Vector Algebra

Prove that the area of a parallelogram formed by taking as its sides the diagonals of a parallelogram ABCD is twice the area of parallelogram ABCD.

Consider a parallelogram $ABCD$.

Let $\overrightarrow{AB} = \overrightarrow{a}$ $\;$ and $\;$ $\overrightarrow{BC} = \overrightarrow{b}$

Then, $\overrightarrow{DC} = \overrightarrow{a}$ $\;$ and $\;$ $\overrightarrow{AD} = \overrightarrow{b}$

Diagonal $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{a} + \overrightarrow{b}$

Diagonal $\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = - \overrightarrow{a} + \overrightarrow{b}$

Area of parallelogram $ABCD = A_1 = \left|\overrightarrow{a} \times \overrightarrow{b}\right|$ $\;\;\; \cdots \; (1)$

Consider parallelogram $PQRS$ where $\overrightarrow{PQ} = \overrightarrow{a} + \overrightarrow{b}$ $\;$ and $\;$ $\overrightarrow{QR} = \overrightarrow{b} - \overrightarrow{a}$

$\begin{aligned} \text{Area of parallelogram PQRS } = A_2 & = \left|\overrightarrow{PQ} \times \overrightarrow{QR}\right| \\\\ & = \left|\left(\overrightarrow{a} + \overrightarrow{b}\right) \times \left(\overrightarrow{b} - \overrightarrow{a}\right)\right| \\\\ & = \left|\overrightarrow{a} \times \overrightarrow{b} - \overrightarrow{a} \times \overrightarrow{a} + \overrightarrow{b} \times \overrightarrow{b} - \overrightarrow{b} \times \overrightarrow{a}\right| \\\\ & = \left|\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{b}\right| \\\\ & = \left|2 \left(\overrightarrow{a} \times \overrightarrow{b}\right)\right| \\\\ & = 2 A_1 \end{aligned}$

Hence proved.

Vector Algebra

Forces $2 \hat{i} + 7 \hat{j}$, $\;$ $2 \hat{i} - 5 \hat{j} + 6 \hat{k}$, $\;$ $- \hat{i} + 2\hat{j} - \hat{k}$ act at a point $P$ whose position vector is $4 \hat{i} - 3 \hat{j} - 2 \hat{k}$. Find the moment of the resultant of the three forces acting at $P$ about the point $Q$ whose position vector is $6 \hat{i} + \hat{j} - 3 \hat{k}$.

Let $\overrightarrow{F_1} = 2 \hat{i} + 7 \hat{j}$; $\;$ $\overrightarrow{F_2} = 2 \hat{i} - 5 \hat{j} + 6 \hat{k}$; $\;$ $\overrightarrow{F_3} = - \hat{i} + 2 \hat{j} - \hat{k}$

$\begin{aligned} \therefore \; \text{Resultant force} = \overrightarrow{F} & = \overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3} \\\\ & = \left(2 \hat{i} + 7 \hat{j}\right) + \left(2 \hat{i} - 5 \hat{j} + 6 \hat{k}\right) + \left(- \hat{i} + 2 \hat{j} - \hat{k}\right) \\\\ & = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \end{aligned}$

Position vector of point $P = 4 \hat{i} - 3 \hat{j} - 2 \hat{k}$ $\;\;\;$ i.e. point $P \left(4, -3, -2\right)$

Position vector of point $Q = 6 \hat{i} + \hat{j} - 3 \hat{k}$ $\;\;\;$ i.e. point $Q \left(6, 1, -3\right)$

$\begin{aligned} \overrightarrow{QP} = \overrightarrow{r} & = \text{position vector of P} - \text{position vector of Q} \\\\ & = \left(4 \hat{i} - 3 \hat{j} - 2 \hat{k}\right) - \left(6 \hat{i} + \hat{j} - 3 \hat{k}\right) \\\\ & = - 2 \hat{i} - 4 \hat{j} + \hat{k} \end{aligned}$

Moment of force (or torque) acting at point $P$ about point $Q$ is

$\begin{aligned} \overrightarrow{M} = \overrightarrow{r} \times \overrightarrow{F} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & - 4 & 1 \\ 3 & 4 & 5 \end{vmatrix} \\\\ & = \hat{i} \left(- 20 - 4\right) - \hat{j} \left(- 10 - 3\right) + \hat{k} \left(- 8 + 12\right) \\\\ & = - 24 \hat{i} + 13 \hat{j} + 4 \hat{k} \end{aligned}$

Vector Algebra

Prove that $\sin \left(\alpha - \beta\right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

Consider an unit circle with center $O \left(0,0\right)$.

