^2\right) + \left(y - 5\right)^2} = \sqrt{x^2 + y^2 - 10 y + 25}$

$\left|z - 5i\right| = \sqrt{\left(x^2\right) + \left(y + 5\right)^2} = \sqrt{x^2 + y^2 + 10 y + 25}$

$\therefore$ $\;$ $\left|z - 5i\right| = \left|z + 5i\right|$ $\implies$ $\sqrt{x^2 + y^2 - 10 y + 25} = \sqrt{x^2 + y^2 + 10 y + 25}$

i.e. $\;$ $x^2 + y^2 - 10 y + 25 = x^2 + y^2 + 10 y + 25$

i.e. $\;$ $y = 0$

$\therefore$ $\;$ The locus of point $P$ is $y = 0$ $\;$ i.e. $\;$ the $X$ axis.

Complex Numbers

$P$ represents the variable complex number $z$.Find the locus of $P$, if $\;$ $Im \left[\dfrac{2z + 1}{iz + 1}\right] = -2$

Let $P$ be the point $z = x + iy$

Then,

$\begin{aligned} \dfrac{2z + 1}{iz + 1} & = \dfrac{\left(2x + 1\right) + i \; 2y}{\left(1 - y\right) + i \; x} \\\\ & = \dfrac{\left[\left(2x + 1\right) + i \; 2y\right] \left[\left(1 - y\right) - i \;x\right]}{\left(1 -y\right)^2 - \left(i \; x\right)^2} \\\\ & = \dfrac{\left(2x + 1\right) \left(1 - y\right) - i \; x \left(2x + 1\right) + i \; 2y \left(1 - y\right) - i^2 \; 2xy}{1 -2y + y^2 - i^2 \; x^2} \\\\ & = \dfrac{\left(2x + 1\right) \left(1 - y\right) + 2xy}{x^2 + y^2 - 2y + 1} + i \;\dfrac{2y \left(1 - y\right) - x \left(2x + 1\right)}{x^2 + y^2 - 2y + 1} \end{aligned}$

$\therefore$ $\;$ $Im \left[\dfrac{2z + 1}{iz + 1}\right] = \dfrac{2y - 2 y^2 -2 x^2 - x}{x^2 + y^2 - 2y + 1}$

Given $\;\;$ $\dfrac{2y - 2 y^2 -2 x^2 - x}{x^2 + y^2 - 2y + 1} = -2$

i.e. $\;$ $2 y - 2 y^2 - 2 x^2 - x = - 2 x^2 - 2 y^2 + 4 y - 2$

i.e. $\;$ $x + 2y - 2 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required equation of locus of $P$.

Complex Numbers

If $\;$ $arg \left(z -1\right) = \dfrac{\pi}{6}$ $\;$ and $\;$ $arg \left(z + 1\right) = \dfrac{2 \pi}{3}$, $\;$ then prove that $\;$ $\left|z\right| = 1$

Let $z = x + iy$

Then, $z -1 = \left(x - 1\right) + iy$ $\;$ and $\;$ $z + 1 = \left(x + 1\right) + iy$

Now, $\;$ $arg \left(z - 1\right) = \tan^{-1} \left(\dfrac{y}{x - 1}\right)$ $\;$ and $\;$ $arg \left(z + 1\right) = \tan^{-1} \left(\dfrac{y}{x + 1}\right)$

Given $\;\;$ $arg \left(z -1\right) = \dfrac{\pi}{6}$ $\;$ and $\;$ $arg \left(z + 1\right) = \dfrac{2 \pi}{3}$

$\implies$ $\tan^{-1} \left(\dfrac{y}{x - 1}\right) = \dfrac{\pi}{6}$; $\;$ $\tan^{-1} \left(\dfrac{y}{x + 1}\right) = \dfrac{2\pi}{3}$

$\implies$ $\dfrac{y}{x - 1} = \tan \left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}}$; $\;$ $\dfrac{y}{x + 1} = \tan \left(\dfrac{2 \pi}{3}\right) = - \sqrt{3}$

i.e. $\;$ $x - \sqrt{3} y = 1$ $\;\;\; \cdots \; (1)$; $\;\;$ $\sqrt{3}x + y = -\sqrt{3}$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously we have,

$4x = -2$ $\implies$ $x = - \dfrac{1}{2}$

Substituting the value of $x$ in equation $(2)$ we have,

$y = - \sqrt{3} + \dfrac{\sqrt{3}}{2} = - \dfrac{\sqrt{3}}{2}$

$\therefore$ $\;$ $z = x + iy = \dfrac{-1}{2} - \dfrac{\sqrt{3}}{2}y$

$\therefore$ $\;$ $\left|z\right| = \sqrt{\left(\dfrac{-1}{2}\right)^2 + \left(\dfrac{-\sqrt{3}}{2}\right)^2}$

i.e. $\;$ $\left|z\right| = \sqrt{\dfrac{1}{4} + \dfrac{3}{4}} = 1$

Complex Numbers

Express the complex number $-1 + i \sqrt{3}$ in its polar form.

The given complex number is $\;$ $z = -1 + i \sqrt{3}$

$z$ $\;$ is of the form $\;$ $x + iy$ $\;$ where $\;$ $x = -1$, $\;$ $y = \sqrt{3}$

Modulus of $z = \left|z\right| = \sqrt{x^2 + y^2} = \sqrt{\left(-1\right)^2 + \left(\sqrt{3}\right)^2} = 2$

$\because$ $\;$ $x$ is negative and $y$ is positive, argument or amplitude $\theta$ of $z$ lies in the second quadrant in the complex plane.

Let $\;$ $\alpha = \tan^{-1} \left(\dfrac{\left|y\right|}{\left|x\right|}\right) = \tan^{-1} \left(\dfrac{\left|\sqrt{3}\right|}{\left|-1\right|}\right) = \tan^{-1} \left(\sqrt{3}\right) = \dfrac{\pi}{3}$

$\therefore$ $\;$ $\theta = \pi - \alpha = \pi - \dfrac{\pi}{3} = \dfrac{2 \pi}{3}$

$\therefore$ $\;$ Polar form of $z = r \left(\cos \theta + i \sin \theta\right) = r \; cis \; \theta = 2 \; cis \; \left(\dfrac{2\pi}{3}\right)$

Complex Numbers

Prove that the triangle formed by the points representing the complex numbers $\left(10 + 8i\right)$, $\left(-2 + 4i\right)$ and $\left(-11 + 31i\right)$ on the Argand plane is right angled.

Let $A$, $B$ and $C$ represent the complex numbers $\left(10 + 8i\right)$, $\left(-2 + 4i\right)$ and $\left(-11 + 31i\right)$ respectively on the Argand diagram.

Now,

$\begin{aligned} AB & = \left|\left(10 + 8i\right) - \left(-2 + 4i\right)\right| \\\\ & = \left|12 + 4i\right| \\\\ & = \sqrt{\left(12\right)^2 + \left(4\right)^2} = \sqrt{160} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} BC & = \left|\left(-2 + 4i\right) - \left(-11 + 31i\right)\right| \\\\ & = \left|9 - 27i\right| \\\\ & = \sqrt{\left(9\right)^2 + \left(-27\right)^2} = \sqrt{810} \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} CA & = \left|\left(-11 + 31i\right) - \left(10 + 8i\right)\right| \\\\ & = \left|-21 + 23i\right| \\\\ & = \sqrt{\left(-21\right)^2 + \left(23\right)^2} = \sqrt{970} \;\;\; \cdots \; (3) \end{aligned}$

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$

$CA^2 = AB^2 + BC^2$

$\implies$ $\angle ABC = 90^{\circ}$

i.e. $\;$ $\triangle ABC$ is a right angled triangle.

Complex Numbers

Find the square root of $\left(-8-6i\right)$

Let $\sqrt{-8 - 6i} = x + iy$ $\;\;\; \cdots \; (1)$

Squaring equation $(1)$ we get,

$-8 - 6i = \left(x^2 - y^2\right) + 2 i xy$ $\;\;\; \cdots \; (2)$

Equating the real parts in equation $(2)$ we have,

$-8 = x^2 - y^2$ $\;\;\; \cdots \; (3a)$

Equating the imaginary parts in equation $(2)$ we have,

$-6 = 2xy$ $\implies$ $xy = -3$ $\;\;\; \cdots \; (3b)$

Now, $\left(x^2 + y^2\right)^2 = \left(x^2 - y^2\right)^2 + 4 x^2 y^2$

i.e. $\;$ $x^2 + y^2 = \sqrt{\left(x^2 - y^2\right)^2 + 4 x^2 y^2}$

$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$

$x^2 + y^2 = \sqrt{\left(-8\right)^2 + 4 \times \left(-3\right)^2} = \sqrt{100}$

i.e. $\;$ $x^2 + y^2 = 10$ $\;\;\; \cdots \; (4)$

Solving equations $(3a)$ and $(4)$ simultaneously we have,

$2x^2 = 2$ $\implies$ $x^2 = 1$ $\implies$ $x = \pm 1$

Substituting the values of $x^2$ in equation $(4)$ we get,

$y^2 = 9$ $\implies$ $y = \pm 3$

$\because$ $\;$ The product $xy$ is negative [equation $(3b)$]

$\implies$ when $x$ is +ve, $y$ is -ve and when $x$ is -ve, $y$ is +ve.

$\therefore$ $\;$ $\sqrt{-8-6i} = 1-3i$ $\;$ or $\;$ $\sqrt{-8-6i} = -1 + 3i$

Complex Numbers

If $\left(1 + i\right) \left(1 + 2i\right) \left(1 + 3i\right) \cdots \left(1 + ni\right) = x + iy$ $\;$ show that $\;$ $2 \cdot 5 \cdot 10 \cdots \left(1 + n^2\right) = x^2 + y^2$

Given: $\;$ $\left(1 + i\right) \left(1 + 2i\right) \left(1 + 3i\right) \cdots \left(1 + ni\right) = x + iy$

i.e. $\;$ $\left|\left(1 + i\right) \left(1 + 2i\right) \left(1 + 3i\right) \cdots \left(1 + ni\right)\right| = \left|x + iy\right|$

i.e. $\;$ $\left|1 + i\right| \; \left|1 + 2i\right| \; \left|1 + 3i\right| \cdots \left|1 + ni\right| = \left|x + iy\right|$

i.e. $\;$ $\sqrt{1^2 + 1^2} \; \sqrt{1^2 + 2^2} \; \sqrt{1^2 + 3^2} \cdots \sqrt{1^2 + n^2} = \sqrt{x^2 + y^2}$

i.e. $\;$ $\sqrt{2} \; \sqrt{5} \; \sqrt{10} \cdots \sqrt{1 + n^2} = \sqrt{x^2 + y^2}$ $\;\;\; \cdots \; (1)$

Squaring both sides of equation $(1)$ we have,

$2 \cdot 5 \cdot 10 \cdots \left(1 + n^2\right) = x^2 + y^2$

Hence proved.

Complex Numbers

For what values of $x$ and $y$, the numbers $-3 + i x^2 y$ and $x^2 + y + 4i$ are complex conjugate of each other?

Let the given numbers be

$z_1 = -3 + i x^2 y$ $\;$ and $\;$ $z_2 = x^2 + y + 4i$

Complex conjugate of $z_1 = \overline{z_1} = -3 - i x^2 y$

Given: $\;$ $\overline{z_1} = z_2$

i.e. $\;$ $-3 - i x^2 y = x^2 + y + 4i$ $\;\;\; \cdots \; (1)$

Equating the real parts on either side of equation $(1)$ we have,

$-3 = x^2 + y$ $\;\;\; \cdots \; (2a)$

Equating the imaginary parts on either side of equation $(1)$ we have,

$- x^2 y = 4$ $\;\;\; \cdots \; (2b)$

From equation $(2a)$, $\;$ $x^2 = -3 - y$ $\;\;\; \cdots \; (3)$

Substituting the value of $x^2$ in equation $(2b)$ we have,

$\left(3 + y\right) y = 4$

i.e. $\;$ $y^2 + 3y-4 = 0$

i.e. $\;$ $\left(y + 4\right) \left(y - 1\right) = 0$

$\implies$ $y = -4$ $\;$ or $\;$ $y = 1$

Substituting $y = -4$ in equation $(3)$ we have,

$x^2 = 1$ $\implies$ $x = \pm 1$

Substituting $y = 1$ in equation $(3)$ we have,

$x^2 = -4$ $\implies$ $x = \pm 2 i$

Complex Numbers

Find the real values of $x$ and $y$ for which the equation $\;$ $\dfrac{\left(1 + i\right)x - 2i}{3 + i} + \dfrac{\left(2 - 3i\right)y + i}{3 - i} = i$ $\;$ is satisfied.

The given equation is $\;\;$ $\dfrac{\left(1 + i\right)x - 2i}{3 + i} + \dfrac{\left(2 - 3i\right)y + i}{3 - i} = i$

i.e. $\;$ $\left[\left(1 + i\right)x - 2i\right] \left(3 - i\right) + \left[\left(2 - 3i\right)y + i\right] \left(3 + i\right) = i\left(3 + i\right) \left(3 - i\right)$

i.e. $\;$ $x \left(3 + 2 i - i^2\right) -6i + 2i^2 + y \left(6 - 7i -3i^2\right) + 3i + i^2 = 9i - i^3$

i.e. $\;$ $\left(4 + 2i\right)x + \left(9 - 7i\right)y -3i -3 = 10i$ $\;\;\;$ [$\because \; i^3 = -i$]

i.e. $\;$ $\left(4x + 9y - 3\right) + i \left(2x - 7y -13\right) = 0$

$\implies$ $4x + 9y - 3 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $2x - 7y -13 = 0$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously, we have

$23 y + 23 = 0$ $\implies$ $y = -1$

Substituting the value of $y$ in equation $(1)$, we get

$4x - 12 = 0$ $\implies$ $x = 3$

$\therefore$ $\;$ The real values of $x$ and $y$ for which the given equation is satisfied are $x = 3$ and $y = -1$

Complex Numbers

Find the least positive integer $n$ such that $\left(\dfrac{1 + i}{1 - i}\right)^n = 1$

$\begin{aligned} \dfrac{1 + i}{1 - i} & = \dfrac{\left(1 + i\right)^2}{\left(1 + i\right) \left(1 - i\right)} \\\\ & = \dfrac{1 + 2i + i^2}{1 - i^2} \\\\ & = \dfrac{2i}{2} = i \;\;\;\; \left[\because \; i^2 = -1\right] \end{aligned}$

$\therefore$ $\;$ $\left(\dfrac{1 + i}{1 - i}\right)^n = i^n$

$\therefore$ $\;$ We have $\;$ $i^n = 1$

Now, $i = \sqrt{-1}$ $\implies$ $i^4 = 1$

$\therefore$ $\;$ The least possible integer value of $n$ is $4$.

Complex Numbers

Find the real and imaginary parts of the complex number $\dfrac{2 + 5i}{4 - 3i}$

The given complex number is

$\begin{aligned} z & = \dfrac{2 + 5i}{4 - 3i} \\\\ & = \left(\dfrac{2 + 5i}{4 - 3i}\right) \times \left(\dfrac{4 + 3i}{4 + 3i}\right) \\\\ & = \dfrac{8 + 26 i + 15 i^2}{16 - 9 i^2} \\\\ & = \dfrac{-7 + 26 i}{25} \;\;\; \left[\because i^2 = -1\right] \\\\ & = \dfrac{-7}{25} + \dfrac{26i}{25} \end{aligned}$

$\therefore$ $\;$ Real part $= \dfrac{-7}{25}$; $\;\;$ Imaginary part $= \dfrac{26}{25}$

Vector Algebra

If $\;$ $A \left(-1, 4, -3\right)$ $\;$ is one end of diameter $AB$ of the sphere $\;$ $x^2 + y^2 + z^2 -3x -2y + 2z - 15 = 0$ $\;$ then find the coordinates of $B$.

Equation of given sphere: $\;$ $x^2 + y^2 + z^2 - 3x -2y + 2z - 15 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the general equation of sphere

$x^2 + y^2 + z^2 + 2ux + 2 vy + 2 wz + d = 0$ $\;$ gives

$u = \dfrac{-3}{2}, \;\; v = -1, \;\; w = 1, \;\; d = -15$

Center of the sphere $= \left(-u, -v, -w\right) = \left(\dfrac{3}{2}, 1, -1\right)$

Let the coordinates of point $B$ be $\left(\alpha, \beta, \gamma\right)$

$\because$ $\;$ $AB$ is the diameter of the sphere, we have

$\dfrac{-1 + \alpha}{2} = \dfrac{3}{2}$ $\implies$ $\alpha = 4$

$\dfrac{4 + \beta}{2} = 1$ $\implies$ $\beta = -2$

$\dfrac{-3 + \gamma}{2} = -1$ $\implies$ $\gamma = 1$

$\therefore$ $\;$ The coordinates of point $B$ are $\left(4, -2, 1\right)$

Vector Algebra

Find the vector and cartesian equation of the sphere on the join of the points $A$ and $B$ having position vectors $\;$ $2 \hat{i} + 6 \hat{j} - 7 \hat{k}$ $\;$ and $\;$ $-2 \hat{i} + 4 \hat{j} - 3 \hat{k}$ $\;$ respectively as diameter. Find also the center and radius of the sphere.

Position vector of point $A = \overrightarrow{a} = 2 \hat{i} + 6 \hat{j} - 7 \hat{k}$

Position vector of point $B = \overrightarrow{b} = -2 \hat{i} + 4 \hat{j} - 3 \hat{k}$

Let position vector of any point on the required sphere be $= \overrightarrow{r}$

Vector equation of the required sphere is

$\left(\overrightarrow{r} - \overrightarrow{a}\right) \cdot \left(\overrightarrow{r} - \overrightarrow{b}\right) = 0$

i.e. $\;$ $\left[\overrightarrow{r} - \left(2 \hat{i} + 6 \hat{j} - 7 \hat{k}\right)\right] \cdot \left[\overrightarrow{r} - \left(-2 \hat{i} + 4 \hat{j} - 3 \hat{k}\right)\right] = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required vector equation of the sphere.

Put $\;$ $\overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k}$ $\;$ on equation $(1)$. We have,

$\left[\left(x \hat{i} + y \hat{j} + z \hat{k}\right) - \left(2 \hat{i} + 6 \hat{j} - 7 \hat{k}\right)\right] \cdot \left[\left(x \hat{i} + y \hat{j} + z \hat{k}\right) - \left(-2 \hat{i} + 4 \hat{j} - 3 \hat{k}\right)\right] = 0$

i.e. $\;$ $\left[\left(x - 2\right) \hat{i} + \left(y - 6\right) \hat{j} + \left(z + 7\right) \hat{k}\right] \cdot \left[\left(x + 2\right) \hat{i} + \left(y - 4\right) \hat{j} + \left(z + 3\right) \hat{k}\right] = 0$

i.e. $\;$ $x^2 - 4 + y^2 - 10 y + 24 + z^2 + 10 z + 21 = 0$

i.e. $\;$ $x^2 + y^2 + z^2 -10 y + 10z + 41 = 0$ $\;\;\; \cdots \; (2)$

Equation $(2)$ is the required cartesian equation of the sphere.

Comparing equation $(2)$ with the general equation of sphere

$x^2 + y^2 + z^2 + 2ux + 2 vy + 2 wz + d = 0$ $\;$ gives

$u = 0, \;\; v = -5, \;\; w = 5, \;\; d = 41$

Center of the sphere $= \left(-u, -v, -w\right) = \left(0, 5, -5\right)$

$\begin{aligned} \text{Radius of sphere} & = \sqrt{u^2 + v^2 + w^2 -d} \\\\ & = \sqrt{\left(0\right)^2 + \left(-5\right)^2 + \left(5\right)^2 - 41} \\\\ & = \sqrt{9} = 3 \; units \end{aligned}$

Vector Algebra

Find the vector and cartesian equation of a sphere with center having position vector $2 \hat{i} - \hat{j} + 3 \hat{k}$ and radius 4 units.

Position vector of center of sphere $= \overrightarrow{c} = 2 \hat{i} - \hat{j} + 3 \hat{k}$

Radius of sphere $= a = 4 \;$ units

Position vector of any point on the sphere $= \overrightarrow{r}$

Vector equation of sphere is: $\;$ $\left|\overrightarrow{r} - \overrightarrow{c}\right| = a$

i.e. $\;$ $\left|\overrightarrow{r} - \left(2 \hat{i} - \hat{j} + 3 \hat{k}\right)\right| = 4$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required vector equation of the sphere.

Put $\;$ $\overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k}$ $\;$ in equation $(1)$. We have,

$\left|\left(x \hat{i} + y \hat{j} + z \hat{k}\right) - \left(2 \hat{i} - \hat{j} + 3 \hat{k}\right) \right| = 4$

i.e. $\;$ $\left|\left(x - 2\right) \hat{i} + \left(y + 1\right) \hat{j} + \left(z - 3\right) \hat{k}\right| = 4$

i.e. $\;$ $\left(x - 2\right)^2 + \left(y + 1\right)^2 + \left(z - 3\right)^2 = 4^2$

i.e. $\;$ $x^2 + y^2 + z^2 - 2x + 2y - 6z - 2 = 0$ $\;\;\; \cdots \; (2)$

Equation $(2)$ is the required cartesian equation of the sphere.

Vector Algebra

Find the angle between the line $\overrightarrow{r} = \hat{i} + \hat{j} + 3 \hat{k} + \lambda \left(2 \hat{i} + \hat{j} - \hat{k}\right)$ and the plane $\overrightarrow{r} \cdot \left(\hat{i} + \hat{j}\right) = 1$

Equation of given line is: $\;$ $\overrightarrow{r} = \hat{i} + \hat{j} + 3 \hat{k} + \lambda \left(2 \hat{i} + \hat{j} - \hat{k}\right)$

It is of the form: $\;$ $\overrightarrow{r}= \overrightarrow{a} + \lambda \overrightarrow{b}$

Here, $\overrightarrow{b} = 2 \hat{i} + \hat{j} - \hat{k}$

Equation of the given plane is: $\;$ $\overrightarrow{r} \cdot \left(\hat{i} + \hat{j}\right) = 1$

Normal to the given plane is $\;$ $\overrightarrow{n} = \hat{i} + \hat{j}$

Let $\theta$ be the angle between the given line and the plane.

Then,

$\begin{aligned} \sin \theta & = \dfrac{\overrightarrow{b} \cdot \overrightarrow{n}}{\left|\overrightarrow{b}\right| \; \left|\overrightarrow{n}\right|} \\\\ & = \dfrac{\left(2 \hat{i} + \hat{j} - \hat{k}\right) \cdot \left(\hat{i} + \hat{j}\right)}{\sqrt{\left(2\right)^2 + \left(1\right)^2 + \left(-1\right)^2} \; \sqrt{\left(1\right)^2 + \left(1\right)^2}} \\\\ & = \dfrac{3}{2 \sqrt{3}} = \dfrac{\sqrt{3}}{2} \end{aligned}$

$\therefore$ $\;$ $\theta = \sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{3}$

Vector Algebra

The planes $\overrightarrow{r} \cdot \left(2 \hat{i} + \lambda \hat{j} - 3 \hat{k}\right) = 10$ $\;$ and $\;$ $\overrightarrow{r} \cdot \left(\lambda \hat{i} + 3 \hat{j} + \hat{k}\right) = 5$ $\;$ are perpendicular. Find $\lambda$.