Consider two points $P$ and $Q$ on this circle.

Then, $\left|\overrightarrow{OP}\right| = 1$ $\;$ and $\;$ $\left|\overrightarrow{OQ}\right| = 1$

Let $OP$ and $OQ$ make angles $\alpha$ and $\beta$ respectively with the $X$ axis.

Then the coordinates of points $P$ and $Q$ are $\left(\cos \alpha, \sin \alpha\right)$ and $\left(\cos \beta, \sin \beta\right)$ respectively.

Let $\hat{i}$ and $\hat{j}$ be unit vectors along the $X$ and $Y$ axes respectively.

Now, $\overrightarrow{OP} = \overrightarrow{OM} + \overrightarrow{MP} = \hat{i} \cos \alpha + \hat{j} \sin \alpha$

and $\overrightarrow{OQ} = \overrightarrow{ON} + \overrightarrow{NQ} = \hat{i} \cos \beta + \hat{j} \sin \beta$

$\begin{aligned} \therefore \; \overrightarrow{OQ} \times \overrightarrow{OP} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \cos \beta & \sin \beta & 0 \\ \cos \alpha & \sin \alpha & 0 \end{vmatrix} \\\\ & = \hat{i} \left(0\right) - \hat{j} \left(0\right) + \hat{k} \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right) \\\\ & = \hat{k} \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right) \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \text{By definition, } \overrightarrow{OQ} \cdot \overrightarrow{OP} & = \left|\overrightarrow{OQ}\right| \left|\overrightarrow{OP}\right| \sin \left(\alpha - \beta\right) \hat{k} \\\\ & = 1 \times 1 \times \sin \left(\alpha - \beta\right) \hat{k} \\\\ & = \sin \left(\alpha - \beta\right) \hat{k} \;\;\; \cdots \; (2) \end{aligned}$

$\therefore$ $\;$ From equations $(1)$ and $(2)$ we have

$\sin \left(\alpha - \beta\right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

Hence proved.

Vector Algebra

Find the area of the triangle whose vertices are $\left(3, -1, 2\right)$, $\left(1, -1, -3\right)$ and $\left(4, -3, 1\right)$

Let $A \left(3, -1, 2\right)$, $B \left(1, -1, -3\right)$ and $C \left(4, -3, 1\right)$

Position vector (p.v) of point $A = 3 \hat{i} - \hat{j} + 2 \hat{k}$

p.v of point $B = \hat{i} - \hat{j} - 3 \hat{k}$

p.v of point $C = 4 \hat{i} - 3 \hat{j} + \hat{k}$

$\begin{aligned} \overrightarrow{AB} & = \text{p.v of point B - p.v of point A} \\\\ & = \left(\hat{i} - \hat{j} - 3 \hat{k}\right) - \left(3 \hat{i} - \hat{j} + 2 \hat{k}\right) \\\\ & = -2 \hat{i} - 5 \hat{k} \end{aligned}$

$\begin{aligned} \overrightarrow{AC} & = \text{p.v of point C - p.v of point A} \\\\ & = \left(4 \hat{i} - 3 \hat{j} + \hat{k}\right) - \left(3 \hat{i} - \hat{j} + 2 \hat{k}\right) \\\\ & = \hat{i} - 2 \hat{j} - \hat{k} \end{aligned}$

Area of $\triangle ABC = \dfrac{1}{2} \left|\overrightarrow{AB} \times \overrightarrow{AC} \right|$

$\begin{aligned} \overrightarrow{AB} \times \overrightarrow{AC} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{vmatrix} \\\\ & = \hat{i} \left(0 - 10\right) - \hat{j} \left(2 + 5\right) + \hat{k} \left(4 - 0\right) \\\\ & = - 10 \hat{i} - 7 \hat{j} + 4 \hat{k} \end{aligned}$

$\therefore$ $\;$ $\left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = \sqrt{\left(-10\right)^2 + \left(-7\right)^2 + \left(4\right)^2} = \sqrt{165}$

$\therefore$ $\;$ Area of $\triangle ABC = \dfrac{1}{2} \sqrt{165}$ sq units

Vector Algebra

Find the area of the parallelogram $ABCD$ whose vertices are $A \left(-5, 2, 5\right)$, $B \left(-3, 6, 7\right)$, $C \left(4, -1, 5\right)$ and $D \left(2, -5, 3\right)$.