The equations of the given planes are

$\overrightarrow{r} \cdot \left(2 \hat{i} + \lambda \hat{j} - 3 \hat{k}\right) = 10$ $\;\;\; \cdots \; (1a)$

$\overrightarrow{r} \cdot \left(\lambda \hat{i} + 3 \hat{j} + \hat{k}\right) = 5$ $\;\;\; \cdots \; (2a)$

The normals to the given planes are

$\overrightarrow{n_1} = 2 \hat{i} + \lambda \hat{j} - 3 \hat{k}$ $\;\;\; \cdots \; (1b)$

$\overrightarrow{n_2} = \lambda \hat{i} + 3 \hat{j} + \hat{k}$ $\;\;\; \cdots \; (2b)$

Let the angle between the planes be $\theta$.

Now, $\cos \theta = \dfrac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right| \left|\overrightarrow{n_2}\right|}$

$\because$ $\;$ The two planes are perpendicular, $\theta = \dfrac{\pi}{2}$ $\;$ and $\;$ $\cos \theta = 0$

$\implies$ $\overrightarrow{n_1} \cdot \overrightarrow{n_2} = 0$

i.e. $\;$ $\left(2 \hat{i} + \lambda \hat{j} - 3 \hat{k}\right) \cdot \left(\lambda \hat{i} + 3 \hat{j} + \hat{k}\right) = 0$

i.e. $\;$ $2 \lambda + 3 \lambda - 3 = 0$ $\implies$ $\lambda = \dfrac{3}{5}$

Vector Algebra

Find the angle between the planes $\;$ $2x + y -z = 9$ $\;$ and $\;$ $x + 2y + z = 7$

The given planes are

$2x + y -z = 9$ $\;\;\; \cdots \; (1a)$

$x + 2y + z = 7$ $\;\;\; \cdots \; (2a)$

The normals to the given planes are

$\overrightarrow{n_1} = 2 \hat{i} + \hat{j} - \hat{k}$ $\;\;\; \cdots \; (1b)$

$\overrightarrow{n_2} = \hat{i} + 2 \hat{j} + \hat{k}$ $\;\;\; \cdots \; (2b)$

If $\;$ $\theta$ $\;$ is the angle between the planes, then

$\begin{aligned} \cos \theta & = \dfrac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right| \left|\overrightarrow{n_2}\right|} \\\\ & = \dfrac{\left(2 \hat{i} + \hat{j} - \hat{k}\right) \cdot \left(\hat{i} + 2 \hat{j} + \hat{k}\right)}{\sqrt{\left(2\right)^2 + \left(1\right)^2 + \left(-1\right)^2} \; \sqrt{\left(1\right)^2 + \left(2\right)^2 + \left(1\right)^2}} \\\\ & = \dfrac{2 + 2 -1}{\sqrt{6} \; \sqrt{6}} \end{aligned}$

$\therefore$ $\;$ $\theta = \cos^{-1} \left(\dfrac{3}{6}\right) = \dfrac{\pi}{3}$

Vector Algebra

Find the distance between the parallel planes $x - y + 3z + 5 = 0$ $\;$ and $\;$ $2x - 2y + 6z + 7 = 0$

Equations of the given planes are

$x - y + 3z + 5 = 0$ $\;\;\; \cdots \; (1)$

$2x - 2y + 6z + 7 = 0$ $\implies$ $x - y + 3z + \dfrac{7}{2} = 0$ $\;\;\; \cdots \; (2)$

Equations $(1)$ and $(2)$ are of the form

$ax + by + cz + d_1 = 0$ $\;$ and $\;$ $ax + by + cz + d_2 = 0$ $\;$ respectively.

Here $\;$ $a = 1, \; b = -1, \; c = 3, \; d_1 = 5, \; d_2 = \dfrac{7}{2}$

Distance between the parallel planes $(1)$ and $(2)$ is

$\begin{aligned} \left|\dfrac{d_1 - d_2}{\sqrt{a^2 + b^2 + c^2}}\right| & = \left|\dfrac{5 - \dfrac{7}{2}}{\sqrt{\left(1\right)^2 + \left(-1\right)^2 + \left(3\right)^2}}\right| \\\\ & = \left|\dfrac{3}{2 \sqrt{1 + 1 + 9}}\right| \\\\ & = \dfrac{3}{2 \sqrt{11}} \;\; units \end{aligned}$

Vector Algebra

Find the distance from the origin to the plane $\overrightarrow{r} \cdot \left(2\hat{i} - \hat{j} + 5 \hat{k}\right) = 7$

Vector equation of given plane is: $\;$ $\overrightarrow{r} \cdot \left(2\hat{i} - \hat{j} + 5 \hat{k}\right) = 7$

$\therefore$ $\;$ The cartesian equation of the given plane is: $\;$ $2x - y + 5z = 7$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is of the form $\;$ $ax + by + cz + d = 0$

where $\;$ $a = 2, \; b = -1, \; c = 5, \; d = -7$

Now, distance from the origin to the plane given by equation $(1)$ is

$\begin{aligned} \left|\dfrac{d}{\sqrt{a^2 + b^2 + c^2}}\right| & = \left|\dfrac{-7}{\sqrt{\left(2\right)^2 + \left(-1\right)^2 + \left(5\right)^2}}\right| \\\\ & = \left|\dfrac{-7}{\sqrt{4 + 1 + 25}}\right| \\\\ & = \dfrac{7}{30} \; units \end{aligned}$

Vector Algebra

Find the point of intersection of the line $\overrightarrow{r} = \left(\overrightarrow{j} - \overrightarrow{k}\right) + \lambda \left(2 \hat{i} - \hat{j} + \hat{k}\right)$ and the $xz$ plane.

Equation of given line in vector form: $\;$ $\overrightarrow{r} = \left(\overrightarrow{j} - \overrightarrow{k}\right) + \lambda \left(2 \hat{i} - \hat{j} + \hat{k}\right)$

$\therefore$ $\;$ Equation of given line in cartesian form is: $\;$ $\dfrac{x}{2} = \dfrac{y - 1}{-1} = \dfrac{z + 1}{1}$ $\;\;\; \cdots \; (1)$

The given line meets the $xz$ plane i.e. $\;$ $y = 0$

Substituting $y = 0$ in equation $(1)$ gives

$\dfrac{x}{2} = 1$ $\implies$ $x = 2$

$\dfrac{z + 1}{1} = 1$ $\implies$ $z = 0$

$\therefore$ $\;$ The point of intersection of the given line and the plane is $\;$ $\left(2, 0, 0\right)$

Vector Algebra

Can a plane be drawn through the lines $\overrightarrow{r} = \left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right) + t \left(2 \hat{i} + 3 \hat{j} + 6 \hat{k}\right)$ and $\overrightarrow{r} = \left(3 \hat{i} + 3 \hat{j} - 5 \hat{k}\right) + s \left(-2 \hat{i} + 3 \hat{j} + 8 \hat{k}\right)$?

The equations of the given lines are

$L_1: \;\; \overrightarrow{r} = \left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right) + t \left(2 \hat{i} + 3 \hat{j} + 6 \hat{k}\right)$

$L_2: \;\; \overrightarrow{r} = \left(3 \hat{i} + 3 \hat{j} - 5 \hat{k}\right) + s \left(-2 \hat{i} + 3 \hat{j} + 8 \hat{k}\right)$

Lines $L_1$ and $L_2$ are of the form

$\overrightarrow{r} = \overrightarrow{a_1} + t \overrightarrow{b_1}$ $\;$ and $\;$ $\overrightarrow{r}= \overrightarrow{a_2} + s \overrightarrow{b_2}$ $\;$ respectively.

Here,

$\overrightarrow{a_1} = \hat{i} + 2 \hat{j} - 4 \hat{k}$, $\;$ $\overrightarrow{b_1} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$

$\overrightarrow{a_2} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k}$, $\;$ $\overrightarrow{b_2} = -2 \hat{i} + 3 \hat{j} + 8 \hat{k}$

Now,

$\begin{aligned} \overrightarrow{a_2} - \overrightarrow{a_1} & = \left(3 \hat{i} + 3 \hat{j} - 5 \hat{k}\right) - \left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right) \\\\ & = 2 \hat{i} + \hat{j} - \hat{k} \end{aligned}$

$\begin{aligned} \overrightarrow{b_1} \times \overrightarrow{b_2} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ -2 & 3 & 8 \end{vmatrix} \\\\ & = \hat{i} \left(24 - 18\right) - \hat{j} \left(16 + 12\right) + \hat{k} \left(6 + 6\right) \\\\ & = 6 \hat{i} - 28 \hat{j} + 12 \hat{k} \end{aligned}$

$\begin{aligned} \left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \cdot \left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) & = \left(2 \hat{i} + \hat{j} - \hat{k}\right) \cdot \left(6 \hat{i} - 28 \hat{j} + 12 \hat{k}\right) \\\\ & = 12 - 28 - 12 = -28 \neq 0 \end{aligned}$

$\because$ $\;$ $\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \cdot \left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \neq 0$ $\implies$ lines $L_1$ and $L_2$ are not coplanar.

$\because$ $\;$ The given lines are not coplanar, a plane cannot be drawn through the lines $L_1$ and $L_2$.

Vector Algebra

Find the cartesian equation of the plane which contains the lines $\dfrac{x+1}{2} = \dfrac{y - 2}{-3} = \dfrac{z - 3}{4}$ and $\dfrac{x - 4}{3} = \dfrac{y - 1}{2} = z - 8$

Equations of the given lines are

$L_1: \;\; \dfrac{x+1}{2} = \dfrac{y - 2}{-3} = \dfrac{z - 3}{4}$

$L_2: \;\; \dfrac{x - 4}{3} = \dfrac{y - 1}{2} = \dfrac{z - 8}{1}$

Lines $L_1$ and $L_2$ are of the form

$\dfrac{x - x_1}{a_1} = \dfrac{y - y_1}{b_1} = \dfrac{z - z_1}{c_1}$ $\;$ and $\;$ $\dfrac{x - x_2}{a_2} = \dfrac{y - y_2}{b_2} = \dfrac{z - z_2}{c_2}$ $\;$ respectively.

Here

$x_1 = -1, \; y_1 = 2, \; z_1 = 3$; $\;$ $x_2 = 4, \; y_2 = 1, \; z_2 = 8$

$a_1 = 2, \; b_1 = -3, \; c_1 = 4$; $\;$ $a_2 = 3, \; b_2 = 2, \; c_2 = 1$

Now,

$\begin{aligned} \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} & = \begin{vmatrix} 5 & -1 & 5 \\ 2 & -3 & 4 \\ 3 & 2 & 1 \end{vmatrix} \\\\ & = 5 \left(-3 - 8\right) + \left(2 - 12\right) + 5 \left(4 + 9\right) \\\\ & = -55 - 10 + 65 = 0 \end{aligned}$

$\therefore$ $\;$ The given lines are coplanar.

$\therefore$ $\;$ The equation of plane containing the given lines is

$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$

i.e. $\;$ $\begin{vmatrix} x +1 & y - 2 & z - 3 \\ 2 & -3 & 4 \\ 3 & 2 & 1 \end{vmatrix} = 0$

i.e. $\;$ $\left(x + 1\right) \left(-3 - 8\right) - \left(y - 2\right) \left(2 - 12\right) + \left(z - 3\right) \left(4 + 9\right) = 0$

i.e. $\;$ $- 11x + 10 y + 13 z - 70 = 0$

i.e. $\;$ $11x - 10 y - 13 z + 70 = 0$

Vector Algebra

Find the parametric form of vector equation and cartesian equation of the plane passing through the points with position vectors $\;$ $3 \hat{i} + 4 \hat{j} + 2 \hat{k}$, $\;$ $2 \hat{i} - 2 \hat{j} - \hat{k}$ $\;$ and $\;$ $7 \hat{i} + \hat{k}$.

Let the position vectors of the three given points be

$\overrightarrow{a} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}$, $\;\;\;$ $\overrightarrow{b} = 2 \hat{i} - 2 \hat{j} - \hat{k}$, $\;\;\;$ $\overrightarrow{c} = 7 \hat{i} + \hat{k}$

Let $\;$ $\lambda$ $\;$ and $\;$ $\mu$ $\;$ be two scalars.

Parametric form of vector equation of a plane passing through three non-collinear points is

$\overrightarrow{r} = \left(1 - \lambda - \mu\right) \overrightarrow{a} + \lambda \overrightarrow{b} + \mu \overrightarrow{c}$

$\begin{aligned} i.e. \; \overrightarrow{r} & = \left(1 - \lambda - \mu\right) \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right) + \lambda \left(2 \hat{i} - 2 \hat{j} - \hat{k}\right) + \mu \left(7 \hat{i} + \hat{k}\right) \\\\ & = \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right) + \lambda \left[\left(2 \hat{i} - 2 \hat{j} - \hat{k}\right) - \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right)\right] \\\\ & \hspace{4cm} + \mu \left[\left(7 \hat{i} + \hat{k}\right) - \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right)\right] \\\\ & = \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right) + \lambda \left(- \hat{i} - 6 \hat{j} - 3 \hat{k}\right) + \mu \left(4 \hat{i} - 4 \hat{j} - \hat{k}\right) \;\;\; \cdots \; (1) \end{aligned}$

Equation $(1)$ is the parametric form of vector equation of the required plane.

Position vector $\overrightarrow{a}$ represents the point $A \left(x_1, y_1, z_1\right) = \left(3, 4, 2\right)$

Position vector $\overrightarrow{b}$ represents the point $B \left(x_2, y_2, z_2\right) = \left(2, -2, -1\right)$

Position vector $\overrightarrow{c}$ represents the point $C \left(x_3, y_3, z_3\right) = \left(7, 0, 1\right)$

Cartesian equation of the required plane is

$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0$

i.e. $\;$ $\begin{vmatrix} x - 3 & y - 4 & z - 2 \\ 2 - 3 & -2 - 4 & -1 - 2 \\ 7 - 3 & 0 - 4 & 1 - 2 \end{vmatrix} = 0$

i.e. $\;$ $\begin{vmatrix} x - 3 & y - 4 & z - 2 \\ -1 & -6 & -3 \\ 4 & -4 & -1 \end{vmatrix} = 0$

i.e. $\;$ $\left(x - 3\right) \left(6 - 12\right) - \left(y - 4\right) \left(1 + 12\right) + \left(z - 2\right) \left(4 + 24\right) = 0$

i.e. $\;$ $-6x - 13 y + 28 z + 14 = 0$

i.e. $\;$ $6x + 13 y - 28 z - 14 = 0$ $\;\;\; \cdots \; (2)$

Equation $(2)$ is the cartesian equation of the required plane.

Vector Algebra

Find the parametric vector equation and cartesian equation of the plane through the points $\left(1,2,3\right)$ and $\left(2,3,1\right)$ and perpendicular to the plane $3x - 2y + 4z - 5 = 0$.

Let the given points be $A \left(x_1, y_1, z_1\right) = \left(1,2,3\right)$ and $B \left(x_2, y_2, z_2\right) = \left(2, 3, 1\right)$

position vector of point $A = \overrightarrow{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ $\;\;\; \cdots \; (1a)$

position vector of point $B = \overrightarrow{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ $\;\;\; \cdots \; (1b)$

Equation of given plane is $3x - 2y + 4z - 5 = 0$ $\;\;\; \cdots \; (2a)$

$\because$ $\;$ The given plane is perpendicular to the required plane, the normal vector to the given plane is parallel to the required plane.

Normal vector to the given plane [equation $(2a)$] is

$\overrightarrow{v} = 3 \hat{i} - 2 \hat{j} + 4 \hat{k}$ $\;\;\; \cdots \; (2b)$

Parametric vector equation of a plane passing through two given points [$(1a)$ and $(1b)$] and parallel to the vector $(2b)$ is

$\overrightarrow{r} = \left(1 - s\right) \overrightarrow{a} + s \overrightarrow{b} + t \overrightarrow{v}$ $\;\;$ where $s$ and $t$ are scalars.

$\begin{aligned} i.e. \;\; \overrightarrow{r} & = \left(1 - s\right) \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right) + s \left(2 \hat{i} + 3 \hat{j} + \hat{k}\right) + t \left(3 \hat{i} - 2 \hat{j} + 4 \hat{k}\right) \\\\ & = \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right) + s \left[\left(2 \hat{i} + 3 \hat{j} + \hat{k}\right) - \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right)\right] + t \left(3 \hat{i} - 2 \hat{j} + 4 \hat{k}\right) \\\\ & = \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right) + s \left(\hat{i} + \hat{j} - 2 \hat{k}\right) + t \left(3 \hat{i} - 2 \hat{j} + 4 \hat{k}\right) \;\;\; \cdots \; (3) \end{aligned}$

Equation $(3)$ is the parametric vector equation of the required plane.

For the given plane [equation $(2a)$], $\;$ $\ell_1 = 3, \; m_1 = -2, \; n_1 = 4$

Cartesian equation of the plane is

$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ \ell_1 & m_1 & n_1 \end{vmatrix} = 0$

i.e. $\;$ $\begin{vmatrix} x - 1 & y - 2 & z - 3 \\ 2 - 1 & 3 - 2 & 1 - 3 \\ 3 & -2 & 4 \end{vmatrix} = 0$

i.e. $\;$ $\begin{vmatrix} x - 1 & y - 2 & z - 3 \\ 1 & 1 & -2 \\ 3 & -2 & 4 \end{vmatrix} = 0$

i.e. $\;$ $\left(x - 1\right) \left(4 - 4\right) - \left(y - 2\right) \left(4 + 6\right) + \left(z - 3\right) \left(-2 -3\right) = 0$

i.e. $\;$ $- 10 \left(y - 2\right) - 5 \left(z - 3\right) = 0$

i.e. $\;$ $2y + z - 7 = 0$ $\;\;\; \cdots \; (4)$

Equation $(4)$ is the cartesian equation of the required plane.

Vector Algebra

Find the non-parametric vector equation, parametric vector equation and cartesian equation of the plane through the point $\left(-1,3,2\right)$ and perpendicular to the planes $x + 2y + 2z = 5$ and $3x + y + 2z = 8$

Non-parametric vector equation of plane

The equation of a plane passing through a given point $\overrightarrow{a}$ and perpendicular to two given planes $\overrightarrow{r} \cdot \overrightarrow{n_1} = d_1$ and $\overrightarrow{r} \cdot \overrightarrow{n_2} = d_2$ is

$\left(\overrightarrow{r} - \overrightarrow{a}\right) \cdot \left(\overrightarrow{n_1} \times \overrightarrow{n_2}\right) = 0$

The required plane pass through the point $A \left(x_1,y_1,z_1\right) = \left(-1,3,2\right)$

$\therefore$ $\;$ position vector of point $A$ is $\overrightarrow{a} = - \hat{i} + 3 \hat{j} + 2 \hat{k}$

Cartesian equations of given planes are

$x + 2y + 2z = 5$ $\;\;\; \cdots \; (1a)$

$3x + y + 2z = 8$ $\;\;\; \cdots \; (2a)$

$\therefore$ $\;$ Vector equations of the given planes are

$\overrightarrow{r} \cdot \left(\hat{i} + 2 \hat{j} + 2 \hat{k}\right) = 5$ $\;\;\; \cdots \; (1b)$

$\overrightarrow{r} \cdot \left(3 \hat{i} + \hat{j} + 2 \hat{k}\right) = 8$ $\;\;\; \cdots \; (2b)$

Let $\;\;$ $\overrightarrow{n_1} = \hat{i} + 2 \hat{j} + 2 \hat{k}$; $\;\;$ $\overrightarrow{n_2} = 3 \hat{i} + \hat{j} + 2 \hat{k}$; $\;\;$ $d_1 = 5$; $\;\;$ $d_2 = 8$

Now,

$\begin{aligned} \overrightarrow{n_1} \times \overrightarrow{n_2} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 1 & 2 \end{vmatrix} \\\\ & = 2 \hat{i} + 4 \hat{j} - 5 \hat{k} \end{aligned}$

$\therefore$ $\;$ The non-parametric vector equation of the required plane is

$\left[\overrightarrow{r} - \left(- \hat{i} + 3 \hat{j} + 2 \hat{k}\right)\right] \cdot \left(2 \hat{i} + 4 \hat{j} - 5 \hat{k}\right) = 0$

i.e. $\;$ $\overrightarrow{r} \cdot \left(2 \hat{i} + 4 \hat{j} - 5 \hat{k}\right) = \left(- \hat{i} + 3 \hat{j} + 2 \hat{k}\right) \cdot \left(2 \hat{i} + 4 \hat{j} - 5 \hat{k}\right)$

i.e. $\;$ $\overrightarrow{r} \cdot \left(2 \hat{i} + 4 \hat{j} - 5 \hat{k}\right) = 0$

Parametric vector equation of plane

The normal vector to plane $(1a)$ is $\;\;$ $\overrightarrow{u} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ $\;\;\; \cdots \; (1c)$

The normal vector to plane $(1a)$ is $\;\;$ $\overrightarrow{v} = 3 \hat{i} + \hat{j} + 2 \hat{k}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ The required plane passes through the point $A$ and is parallel to the vectors $(1c)$ and $(2c)$.

$\therefore$ $\;$ Parametric vector equation of the plane is

$\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{u} + \mu \overrightarrow{v}$ $\;\;$ [$\lambda$ and $\mu$ are scalars]

i.e. $\;$ $\overrightarrow{r} = \left(- \hat{i} + 3 \hat{j} + 2 \hat{k}\right) + \lambda \left(\hat{i} + 2 \hat{j} + 2 \hat{k}\right) + \mu \left(3 \hat{i} + \hat{j} + 2 \hat{k}\right)$

Cartesian equation of plane

Equations of given planes are $(1a)$ and $(2a)$.

They are of the form $\;\;$ $a_1 x + b_y + c_1z + d_1 = 0$ $\;$ and $\;$ $a_2 x + b_2 y + c_2 z + d_2 = 0$ respectively.

$\therefore$ $\;$ $a_1 = 1, \; b_1 = 2, \; c_1 = 2, \; d_1 = -5$ $\;$ and $\;$ $a_2 = 3, \; b_2 = 1, \; c_2 = 2, \; d_2 = - 8$

Equation of the required plane through point $A$ is

$a \left(x + 1\right) + b \left(y - 3\right) + c \left(z - 2\right) = 0$ $\;\;\; \cdots \; (3a)$

$\because$ $\;$ The required plane is perpendicular to the given planes, we have

$aa_1 + bb_1 + c c_1 = 0$

i.e. $\;$ $a + 2b + 2c = 0$ $\;\;\; \cdots \; (3b)$

and $\;$ $a a_2 + b b_2 + c c_2 = 0$

i.e. $\;$ $3a + b + 2c = 0$ $\;\;\; \cdots \; (3c)$

Eliminating $a$, $b$ and $c$ from equations $(3a)$, $(3b)$ and $(3c)$, the required equation of plane is

$\begin{vmatrix} x + 1 & y - 3 & z - 2 \\ 1 & 2 & 2 \\ 3 & 1 & 2 \end{vmatrix} = 0$

i.e. $\;$ $2 \left(x + 1\right) + 4 \left(y - 3\right) - 5 \left(z - 2\right) = 0$

i.e. $\;$ $2x + 4y - 5z = 0$

Vector Algebra

Find the vector and cartesian equation of the plane through the point $\left(1,3,2\right)$ and parallel to the lines $\dfrac{x + 1}{2} = \dfrac{y + 2}{-1} = \dfrac{z+3}{3}$ and $\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z + 2}{2}$

The required plane passes through the point $A \left(1,3,2\right)$

$\therefore$ $\;$ position vector of point $A$ is $\overrightarrow{a} = \hat{i} + 3 \hat{j} + 2 \hat{k}$

The given equations of lines are

$\dfrac{x + 1}{2} = \dfrac{y + 2}{-1} = \dfrac{z+3}{3}$ $\;\;\; \cdots \; (1a)$

$\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z + 2}{2}$ $\;\;\; \cdots \; (2a)$