Given: $\;$ Vertices $\;$ $A \left(-5, 2, 5\right)$, $B \left(-3, 6, 7\right)$, $C \left(4, -1, 5\right)$, $D \left(2, -5, 3\right)$

position vector of $A$ $= - 5 \hat{i} + 2 \hat{j} + 5 \hat{k}$

position vector of $B$ $= - 3 \hat{i} + 6 \hat{j} + 7 \hat{k}$

position vector of $C$ $= 4 \hat{i} - \hat{j} + 5 \hat{k}$

position vector of $D$ $= 2 \hat{i} - 5 \hat{j} + 3 \hat{k}$

$\begin{aligned} \text{Diagonal } \overrightarrow{AC} = \overrightarrow{d_1} & = \text{position vector of C} - \text{position vector of A} \\\\ & = \left(4 \hat{i} - \hat{j} + 5 \hat{k}\right) - \left(- 5 \hat{i} + 2 \hat{j} + 5 \hat{k}\right) \\\\ & = 9 \hat{i} - 3 \hat{j} \end{aligned}$

$\begin{aligned} \text{Diagonal } \overrightarrow{BD} = \overrightarrow{d_2} & = \text{position vector of D} - \text{position vector of B} \\\\ & = \left(2 \hat{i} - 5 \hat{j} + 3 \hat{k}\right) - \left(- 3 \hat{i} + 6 \hat{j} + 7 \hat{k}\right) \\\\ & = 5 \hat{i} - 11 \hat{j} - 4 \hat{k} \end{aligned}$

Area of parallelogram $= \dfrac{1}{2} \left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right|$

$\begin{aligned} \overrightarrow{d_1} \times \overrightarrow{d_2} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 9 & - 3 & 0 \\ 5 & -11 & -4 \end{vmatrix} \\\\ & = \hat{i} \left(12\right) - \hat{j} \left(-36\right) + \hat{k} \left(-99 + 15\right) \\\\ & = 12 \hat{i} + 36 \hat{j} - 84 \hat{k} \\\\ & = 12 \left(\hat{i} + 3 \hat{j} - 7 \hat{k}\right) \end{aligned}$

$\therefore$ $\;$ $\left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right| = 12 \times \sqrt{\left(1\right)^2 + \left(3\right)^2 + \left(-7\right)^2} = 12 \sqrt{59}$

$\therefore$ $\;$ Area of parallelogram $= \dfrac{1}{2} \times 12 \times \sqrt{59} = 6 \sqrt{59}$ sq units

Vector Algebra

If $\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{b} \times \overrightarrow{d}$, show that $\overrightarrow{a} - \overrightarrow{d} \parallel \overrightarrow{b} - \overrightarrow{c}$

Given: $\;$ $\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{d}$ $\;\;\; \cdots \; (1)$; $\;\;\;$ $\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{b} \times \overrightarrow{d}$ $\;\;\; \cdots \; (2)$

To prove that $\;$ $\overrightarrow{a} - \overrightarrow{d} \parallel \overrightarrow{b} - \overrightarrow{c}$

i.e. $\;$ To prove that $\;$ $\left(\overrightarrow{a} - \overrightarrow{d}\right) \times \left(\overrightarrow{b} - \overrightarrow{c}\right) = \overrightarrow{0}$

Now, $\;$ $\left(\overrightarrow{a} - \overrightarrow{d}\right) \times \left(\overrightarrow{b} - \overrightarrow{c}\right)$

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

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

$= \left(\overrightarrow{c} \times \overrightarrow{d}\right) - \left(\overrightarrow{b} \times \overrightarrow{d}\right) - \left(\overrightarrow{d} \times \overrightarrow{b}\right) + \left(\overrightarrow{d} \times \overrightarrow{c}\right)$ $\;\;\;$ [from $(1)$ and $(2)$]

$= \left(\overrightarrow{c} \times \overrightarrow{d}\right) - \left(\overrightarrow{b} \times \overrightarrow{d}\right) + \left(\overrightarrow{b} \times \overrightarrow{d}\right) - \left(\overrightarrow{c} \times \overrightarrow{d}\right)$

$= \overrightarrow{0}$

$\implies$ $\overrightarrow{a} - \overrightarrow{d} \parallel \overrightarrow{b} - \overrightarrow{c}$ $\;\;\;$ Hence proved.

Vector Algebra

If $\overrightarrow{a} =2$, $\overrightarrow{b} = 7$ and $\overrightarrow{a} \times \overrightarrow{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$, find the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.

Given: $\left|\overrightarrow{a}\right| = 2$; $\;$ $\left|\overrightarrow{b}\right| = 7$; $\;$ $\overrightarrow{a} \times \overrightarrow{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$

By definition, $\;$ $\left|\overrightarrow{a} \times \overrightarrow{b}\right| = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \sin \theta$

$\left|\overrightarrow{a} \times \overrightarrow{b}\right| = \sqrt{\left(3\right)^2 + \left(-2\right)^2 + \left(6\right)^2} = \sqrt{49} = 7$

$\therefore$ $\;$ $\sin \theta = \dfrac{\left|\overrightarrow{a} \times \overrightarrow{b}\right|}{\left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right|} = \dfrac{7}{2 \times 7} = \dfrac{1}{2}$

$\implies$ $\theta = \sin^{-1} \left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}$

Vector Algebra

Find the vectors whose magnitude is $5$ and which are perpendicular to the vectors $3 \hat{i} + \hat{j} - 4 \hat{k}$ and $6 \hat{i} + 5 \hat{j} - 2 \hat{k}$

Let $\;$ $\overrightarrow{a} = 3 \hat{i} + \hat{j} - 4 \hat{k}$; $\;$ $\overrightarrow{b} = 6 \hat{i} + 5 \hat{j} - 2 \hat{k}$

Let unit vector perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}$ be $\hat{n}$.

Then, $\hat{n} = \pm \left(\dfrac{\overrightarrow{a} \times \overrightarrow{b}}{\left|\overrightarrow{a} \times \overrightarrow{b}\right|}\right)$

$\begin{aligned} \overrightarrow{a} \times \overrightarrow{b} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{vmatrix} \\\\ & = \hat{i} \left(-2 + 20\right) - \hat{j} \left(-6 + 24\right) + \hat{k} \left(15 - 6\right) \\\\ & = 18 \hat{i} - 18 \hat{j} + 9 \hat{k} \end{aligned}$

$\left|\overrightarrow{a} \times \overrightarrow{b}\right| = \sqrt{\left(18\right)^2 + \left(-18\right)^2 + \left(9\right)^2} = \sqrt{729} = 27$

$\therefore$ $\;$ $\hat{n} = \pm \left(\dfrac{18 \hat{i} - 18 \hat{j} + 9 \hat{k}}{27}\right) = \pm \left(\dfrac{2 \hat{i} - 2 \hat{j} + \hat{k}}{3}\right)$

$\therefore$ $\;$ Vectors of magnitude $5$ perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}$

$= \pm 5 \times \left(\dfrac{2 \hat{i} - 2 \hat{j} + \hat{k}}{3}\right) = \pm \left(\dfrac{10 \hat{i} - 10 \hat{j} + 5 \hat{k}}{3}\right)$

Vector Algebra

If $\left|\overrightarrow{a}\right| = 3$, $\;$ $\left|\overrightarrow{b}\right| = 4$ $\;$ and $\;$ $\overrightarrow{a} \cdot \overrightarrow{b} = 9$, then find $\left|\overrightarrow{a} \times \overrightarrow{b}\right|$

Given: $\;$ $\left|\overrightarrow{a}\right| = 3$, $\;$ $\left|\overrightarrow{b}\right| = 4$, $\;$ $\overrightarrow{a} \cdot \overrightarrow{b} = 9$

By definition, $\;$ $\overrightarrow{a} \cdot \overrightarrow{b} = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \cos \theta$

i.e. $\;$ $9 = 3 \times 4 \times \cos \theta$ $\implies$ $\cos \theta = \dfrac{3}{4}$

$\therefore$ $\;$ $\sin \theta = \sqrt{1 - \cos ^2 \theta} = \sqrt{1 - \dfrac{9}{16}} = \dfrac{\sqrt{7}}{4}$

$\begin{aligned} \text{Now, } \; \left|\overrightarrow{a} \times \overrightarrow{b}\right| & = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \sin \theta \\\\ & = 3 \times 4 \times \dfrac{\sqrt{7}}{4} = 3 \sqrt{7} \end{aligned}$

Vector Algebra

Forces of magnitudes 3 and 4 units acting in the directions $\;$ $6 \hat{i} + 2 \hat{j} + 3 \hat{k}$ $\;$ and $\;$ $3 \hat{i} - 2 \hat{j} + 6 \hat{k}$ $\;$ respectively act on a particle which is displaced from $\left(2, 2, -1\right)$ to $\left(4, 3, 1\right)$. Find the work done by the forces.