Vector form of equation $(1a)$ is

$\overrightarrow{r} = \left(-\hat{i}-2 \hat{j} -3 \hat{k}\right) + \lambda \left(2 \hat{i} - \hat{j} + 3 \hat{k}\right)$ $\;\;\; \cdots \; (1b)$

Vector form of equation $(2a)$ is

$\overrightarrow{r} = \left(2 \hat{i} - \hat{j} - 2 \hat{k}\right) + \mu \left(\hat{i} + 2 \hat{j} + 2 \hat{k}\right)$ $\;\;\; \cdots \; (2b)$

Required vector equation of plane in vector form is

$\overrightarrow{r} = \overrightarrow{a} + s \overrightarrow{u} + t \overrightarrow{v}$

where $\;$ $\overrightarrow{u} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ $\;$ and $\;$ $\overrightarrow{v} = \hat{i} + 2 \hat{j} + 2 \hat{k}$; $\;\;\;$ $s$ and $t$ are scalars.

i.e. $\;$ $\overrightarrow{r} = \hat{i} + 3 \hat{j} + 2 \hat{k} + s \left(2 \hat{i} - \hat{j} + 3 \hat{k}\right) + t \left(\hat{i} + 2 \hat{j} + 2 \hat{k}\right)$

Now,

$A \left(x_1, y_1, z_1\right) = \left(1, 3, 2\right)$

direction ratios of line $(1a)$ are $\; \ell_1 = 2, \; m_1 = -1, \; n_1 = 3$

direction ratios of line $(2a)$ are $\; \ell_2 = 1, \; m_2 = 2, \; n_2 = 2$

Equation of the required plane in cartesian coordinates is

$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ \ell_1 & m_1 & n_1 \\ \ell_2 & m_2 & n_2 \end{vmatrix} = 0$

i.e. $\;$ $\begin{vmatrix} x - 1 & y - 3 & z - 2 \\ 2 & -1 & 3 \\ 1 & 2 & 2 \end{vmatrix} = 0$

i.e. $\;$ $\left(x - 1\right) \left(-2 - 6\right) - \left(y - 3\right) \left(4 - 3\right) + \left(z - 2\right) \left(4 + 1\right) = 0$

i.e. $\;$ $-8x + 8 - y + 3 + 5z - 10 =0$

i.e. $\;$ $8x + y - 5z = 1$

Vector Algebra

Find the cartesian equation of the plane containing the line $\dfrac{x - 2}{2} = \dfrac{y - 2}{3} = \dfrac{z - 1}{3}$ and parallel to the line $\dfrac{x + 1}{3} = \dfrac{y - 1}{2} = \dfrac{z + 1}{1}$

The equation of any plane through the line $\;$ $\dfrac{x - 2}{2} = \dfrac{y - 2}{3} = \dfrac{z -1}{3}$ $\;$ is

$A \left(x - 2\right) + B \left(y - 2\right) + C \left(z - 1\right) = 0$ $\;\;\; \cdots \; (1)$

where $\;$ $2 A + 3 B + 3 C = 0$ $\;\;\; \cdots \; (2)$

$\because$ $\;$ Plane given by equation $(1)$ is parallel to the line $\;$ $\dfrac{x + 1}{3} = \dfrac{y - 1}{2} = \dfrac{z + 1}{1}$,

the normal to equation $(1)$ is perpendicular to $(1)$ and has direction ratios $3, \; 2, \; 1$.

$\therefore$ $\;$ $3 A + 2 B + C = 0$ $\;\;\; \cdots \; (3)$

Solving equations $(2)$ and $(3)$ we get

$\dfrac{A}{\begin{vmatrix} 3 & 3 \\ 2 & 1 \end{vmatrix}} = \dfrac{-B}{\begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix}} = \dfrac{C}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}} = k$ (say)

i.e. $\dfrac{A}{3 - 6} = \dfrac{B}{9 - 2} = \dfrac{C}{4 - 9} = k$

i.e. $A = - 3k$, $\;$ $B = 7 k$, $\;$ $C = -5k$

Substituting the values of $A$, $B$ and $C$ in equation $(1)$, we have

$-3k \left(x - 2\right) + 7 k \left(y - 2\right) - 5 k \left(z - 1\right) = 0$

$\implies$ $-3x + 7y - 5z = 3$

i.e. $\;$ $3x - 7y + 5z + 3 = 0$ $\;\;\; \cdots \; (4)$

Equation $(4)$ is the required equation of the plane in cartesian form.

Vector Algebra

The foot of the perpendicular drawn from the origin to a plane is $\left(8, -4, 3\right)$. Find the equation of the plane.

Let $O$ be the origin and $N \left(8, -4, 3\right)$ be the foot of perpendicular from $O$ to the given plane.

Let $P \left(x, y, z\right)$ be an arbitrary point on the plane.

Then,

direction ratios of $\overrightarrow{NP}$ are: $\;$ $\left(x - 8\right)$, $\left(y + 4\right)$, $\left(z - 3\right)$

direction ratios of $\overrightarrow{ON}$ are: $\;$ $8, \; -4, \; 3$

But $\overrightarrow{ON} \perp \overrightarrow{NP}$.

$\therefore$ $\;$ $\overrightarrow{ON} \cdot \overrightarrow{NP} = 0$

i.e. $\;$ $8 \left(x - 8\right) - 4 \left(y + 4\right) + 3 \left(z - 3\right) = 0$

$\therefore$ $\;$ the required equation of the plane is

$8x - 4y + 3z = 89$

Vector Algebra

Find the length of the perpendicular from the origin to the plane $\;$ $\overrightarrow{r} \cdot \left(3 \hat{i} + 4 \hat{j} + 12 \hat{k}\right) = 26$

Equation of the given plane is $\;$ $\overrightarrow{r} \cdot \left(3 \hat{i} + 4 \hat{j} + 12 \hat{k}\right) = 26$

which is of the form $\;$ $\overrightarrow{r} \cdot \overrightarrow{n} = d$

where $\;$ $\overrightarrow{n} = 3 \hat{i} + 4 \hat{j} + 12 \hat{k}$, $\;$ $d = 26 > 0$

Length of the perpendicular from the origin to the plane is

$\dfrac{d}{\left|\overrightarrow{n}\right|} = \dfrac{26}{\sqrt{\left(3\right)^2 + \left(4\right)^2 + \left(12\right)^2}} = \dfrac{26}{13} = 2$ units

Vector Algebra

Find the unit normal vectors to the plane $2x - y + 2z = 5$.

Equation of the plane is

$2x - y + 2z = 5$

Its vector equation is

$\overrightarrow{r} \cdot \left(2 \hat{i} - \hat{j} + 2 \hat{k}\right) = 5$ $\;\;\;$ [$\overrightarrow{r}$ is a vector lying in the plane.]

A normal vector to the plane is

$\overrightarrow{n} = 2 \hat{i} - \hat{j} + 2 \hat{k}$

$\therefore$ $\;$ Unit normal vectors to the plane are

$\begin{aligned} \hat{n} & = \pm \left(\dfrac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\right) \\\\ & = \pm \left(\dfrac{2 \hat{i} - \hat{j} + 2 \hat{k}}{\sqrt{\left(2\right)^2 + \left(-1\right)^2 + \left(2\right)^2}}\right) \\\\ & = \pm \left(\dfrac{2 \hat{i} - \hat{j} + 2 \hat{k}}{3}\right) \end{aligned}$

Vector Algebra

Find the vector and cartesian equations of a plane which is at a distance of 18 units from the origin and which is normal to the vector $2 \hat{i} + 7 \hat{j} + 8 \hat{k}$.

Vector equation of a plane at a distance $p$ from the origin and normal to unit vector $\hat{n}$ is

$\overrightarrow{r} \cdot \hat{n} = p$ $\;\;\; \cdots \; (1)$

Here, $p = 18$ units;

$\overrightarrow{n} = 2 \hat{i} + 7 \hat{j} + 8 \hat{k}$

Now,

Unit vector $\hat{n} = \dfrac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}$

i.e. $\;$ $\hat{n} = \dfrac{2 \hat{i} + 7 \hat{j} + 8 \hat{k}}{\sqrt{\left(2\right)^2 + \left(7\right)^2 + \left(8\right)^2}} = \dfrac{2 \hat{i} + 7 \hat{j} + 8 \hat{k}}{\sqrt{117}} = \dfrac{2 \hat{i} + 7 \hat{j} + 8 \hat{k}}{3 \sqrt{13}}$

$\therefore$ $\;$ The required vector equation of the plane is [from equation $(1)$]

$\overrightarrow{r} \cdot \left(\dfrac{2 \hat{i} + 7 \hat{j} + 8 \hat{k}}{3\sqrt{13}}\right) = 18$ $\;\;\; \cdots \; (2)$

Putting $\overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k}$ $\;$ in equation $(2)$, the cartesian equation of the plane is

$\left(x \hat{i} + y \hat{j} + z \hat{k}\right) \cdot \left(2 \hat{i} + 7 \hat{j} + 8 \hat{k}\right) = 54 \sqrt{13}$

i.e. $\;$ $2x + 7 y + 8 z = 54 \sqrt{13}$

Vector Algebra

If the points $\left(\lambda, 0, 3\right)$, $\left(1, 3 , -1\right)$ and $\left(-5, -3, 7\right)$ are collinear, then find $\lambda$.

Let $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be the position vectors of the points $\left(\lambda, 0, 3\right)$, $\left(1, 3 , -1\right)$ and $\left(-5, -3, 7\right)$ respectively.

Then,

$\overrightarrow{a} = \lambda \hat{i} + 3 \hat{k}$

$\overrightarrow{b} = \hat{i} + 3 \hat{j} - \hat{k}$

$\overrightarrow{c} = -5 \hat{i} - 3 \hat{j} + 7 \hat{k}$

$\because$ $\;$ The points are collinear, the position vectors of the points are coplanar.

The position vectors are coplanar $\implies$ $\left[\overrightarrow{a} \; \overrightarrow{b} \; \overrightarrow{c}\right] = 0$

Now,

$\left[\overrightarrow{a} \; \overrightarrow{b} \; \overrightarrow{c}\right] = 0$ $\implies$ $\begin{vmatrix} \lambda & 0 & 3 \\ 1 & 3 & -1 \\ -5 & -3 & 7 \end{vmatrix} = 0$

i.e. $\;$ $\lambda \left(21 - 3\right) + 3 \left(- 3 + 15\right) = 0$

i.e. $\;$ $18 \lambda = -36$ $\implies$ $\lambda = -2$

Vector Algebra

Show that the lines $\;$ $\dfrac{x-1}{1} = \dfrac{y + 1}{-1} = \dfrac{z}{3}$ $\;$ and $\;$ $\dfrac{x-2}{1} = \dfrac{y - 1}{2} = \dfrac{-z -1}{1}$ $\;$ intersect and find their point of intersection.

Equations of given lines are

$\dfrac{x-1}{1} = \dfrac{y + 1}{-1} = \dfrac{z}{3}$ $\;\;\; \cdots \; (1a)$

$\dfrac{x-2}{1} = \dfrac{y - 1}{2} = \dfrac{-z -1}{1}$ $\;\;\; \cdots \; (1b)$

Let $\ell_1$, $m_1$ and $n_1$ be the direction ratios of equation $(1a)$.

Let $\ell_2$, $m_2$ and $n_2$ be the direction ratios of equation $(1b)$.

Comparing equations $(1a)$ and $(1b)$ with

$\dfrac{x - x_1}{\ell_1} = \dfrac{y - y_1}{m_1} = \dfrac{z - z_1}{n_1}$ $\;$ and $\;$ $\dfrac{x - x_2}{\ell_2} = \dfrac{y - y_2}{m_2} = \dfrac{z - z_2}{n_2}$ $\;$ respectively, we have

$x_1 = 1$, $y_1 = -1$, $z_1 = 0$; $\;\;$ $x_2 = 2$, $y_2 = 1$, $z_2 = -1$

$\ell_1 = 1$, $m_1 = -1$, $n_1 = 3$; $\;\;$ $\ell_2 = 1$, $m_2 = 2$, $n_2 = -1$

The condition for intersecting lines is

$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ \ell_1 & m_1 & n_1 \\ \ell_2 & m_2 & n_2 \end{vmatrix} = 0$ $\;\;\; \cdots \; (2)$

Now,

$\begin{aligned} \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ \ell_1 & m_1 & n_1 \\ \ell_2 & m_2 & n_2 \end{vmatrix} & = \begin{vmatrix} 2 - 1 & 1 + 1 & -1 - 0 \\ 1 & -1 & 3 \\ 1 & 2 & -1 \end{vmatrix} \\\\ & = \begin{vmatrix} 1 & 2 & -1 \\ 1 & -1 & 3 \\ 1 & 2 & -1 \end{vmatrix} \\\\ & = 1 \left(1 - 6\right) - 2 \left(-1 -3\right) -1 \left(2 + 1\right) \\\\ & = -5 + 8 -3 = 0 \end{aligned}$

$\because$ $\;$ equation $(2)$ is satisfied by the given lines, the given lines are intersecting.

Let $\;$ $\dfrac{x-1}{1} = \dfrac{y + 1}{-1} = \dfrac{z}{3} = \lambda$ $\;\;\; \cdots \; (3a)$

$\therefore$ $\;$ Any point on line $(3a)$ is of the form $\;$ $\left(\lambda + 1, \; - \lambda - 1, \; 3 \lambda\right)$

Let $\dfrac{x-2}{1} = \dfrac{y - 1}{2} = \dfrac{-z -1}{1} = \mu$ $\;\;\; \cdots \; (3b)$

$\therefore$ $\;$ Any point on line $(3b)$ is of the form $\;$ $\left(\mu + 2, \; 2\mu + 1, \; -\mu - 1\right)$

$\because$ $\;$ The two given lines are intersecting, for some $\lambda$, $\mu$

$\left(\lambda + 1, \; - \lambda - 1, \; 3 \lambda\right) = \left(\mu + 2, \; 2\mu + 1, \; -\mu - 1\right)$

$\implies$ $\lambda + 1 = \mu + 2$ $\;$ i.e. $\;$ $\lambda - \mu = 1$ $\;\;\; \cdots \; (4a)$

and $\;$ $3 \lambda = -\mu - 1$ $\;$ i.e. $\;$ $3 \lambda + \mu = -1$ $\;\;\; \cdots \; (4b)$

From equations $(4a)$ and $(4b)$, $\;$ $\lambda = 0$, $\;$ $\mu = -1$

Taking either $\lambda = 0$ or $\mu = -1$, the point of intersection of the two lines is $\;$ $\left(1, -1, 0\right)$

Vector Algebra

Show that the lines $\;$ $\overrightarrow{r} = \left(3 \hat{i} + 5 \hat{j} + 7 \hat{k}\right) + t \left(\hat{i} - 2 \hat{j} + \hat{k}\right)$ $\;$ and $\;$ $\overrightarrow{r} = \left(\hat{i} + \hat{j} + \hat{k}\right) + s \left(7 \hat{i} + 6 \hat{j} + 7 \hat{k}\right)$ $\;$ are skew lines and find the shortest distance between the lines.

The given lines are

$\overrightarrow{r} = \left(3 \hat{i} + 5 \hat{j} + 7 \hat{k}\right) + t \left(\hat{i} - 2 \hat{j} + \hat{k}\right)$ $\;\;\; \cdots \; (1a)$

$\overrightarrow{r} = \left(\hat{i} + \hat{j} + \hat{k}\right) + s \left(7 \hat{i} + 6 \hat{j} + 7 \hat{k}\right)$ $\;\;\; \cdots \; (1b)$

Comparing equations $(1a)$ and $(1b)$ with

$\overrightarrow{r} = \overrightarrow{a_1} + t \; \overrightarrow{u}$ $\;$ and $\;$ $\overrightarrow{r} = \overrightarrow{a_2} + s \; \overrightarrow{v}$ $\;$ respectively, we have

$\overrightarrow{a_1} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k}$ $\;\;\; \cdots \; (2a)$

$\overrightarrow{a_2} = \hat{i} + \hat{j} + \hat{k}$ $\;\;\; \cdots \; (2b)$

$\overrightarrow{u} = \hat{i} - 2 \hat{j} + \hat{k}$ $\;\;\; \cdots \; (2c)$

$\overrightarrow{v} = 7 \hat{i} + 6 \hat{j} + 7 \hat{k}$ $\;\;\; \cdots \; (2d)$

Condition for two lines are skew lines to be skew: $\;$ $\left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right] \neq 0$

Now,

$\begin{aligned} \overrightarrow{a_2} - \overrightarrow{a_1} & = \left(\hat{i} + \hat{j} + \hat{k}\right) - \left(3 \hat{i} + 5 \hat{j} + 7 \hat{k}\right) \\\\ & = - 2 \hat{i} - 4 \hat{j} - 6 \hat{k} \end{aligned}$

$\begin{aligned} \therefore \; \left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right] & = \begin{vmatrix} -2 & -4 & -6 \\ 1 & -2 & 1 \\ 7 & 6 & 7 \end{vmatrix} \\\\ & = -2 \left(-14 - 6\right) + 4 \left(7 - 7\right) - 6 \left(6 + 14\right) \\\\ & = 40 - 120 \\\\ & = - 80 \neq 0 \end{aligned}$

$\because$ $\;$ $\left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right] \neq 0$, the given lines are skew lines.

Shortest distance between the lines $= d = \dfrac{\left|\left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right]\right|}{\left|\overrightarrow{u} \times \overrightarrow{v}\right|}$ $\;\;\; \cdots \; (3)$

Now, $\left|\left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right]\right| = \left|-80\right| = 80$ $\;\;\; \cdots \; (4a)$

$\begin{aligned} \overrightarrow{u} \times \overrightarrow{v} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 7 & 6 & 7 \end{vmatrix} \\\\ & = \hat{i} \left(-14 - 6\right) - \hat{j} \left(7 - 7\right) + \hat{k} \left(6 + 14\right) \\\\ & = - 20 \hat{i} + 20 \hat{k} \end{aligned}$

$\therefore$ $\;$ $\left|\overrightarrow{u} \times \overrightarrow{v}\right| = \sqrt{\left(-20\right)^2 + \left(20\right)^2} = \sqrt{800} = 20 \sqrt{2}$ $\;\;\; \cdots \; (4b)$

$\therefore$ $\;$ From equations $(3)$, $(4a)$ and $(4b)$, the shortest distance between the lines is

$d = \dfrac{80}{20 \sqrt{2}} = 2 \sqrt{2}$

Vector Algebra

Find the shortest distance between the parallel lines $\;$ $\dfrac{x - 1}{-1} = \dfrac{y}{3} = \dfrac{z + 3}{2}$ $\;$ and $\;$ $\dfrac{x - 3}{-1} = \dfrac{y + 1}{3} = \dfrac{z - 1}{2}$

The equations of the lines in Cartesian form are

$\dfrac{x - 1}{-1} = \dfrac{y}{3} = \dfrac{z + 3}{2}$ $\;\;\; \cdots \; (1a)$

$\dfrac{x - 3}{-1} = \dfrac{y + 1}{3} = \dfrac{z - 1}{2}$ $\;\;\; \cdots \; (1b)$

Vector form of equation $(1a)$ is

$\overrightarrow{r} = \left(\hat{i} - 3 \hat{k}\right) + \lambda \left(-\hat{i} + 3 \hat{j} + 2 \hat{k}\right)$ $\;\;\; \cdots \; (2a)$

Vector form of equation $(1b)$ is

$\overrightarrow{r} = \left(3 \hat{i} - \hat{j} + \hat{k} \right) + \mu \left(- \hat{i} + 3 \hat{j} + 2 \hat{k}\right)$ $\;\;\; \cdots \; (2b)$

Comparing equations $(2a)$ and $(2b)$ with

$\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{u}$ $\;$ and $\;$ $\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{u}$ $\;$ respectively, we have

$\overrightarrow{a_1} = \hat{i} - 3 \hat{k}$ $\;\;\; \cdots \; (3a)$

$\overrightarrow{a_2} = 3 \hat{i} - \hat{j} + \hat{k}$ $\;\;\; \cdots \; (3b)$

$\overrightarrow{u} = - \hat{i} + 3 \hat{j} + 2 \hat{k}$ $\;\;\; \cdots \; (3c)$

Shortest distance between parallel lines $= d = \dfrac{\left|\overrightarrow{u} \times \left(\overrightarrow{a_2} - \overrightarrow{a_1}\right)\right|}{\left|\overrightarrow{u}\right|}$ $\;\;\; \cdots \; (4)$

$\begin{aligned} \overrightarrow{a_2} - \overrightarrow{a_1} & = \left(3 \hat{i} - \hat{j} + \hat{k}\right) - \left(\hat{i} - 3 \hat{k}\right) \\\\ & = 2 \hat{i} - \hat{j} + 4 \hat{k} \end{aligned}$

$\begin{aligned} \therefore \; \overrightarrow{u} \times \left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 2 \\ 2 & -1 & 4 \end{vmatrix} \\\\ & = \hat{i} \left(12 + 2\right) - \hat{j} \left(-4 - 4\right) + \hat{k} \left(1 - 6\right) \\\\ & = 14 \hat{i} + 8 \hat{j} - 5 \hat{k} \end{aligned}$

$\therefore$ $\;$ $\left|\overrightarrow{u} \times \left(\overrightarrow{a_2} - \overrightarrow{a_1}\right)\right| = \sqrt{\left(14\right)^2 + \left(8\right)^2 + \left(-5\right)^2} = \sqrt{196 + 64 + 25} = \sqrt{285}$

$\left|\overrightarrow{u}\right| = \sqrt{\left(-1\right)^2 + \left(3\right)^2 + \left(2\right)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$ $\;\;\; \cdots \; (5b)$

Substituting equations $(5a)$ and $(5b)$ in equation $(4)$, we have

$d = \sqrt{\dfrac{285}{14}}$

Vector Algebra

Find the angle between the lines $\;$ $\overrightarrow{r} = 5 \hat{i} - 7 \hat{j} + \mu \left(- \hat{i} + 4 \hat{j} + 2 \hat{k}\right)$ $\;$ and $\;$ $\overrightarrow{r} = -2 \hat{i} + \hat{k} + \lambda \left(3 \hat{i} + 4 \hat{k}\right)$

Let the given lines be in the direction of $\overrightarrow{u}$ and $\overrightarrow{v}$

Then, $\;$ $\overrightarrow{u} = - \hat{i} + 4 \hat{j} + 2 \hat{k}$ $\;$ and $\;$ $\overrightarrow{v} = 3 \hat{i} + 4 \hat{k}$

Let $\theta$ be the angle between the two lines.

[Note: Angle between two lines is defined as the angle between their directions.]

Now, $\;$ $\cos \theta = \left(\dfrac{\overrightarrow{u} \cdot \overrightarrow{v}}{\left|\overrightarrow{u}\right| \left|\overrightarrow{v}\right|}\right)$

$\begin{aligned} \overrightarrow{u} \cdot \overrightarrow{v} & = \left(- \hat{i} + 4 \hat{j} + 2 \hat{k}\right) \cdot \left(3 \hat{i} + 4 \hat{k}\right) \\\\ & = -3 + 8 = 5 \end{aligned}$

$\left|\overrightarrow{u}\right| = \sqrt{\left(-1\right)^2 + \left(4\right)^2 + \left(2\right)^2} = \sqrt{21}$

$\left|\overrightarrow{v}\right| = \sqrt{\left(3\right)^2 + \left(4\right)^2} = \sqrt{25} = 5$

$\therefore$ $\;$ $\cos \theta = \left(\dfrac{5}{5 \sqrt{21}}\right)$

$\therefore$ $\;$ $\theta = \cos^{-1} \left(\dfrac{1}{\sqrt{21}}\right)$

Vector Algebra

Find the angle between the lines $\;$ $\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z - 4}{6}$ $\;$ and $\;$ $x + 1 = \dfrac{y + 2}{2} = \dfrac{z - 4}{2}$

The two lines are

$\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z - 4}{6}$ $\;\;\; \cdots \; (1)$

$\dfrac{x + 1}{1} = \dfrac{y + 2}{2} = \dfrac{z - 4}{2}$ $\;\;\; \cdots \; (2)$

Let $a_1$, $b_1$ and $c_1$ be the direction ratios of line $(1)$.