Let $\overrightarrow{p} = 6 \hat{i} + 2 \hat{j} + 3 \hat{k}$; $\;$ $\overrightarrow{q} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$

$\left|\overrightarrow{p}\right| = \sqrt{\left(6\right)^2 + \left(2\right)^2 + \left(3\right)^2} = \sqrt{49} = 7$

$\left|\overrightarrow{q}\right| = \sqrt{\left(3\right)^2 + \left(-2\right)^2 + \left(6\right)^2} = \sqrt{49} = 7$

Let $F_1 =$ force of magnitude $3$ units in the direction of $\overrightarrow{p}$

Then, $\overrightarrow{F_1} = \dfrac{3}{7} \left(6 \hat{i} + 2 \hat{j} + 3 \hat{k}\right) = \dfrac{18}{7} \hat{i} + \dfrac{6}{7} \hat{j} + \dfrac{9}{7} \hat{k}$

Let $F_2 =$ force of magnitude $4$ units in the direction of $\overrightarrow{q}$

Then, $\overrightarrow{F_2} = \dfrac{4}{7} \left(3 \hat{i} - 2 \hat{j} + 6 \hat{k}\right) = \dfrac{12}{7} \hat{i} - \dfrac{8}{7} \hat{j} + \dfrac{24}{7} \hat{k}$

$\begin{aligned} \therefore \; \text{Resultant force } = \overrightarrow{F} & = \overrightarrow{F_1} + \overrightarrow{F_2} \\\\ & = \left(\dfrac{18}{7} \hat{i} + \dfrac{6}{7} \hat{j} + \dfrac{9}{7} \hat{k}\right) + \left(\dfrac{12}{7} \hat{i} - \dfrac{8}{7} \hat{j} + \dfrac{24}{7} \hat{k}\right) \\\\ & = \dfrac{30}{7} \hat{i} - \dfrac{2}{7} \hat{j} + \dfrac{33}{7} \hat{k} \end{aligned}$

Initial position of the particle $= \left(2, 2, -1\right)$

$\therefore$ $\;$ Initial position vector of the particle $= \overrightarrow{s_1} = 2 \hat{i} + 2 \hat{j} - \hat{k}$

Final position of the particle $= \left(4, 3, 1\right)$

$\therefore$ $\;$ Final position vector of the particle $= \overrightarrow{s_2} = 4 \hat{i} + 3 \hat{j} + \hat{k}$

$\begin{aligned} \therefore \; \text{Displacement of the particle } = \overrightarrow{s} & = \overrightarrow{s_2} - \overrightarrow{s_1} \\\\ & = \left(4 \hat{i} + 3 \hat{j} + \hat{k}\right) - \left(2 \hat{i} + 2 \hat{j} - \hat{k}\right) \\\\ & = 2 \hat{i} + \hat{j} + 2 \hat{k} \end{aligned}$

$\begin{aligned} \text{Now, work done by the forces } & = \overrightarrow{F} \cdot \overrightarrow{s} \\\\ & = \left(\dfrac{30}{7} \hat{i} - \dfrac{2}{7} \hat{j} + \dfrac{33}{7} \hat{k}\right) \cdot \left(2 \hat{i} + \hat{j} + 2 \hat{k}\right) \\\\ & = \dfrac{60}{7} - \dfrac{2}{7} + \dfrac{66}{7} = \dfrac{124}{7} \; \text{ units} \end{aligned}$

Vector Algebra

Prove by vector method $\;$ $\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Consider an unit circle with center $O \left(0,0\right)$.

Consider two points $P$ and $Q$ on this circle.

Then, $\left|\overrightarrow{OP}\right| = 1$ $\;$ and $\;$ $\left|\overrightarrow{OQ}\right| = 1$

Let $OP$ and $OQ$ make angles $\alpha$ and $\beta$ respectively with the $X$ axis.

Then the coordinates of points $P$ and $Q$ are $\left(\cos \alpha, \sin \alpha\right)$ and $\left(\cos \beta, - \sin \beta\right)$ respectively.

Let $\hat{i}$ and $\hat{j}$ be unit vectors along the $X$ and $Y$ axes respectively.

Now, $\overrightarrow{OP} = \overrightarrow{OM} + \overrightarrow{MP} = \hat{i} \cos \alpha + \hat{j} \sin \alpha$

and $\overrightarrow{OQ} = \overrightarrow{ON} + \overrightarrow{NQ} = \hat{i} \cos \beta - \hat{j} \sin \beta$

$\begin{aligned} \therefore \; \overrightarrow{OP} \cdot \overrightarrow{OQ} & = \left(\hat{i} \cos \alpha + \hat{j} \sin \alpha\right) \cdot \left(\hat{i} \cos \beta - \hat{j} \sin \alpha\right) \\\\ & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \text{By definition, } \overrightarrow{OP} \cdot \overrightarrow{OQ} & = \left|\overrightarrow{OP}\right| \cdot \left|\overrightarrow{OQ}\right| \cos \left(\alpha + \beta\right) \\\\ & = 1 \times 1 \times \cos \left(\alpha + \beta\right) \\\\ & = \cos \left(\alpha + \beta\right) \;\;\; \cdots \; (2) \end{aligned}$

$\therefore$ $\;$ From equations $(1)$ and $(2)$ we have

$\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Hence proved.

Vector Algebra

Prove by vector method that if the diagonals of a parallelogram are equal, then the parallelogram is a rectangle.

Let $ABCD$ be a parallelogram.

Let $\overrightarrow{AB} = \overrightarrow{a}$, $\;$ $\overrightarrow{BC} = \overrightarrow{b}$

Then $\overrightarrow{CD} = - \overrightarrow{a}$, $\;$ $\overrightarrow{DA} = - \overrightarrow{b}$

Also, $\left|\overrightarrow{AB}\right| = \left|\overrightarrow{CD}\right|$; $\;$ $\left|\overrightarrow{BC}\right| = \left|\overrightarrow{DA}\right|$ $\;\;\; \cdots \; (1)$

Diagonal $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{a} + \overrightarrow{b}$

Diagonal $\overrightarrow{DB} = \overrightarrow{DA} + \overrightarrow{AB} = \overrightarrow{a} - \overrightarrow{b}$

Given: $\;$ $\left|\overrightarrow{AC}\right| = \left|\overrightarrow{DB}\right|$

i.e. $\;$ $\left|\overrightarrow{a} + \overrightarrow{b}\right| = \left|\overrightarrow{a} - \overrightarrow{b}\right|$

i.e. $\;$ $\left|\overrightarrow{a} + \overrightarrow{b}\right|^2 = \left|\overrightarrow{a} - \overrightarrow{b}\right|^2$

i.e. $\;$ $\left|\overrightarrow{a}\right|^2 + \left|\overrightarrow{b}\right|^2 + 2 \left|\overrightarrow{a}\right| \cdot \left|\overrightarrow{b}\right| = \left|\overrightarrow{a}\right|^2 + \left|\overrightarrow{b}\right|^2 - 2 \left|\overrightarrow{a}\right| \cdot \left|\overrightarrow{b}\right|$

i.e. $\;$ $4 \left|\overrightarrow{a}\right| \cdot \left|\overrightarrow{b}\right| = 0$

i.e. $\;$ $\left|\overrightarrow{a}\right| \cdot \left|\overrightarrow{b}\right| = 0$

$\implies$ $\overrightarrow{a} \perp \overrightarrow{b}$ (i.e. the adjacent sides are perpendicular to each other) $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ We have from $(1)$ and $(2)$, if the diagonals of a parallelogram are equal, then the parallelogram is a rectangle.