Let $a_2$, $b_2$ and $c_2$ be the direction ratios of line $(2)$.

Then, $\;$ $a_1 = 2$, $b_1 = 3$, $c_1 = 6$; $\;$ $a_2 = 1$, $b_2 = 2$, $c_2 = 2$

Let $\theta$ be the angle between the lines.

Then,

$\begin{aligned} \theta & = \cos^{-1} \left[\dfrac{a_1a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right] \\\\ & = \cos^{-1} \left[\dfrac{2 \times 1 + 3 \times 2 + 6 \times 2}{\sqrt{\left(2\right)^2 + \left(3\right)^2 + \left(6\right)^2 } \sqrt{\left(1\right)^2 + \left(2\right)^2 + \left(2\right)^2}}\right] \\\\ & = \cos^{-1} \left[\dfrac{2 + 6 + 12}{\sqrt{4 + 9+ 36} \sqrt{1 + 4 + 4}}\right] \\\\ & = \cos^{-1} \left[\dfrac{20}{\sqrt{49} \sqrt{9}}\right] \\\\ & = \cos^{-1} \left[\dfrac{20}{21}\right] \end{aligned}$

Vector Algebra

Find the vector and cartesian equation of the line through the point $\left(3, -4, -2\right)$ and parallel to the vector $9 \hat{i} + 6 \hat{j} + 2 \hat{k}$.

Let $\overrightarrow{a}$ be the position vector of the point $\left(3, -4, -2\right)$.

Then, $\;$ $\overrightarrow{a} = 3 \hat{i} - 4 \hat{j} - 2 \hat{k}$

Let $\;$ $\overrightarrow{m} = 9 \hat{i} + 6 \hat{j} + 2 \hat{k}$

Then, the vector equation of the required line is $\;\;$ $\overrightarrow{r} = a + \lambda \overrightarrow{m}$ $\;\;$ where $\lambda$ is a constant.

i.e. $\;$ $\overrightarrow{r} = \left(3 \hat{i} - 4 \hat{j} - 2 \hat{k}\right) + \lambda \left(9 \hat{i} + 6 \hat{j} + 2 \hat{k}\right)$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required vector equation of the line.

Now, $\overrightarrow{r}$ is the position vector of any point $P \left(x, y, z\right)$ on the line.

i.e. $\;$ $\overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k}$

$\begin{aligned} \therefore \; x \hat{i} + y \hat{j} + z \hat{k} & = \left(3 \hat{i} - 4 \hat{j} - 2 \hat{k}\right) + \lambda \left(9 \hat{i} + 6 \hat{j} + 2 \hat{k}\right) \\\\ & = \left(3 + 9 \lambda\right) \hat{i} + \left(-4 + 6 \lambda\right) \hat{j} + \left(-2 + 2 \lambda\right) \hat{k} \end{aligned}$

Comparing the coefficients of $\hat{i}$, $\hat{j}$ and $\hat{k}$ on both sides, we get

$x = 3 + 9 \lambda$, $\;\;$ $y = -4 + 6 \lambda$ $\;\;$ $z = -2 + 2 \lambda$

Eliminating $\lambda$,

$\dfrac{x - 3}{9} = \dfrac{y + 4}{6} = \dfrac{z + 2}{2}$ $\;\;\; \cdots \; (2)$

Equation $(2)$ gives the equation of the line in cartesian form.

Vector Algebra

Find the direction cosines of the line joining $\left(2, -3, 1\right)$ and $\left(3, 1, -2\right)$.

Let the given points be $\;$ $A \left(2, -3, 1\right)$ and $B \left(3, 1, -2\right)$.

Then,

$\begin{aligned} \overrightarrow{AB} & = \text{position vector of B} - \text{position vector of A} \\\\ & = \left(3 \hat{i} + \hat{j} - 2 \hat{k}\right) - \left(2 \hat{i} - 3 \hat{j} + \hat{k}\right) \\\\ & = \hat{i} + 4 \hat{j} -3 \hat{k} \end{aligned}$

$\therefore$ $\;$ direction ratios of $\overrightarrow{AB}$ are $\;\;$ $1, \; 4, \; -3$

Let the direction cosines of $\overrightarrow{AB}$ be $\;\;$ $k, \; 4k, \; -3k$ $\;$ where $k$ is a constant.

Then, $\;$ $\left(k\right)^2 + \left(4k\right)^2 + \left(-3k\right)^2 = 1$

i.e. $\;$ $26 k^2 = 1$ $\implies$ $k = \pm \dfrac{1}{\sqrt{26}}$

$\therefore$ $\;$ The direction cosines of $\overrightarrow{AB}$ are $\;$ $\pm \left(\dfrac{1}{\sqrt{26}}, \; \dfrac{4}{\sqrt{26}}, \; \dfrac{-3}{\sqrt{26}}\right)$

Vector Algebra

A vector $\overrightarrow{r}$ has length $35 \sqrt{2}$ and direction ratios $\left(3, 4, 5\right)$. Find the direction cosines and the components of $\overrightarrow{r}$, assuming that $\overrightarrow{r}$ makes an acute angle with the $X$ axis.

Given: $\;$ $\left|\overrightarrow{r}\right| = 35 \sqrt{2}$

Let $\ell$, $m$ and $n$ be the direction cosines of $\overrightarrow{r}$.

$\because$ $\;$ $\overrightarrow{r}$ makes an acute angle with the $X$ axis, $\ell > 0$.

Direction ratios are $\;$ $3$, $4$, $5$.

We have $\;$ $\sqrt{\left(3\right)^2 + \left(4\right)^2 + \left(5\right)^2} = \sqrt{50} = 5 \sqrt{2}$

$\therefore$ $\;$ direction cosines of $\overrightarrow{r}$ are $\;$ $\ell = \dfrac{3}{5 \sqrt{2}}$, $\;$ $m = \dfrac{4}{5 \sqrt{2}}$, $\;$ $n = \dfrac{5}{5 \sqrt{2}}$

Now,

$\begin{aligned} \overrightarrow{r} & = \left|\overrightarrow{r}\right| \left(\ell \hat{i} + m \hat{j} + n \hat{k}\right) \\\\ & = 35 \sqrt{2} \left(\dfrac{3}{5 \sqrt{2}} \hat{i} + \dfrac{4}{5 \sqrt{2}} \hat{j} + \dfrac{5}{5 \sqrt{2}} \hat{j}\right) \\\\ & = 21 \hat{i} + 28 \hat{j} + 35 \hat{k} \end{aligned}$

$\therefore$ $\;$ Components of $\overrightarrow{r}$ along the $X$, $Y$ and $Z$ axes are: $\;$ $21 \hat{i}$, $28 \hat{j}$ and $35 \hat{k}$ respectively.

Vector Algebra

Can a vector have direction angles $30^{\circ}$, $45^{\circ}$, $60^{\circ}$?

Let $\alpha = 30^{\circ}$, $\beta = 45^{\circ}$, $\gamma = 60^{\circ}$

direction cosines are: $\;\;$ $\ell = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$

$\therefore$ $\;$ $\ell = \cos 30^{\circ} = \dfrac{\sqrt{3}}{2}$, $\;$ $m = \cos 45^{\circ} = \dfrac{1}{\sqrt{2}}$, $\;$ $n = \cos 60^{\circ} = \dfrac{1}{2}$

Now, $\;$ $\ell^2 + m^2 + n^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 + \left(\dfrac{1}{\sqrt{2}}\right)^2 + \left(\dfrac{1}{2}\right)^2 = \dfrac{3}{4} + \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{2}$

$\because$ $\;$ $\ell^2 + m^2 + n^2 \neq 1$, $\;$ $\therefore$ $\;$ the vector cannot have direction angles $30^{\circ}$, $45^{\circ}$ and $60^{\circ}$.

Vector Algebra

Verify that $\left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \left(\overrightarrow{c} \times \overrightarrow{d}\right) = \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] \overrightarrow{c} - \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] \overrightarrow{d}$ if $\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$, $\overrightarrow{b} = 2 \hat{i} + \hat{k}$, $\overrightarrow{c} = 2 \hat{i} + \hat{j} + \hat{k}$ and $\overrightarrow{d} = \hat{i} + \hat{j} + 2 \hat{k}$

$\begin{aligned} \overrightarrow{a} \times \overrightarrow{b} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 0 & 1 \end{vmatrix} \\\\ & = \hat{i} \left(1 - 0\right) - \hat{j} \left(1 - 2\right) + \hat{k} \left(0 - 2\right) \\\\ & = \hat{i} + \hat{j} - 2 \hat{k} \end{aligned}$

$\begin{aligned} \overrightarrow{c} \times \overrightarrow{d} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{vmatrix} \\\\ & = \hat{i} \left(2 - 1\right) - \hat{j} \left(4 - 1\right) + \hat{k} \left(2 - 1\right) \\\\ & = \hat{i} - 3 \hat{j} + \hat{k} \end{aligned}$

$\begin{aligned} \therefore \; \left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \left(\overrightarrow{c} \times \overrightarrow{d}\right) & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 1 & -3 & 1 \end{vmatrix} \\\\ & = \hat{i} \left(1 - 6\right) - \hat{j} \left(1 + 2\right) + \hat{k} \left(-3 - 1\right) \\\\ & = - 5 \hat{i} - 3 \hat{j} - 4 \hat{k} \;\;\; \cdots \; (1) \end{aligned}$

$\left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] = \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right)$

$\begin{aligned} \overrightarrow{b} \times \overrightarrow{d} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 2 \end{vmatrix} \\\\ & = \hat{i} \left(0 - 1\right) - \hat{j} \left(4 - 1\right) + \hat{k} \left(2 - 0\right) \\\\ & = - \hat{i} - 3 \hat{j} + 2 \hat{k} \end{aligned}$

$\begin{aligned} \therefore \; \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right) & = \left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(- \hat{i} - 3 \hat{j} + 2 \hat{k}\right) \\\\ & = -1 - 3 + 2 \\\\ & = - 2 \end{aligned}$

$\begin{aligned} \therefore \; \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] \overrightarrow{c} & = -2 \left(2 \hat{i} + \hat{j} + \hat{k}\right) \\\\ & = - 4 \hat{i} - 2 \hat{j} - 2 \hat{k} \;\;\; \cdots \; (2a) \end{aligned}$

$\left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] = \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right)$

$\begin{aligned} \overrightarrow{b} \times \overrightarrow{c} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 2 & 1 & 1 \end{vmatrix} \\\\ & = \hat{i} \left(0 - 1\right) - \hat{j} \left(2 - 2\right) + \hat{k} \left(2 - 0\right) \\\\ & = - \hat{i} + 2 \hat{k} \end{aligned}$

$\begin{aligned} \therefore \; \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right) & = \left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(- \hat{i} + 2 \hat{k}\right) \\\\ & = -1 + 2 \\\\ & = 1 \end{aligned}$

$\begin{aligned} \therefore \; \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] \overrightarrow{d} & = 1 \left(\hat{i} + \hat{j} + 2 \hat{k}\right) \\\\ & = \hat{i} + \hat{j} + 2 \hat{k} \;\;\; \cdots \; (2b) \end{aligned}$

$\therefore$ $\;$ We have from equations $(2a)$ and $(2b)$,

$\begin{aligned} \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] \overrightarrow{c} - \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] \overrightarrow{d} & = \left(-4 \hat{i} - 2 \hat{j} - 2 \hat{k}\right) - \left(\hat{i} + \hat{j} + 2 \hat{k}\right) \\\\ & = - 5 \hat{i} - 3 \hat{j} - 4 \hat{k} \;\;\; \cdots \; (3) \end{aligned}$

$\therefore$ $\;$ We have from equations $(1)$ and $(3)$,

$\left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \left(\overrightarrow{c} \times \overrightarrow{d}\right) = \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] \overrightarrow{c} - \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] \overrightarrow{d}$

Hence proved.

Vector Algebra

Prove that $\left(\overrightarrow{a} \times \overrightarrow{b}\right) \cdot \left(\overrightarrow{c} \times \overrightarrow{d}\right) + \left(\overrightarrow{b} \times \overrightarrow{c}\right) \cdot \left(\overrightarrow{a} \times \overrightarrow{d}\right) + \left(\overrightarrow{c} \times \overrightarrow{a}\right) \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right) = 0$

By definition,

$\begin{aligned} \left(\overrightarrow{a} \times \overrightarrow{b}\right) \cdot \left(\overrightarrow{c} \times \overrightarrow{d}\right) & = \begin{vmatrix} \overrightarrow{a} \cdot \overrightarrow{c} & \overrightarrow{a} \cdot \overrightarrow{d} \\\\ \overrightarrow{b} \cdot \overrightarrow{c} & \overrightarrow{b} \cdot \overrightarrow{d} \end{vmatrix} \\\\ & = \left(\overrightarrow{a} \cdot \overrightarrow{c}\right) \left(\overrightarrow{b} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{b} \cdot \overrightarrow{c}\right) \left(\overrightarrow{a} \cdot \overrightarrow{d}\right) \end{aligned}$

$\begin{aligned} \left(\overrightarrow{b} \times \overrightarrow{c}\right) \cdot \left(\overrightarrow{a} \times \overrightarrow{d}\right) & = \begin{vmatrix} \overrightarrow{b} \cdot \overrightarrow{a} & \overrightarrow{b} \cdot \overrightarrow{d} \\\\ \overrightarrow{c} \cdot \overrightarrow{a} & \overrightarrow{c} \cdot \overrightarrow{d} \end{vmatrix} \\\\ & = \left(\overrightarrow{b} \cdot \overrightarrow{a}\right) \left(\overrightarrow{c} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{c} \cdot \overrightarrow{a}\right) \left(\overrightarrow{b} \cdot \overrightarrow{d}\right) \end{aligned}$

$\begin{aligned} \left(\overrightarrow{c} \times \overrightarrow{a}\right) \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right) & = \begin{vmatrix} \overrightarrow{c} \cdot \overrightarrow{b} & \overrightarrow{c} \cdot \overrightarrow{d} \\\\ \overrightarrow{a} \cdot \overrightarrow{b} & \overrightarrow{a} \cdot \overrightarrow{d} \end{vmatrix} \\\\ & = \left(\overrightarrow{c} \cdot \overrightarrow{b}\right) \left(\overrightarrow{a} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{a} \cdot \overrightarrow{b}\right) \left(\overrightarrow{c} \cdot \overrightarrow{d}\right) \end{aligned}$

$\therefore$ $\;$ $\left(\overrightarrow{a} \times \overrightarrow{b}\right) \cdot \left(\overrightarrow{c} \times \overrightarrow{d}\right) + \left(\overrightarrow{b} \times \overrightarrow{c}\right) \cdot \left(\overrightarrow{a} \times \overrightarrow{d}\right) + \left(\overrightarrow{c} \times \overrightarrow{a}\right) \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right) =$

$\;\;\;\;$ $\left(\overrightarrow{a} \cdot \overrightarrow{c}\right) \left(\overrightarrow{b} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{b} \cdot \overrightarrow{c}\right) \left(\overrightarrow{a} \cdot \overrightarrow{d}\right) + \left(\overrightarrow{b} \cdot \overrightarrow{a}\right) \left(\overrightarrow{c} \cdot \overrightarrow{d}\right)$
$\;\;\;\;\;\;\;$ $- \left(\overrightarrow{c} \cdot \overrightarrow{a}\right) \left(\overrightarrow{b} \cdot \overrightarrow{d}\right) + \left(\overrightarrow{c} \cdot \overrightarrow{b}\right) \left(\overrightarrow{a} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{a} \cdot \overrightarrow{b}\right) \left(\overrightarrow{c} \cdot \overrightarrow{d}\right)$

$= \left\{\left(\overrightarrow{a} \cdot \overrightarrow{c}\right) - \left(\overrightarrow{c} \cdot \overrightarrow{a}\right)\right\} \left(\overrightarrow{b} \cdot \overrightarrow{d}\right) + \left\{\left(\overrightarrow{c} \cdot \overrightarrow{b}\right) - \left(\overrightarrow{b} \cdot \overrightarrow{c}\right)\right\} \left(\overrightarrow{a} \cdot \overrightarrow{d}\right)$
$\;\;\;\;\;\;\;\;\;\;\;\;\;$ $+ \left\{\left(\overrightarrow{b} \cdot \overrightarrow{a}\right) - \left(\overrightarrow{a} \cdot \overrightarrow{b}\right)\right\} \left(\overrightarrow{c} \cdot \overrightarrow{d}\right)$

$= 0$ $\;\;\;$ $\because$ $\;$ $\overrightarrow{a} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{a}$; $\;\;$ $\overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{b}$; $\;\;$ $\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{a}$

Hence proved.

Vector Algebra

Prove that $\left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \overrightarrow{c} = \overrightarrow{a} \times \left(\overrightarrow{b} \times \overrightarrow{c}\right)$ iff $\overrightarrow{a}$ and $\overrightarrow{c}$ are collinear.

If $\;$ $\left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \overrightarrow{c} = \overrightarrow{a} \times \left(\overrightarrow{b} \times \overrightarrow{c}\right)$ $\;$ then

$\left(\overrightarrow{a} \cdot \overrightarrow{c}\right) \overrightarrow{b} - \left(\overrightarrow{b} \cdot \overrightarrow{c}\right) \overrightarrow{a} = \left(\overrightarrow{a} \cdot \overrightarrow{c}\right) \overrightarrow{b} - \left(\overrightarrow{a} \cdot \overrightarrow{b}\right) \overrightarrow{c}$

i.e. $\;$ $\left(\overrightarrow{b} \cdot \overrightarrow{c}\right) \overrightarrow{a} = \left(\overrightarrow{a} \cdot \overrightarrow{b}\right) \overrightarrow{c}$

Let $\left(\overrightarrow{b} \cdot \overrightarrow{c}\right) = p$; $\;$ $\left(\overrightarrow{a} \cdot \overrightarrow{b}\right) = q$ $\;$ where $p$ and $q$ are scalar quantities.

$\therefore$ $\;$ We have $\;$ $p \overrightarrow{a} = q \overrightarrow{c}$

$\implies$ $\overrightarrow{a} = \left(\dfrac{p}{q}\right) \overrightarrow{c}$

$\implies$ $\overrightarrow{a}$ and $\overrightarrow{c}$ are collinear vectors. $\;\;\; \cdots \; (1)$

[Note: Two vectors are collinear if one vector is a scalar multiple of the other.]

Let $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be three vectors.

Given: $\;$ $\overrightarrow{a}$ and $\overrightarrow{c}$ are collinear vectors, then

$\overrightarrow{a} = k \overrightarrow{c}$ $\;$ where $k$ is a scalar quantity.

Then,

$\begin{aligned} \left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \overrightarrow{c} & = \left(k \overrightarrow{c} \times \overrightarrow{b}\right) \times \overrightarrow{c} \\\\ & = \left(k \overrightarrow{c} \cdot \overrightarrow{c}\right) \overrightarrow{b} - \left(\overrightarrow{b} \cdot \overrightarrow{c}\right) k \overrightarrow{c} \;\;\; \cdots \; (2a) \end{aligned}$

$\begin{aligned} \overrightarrow{a} \times \left(\overrightarrow{b} \times \overrightarrow{c}\right) & = k \overrightarrow{c} \times \left(\overrightarrow{b} \times \overrightarrow{c}\right) \\\\ & = \left(k \overrightarrow{c} \cdot \overrightarrow{c}\right) \overrightarrow{b} - \left(k \overrightarrow{c} \cdot \overrightarrow{b}\right) \overrightarrow{c} \\\\ & = \left(k \overrightarrow{c} \cdot \overrightarrow{c}\right) \overrightarrow{b} - \left(\overrightarrow{c} \cdot \overrightarrow{b}\right) k \overrightarrow{c} \;\;\; \cdots \; (2b) \end{aligned}$

$\therefore$ $\;$ From equations $(2a)$ and $(2b)$ we have

$\left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \overrightarrow{c} = \overrightarrow{a} \times \left(\overrightarrow{b} \times \overrightarrow{c}\right)$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ We have from equations $(1)$ and $(2)$

$\left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \overrightarrow{c} = \overrightarrow{a} \times \left(\overrightarrow{b} \times \overrightarrow{c}\right)$ iff $\overrightarrow{a}$ and $\overrightarrow{c}$ are collinear.

Vector Algebra

Prove that $\;$ $\overrightarrow{a} \times \left(\overrightarrow{b} \times \overrightarrow{c}\right) + \overrightarrow{b} \times \left(\overrightarrow{c} \times \overrightarrow{a}\right) + \overrightarrow{c} \times \left(\overrightarrow{a} \times \overrightarrow{b}\right) = \overrightarrow{0}$

$\overrightarrow{a} \times \left(\overrightarrow{b} \times \overrightarrow{c}\right) = \left(\overrightarrow{a} \cdot \overrightarrow{c}\right) \overrightarrow{b} - \left(\overrightarrow{a} \cdot \overrightarrow{b}\right) \overrightarrow{c}$ $\;\;\; \cdots \; (1a)$

$\overrightarrow{b} \times \left(\overrightarrow{c} \times \overrightarrow{a}\right) = \left(\overrightarrow{b} \cdot \overrightarrow{a}\right) \overrightarrow{c} - \left(\overrightarrow{b} \cdot \overrightarrow{c}\right) \overrightarrow{a}$ $\;\;\; \cdots \; (1b)$

$\overrightarrow{c} \times \left(\overrightarrow{a} \times \overrightarrow{b}\right) = \left(\overrightarrow{c} \cdot \overrightarrow{b}\right) \overrightarrow{a} - \left(\overrightarrow{c} \cdot \overrightarrow{a}\right) \overrightarrow{b}$ $\;\;\; \cdots \; (1c)$

Now, $\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{a}$ $\;\;\; \cdots \; (2a)$;

$\overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{b}$ $\;\;\; \cdots \; (2b)$;

$\overrightarrow{c} \cdot \overrightarrow{a} = \overrightarrow{a} \cdot \overrightarrow{c}$ $\;\;\; \cdots \; (2c)$

Adding equations $(1a)$, $(1b)$ and $(1c)$, we have

$\overrightarrow{a} \times \left(\overrightarrow{b} \times \overrightarrow{c}\right) + \overrightarrow{b} \times \left(\overrightarrow{c} \times \overrightarrow{a}\right) + \overrightarrow{c} \times \left(\overrightarrow{a} \times \overrightarrow{b}\right)$

$= \left(\overrightarrow{a} \cdot \overrightarrow{c}\right) \overrightarrow{b} - \left(\overrightarrow{a} \cdot \overrightarrow{b}\right) \overrightarrow{c} + \left(\overrightarrow{b} \cdot \overrightarrow{a}\right) \overrightarrow{c}$
$\;\;\;\;\;$ $- \left(\overrightarrow{b} \cdot \overrightarrow{c}\right) \overrightarrow{a} + \left(\overrightarrow{c} \cdot \overrightarrow{b}\right) \overrightarrow{a} - \left(\overrightarrow{c} \cdot \overrightarrow{a}\right) \overrightarrow{b}$

$= \overrightarrow{0}$ $\;\;$ [by equations $(2a)$, $(2b)$ and $(2c)$]