Vector Algebra

Find the projection of $\hat{i} + 2 \hat{j} - 2 \hat{k}$ $\;$ on $\;$ $2 \hat{i} - \hat{j} + 5 \hat{k}$

Let $\overrightarrow{a} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ $\;$ and $\;$ $\overrightarrow{b} = 2 \hat{i} - \hat{j} + 5 \hat{k}$

$\begin{aligned} \text{Projection of } \overrightarrow{a} \text{ on } \overrightarrow{b} & = \dfrac{\overrightarrow{a} \cdot \overrightarrow{b}}{\left|\overrightarrow{b}\right|} \\\\ & = \dfrac{\left(\hat{i} + 2 \hat{j} - 2 \hat{k}\right) \cdot \left(2 \hat{i} - \hat{j} + 5 \hat{k}\right)}{\sqrt{\left(2\right)^2 + \left(-1\right)^2 + \left(5\right)^2}} \\\\ & = \dfrac{2 - 2 - 10}{\sqrt{4 + 1 + 25}} \\\\ & = \dfrac{-10}{\sqrt{30}} \end{aligned}$

Vector Algebra

Show that the vectors $3 \hat{i} - 2 \hat{j} + \hat{k}$, $\hat{i} - 3 \hat{j} + 5 \hat{k}$ and $2 \hat{i} + \hat{j} - 4 \hat{k}$ form a right angled triangle.

Let $\overrightarrow{a} = 3 \hat{i} - 2 \hat{j} + \hat{k}$, $\;$ $\overrightarrow{b} = \hat{i} - 3 \hat{j} + 5 \hat{k}$, $\;$ $\overrightarrow{c} = 2 \hat{i} + \hat{j} - 4 \hat{k}$

$\left|\overrightarrow{a}\right| = \sqrt{\left(3\right)^2 + \left(-2\right)^2 + \left(1\right)^2} = \sqrt{14}$

$\left|\overrightarrow{b}\right| = \sqrt{\left(1\right)^2 + \left(-3\right)^2 + \left(5\right)^2} = \sqrt{35}$

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

Now, $\;$ $\left|\overrightarrow{a}\right|^2 + \left|\overrightarrow{c}\right|^2 = 14 + 21 = 35 = \left|\overrightarrow{b}\right|^2$ $\;\;\; \cdots \; (1)$

and $\;$ $\overrightarrow{a} \cdot \overrightarrow{c} = \left(3 \hat{i} - 2 \hat{j} + \hat{k}\right) \cdot \left(2 \hat{i} + \hat{j} - 4 \hat{k}\right) = 6 - 2 - 4 = 0$

$\implies$ $\overrightarrow{a} \perp \overrightarrow{c}$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ From equations $(1)$ and $(2)$, the vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ form a right angled triangle.

Vector Algebra

Let $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ be three vectors such that $\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} = \overrightarrow{0}$.
If $\left|\overrightarrow{u}\right| = 3$, $\left|\overrightarrow{v}\right| = 4$ and $\left|\overrightarrow{w}\right| = 5$, then find $\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}$

Let

$\alpha$ be the angle between $\overrightarrow{u}$ and $\overrightarrow{v}$

$\beta$ be the angle between $\overrightarrow{v}$ and $\overrightarrow{w}$

$\gamma$ be the angle between $\overrightarrow{w}$ and $\overrightarrow{u}$

Given: $\left|\overrightarrow{u}\right| = 3$, $\left|\overrightarrow{v}\right| = 4$ and $\left|\overrightarrow{w}\right| = 5$

Given: $\;$ $\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} = \overrightarrow{0}$

$\implies$ $\overrightarrow{u} + \overrightarrow{v} = - \overrightarrow{w}$

Then, $\;$ $\left(\overrightarrow{u} + \overrightarrow{v}\right)^2 = \left(- \overrightarrow{w}\right)^2$

i.e. $\;$ $\left|\overrightarrow{u}\right|^2 + \left|\overrightarrow{v}\right|^2 + 2 \left|\overrightarrow{u}\right| \left|\overrightarrow{v}\right| \cos \alpha = \left|w\right|^2$

i.e. $\;$ $\left(3\right)^2 + \left(4\right)^2 + 2 \times 3 \times 4 \times \cos \alpha = \left(5\right)^2$

i.e. $\;$ $\cos \alpha = \dfrac{25 - 9 - 16}{24} = 0$

$\because$ $\;$ $\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} = \overrightarrow{0}$

$\implies$ $\overrightarrow{v} + \overrightarrow{w} = - \overrightarrow{u}$

Then, $\;$ $\left(\overrightarrow{v} + \overrightarrow{w}\right)^2 = \left(- \overrightarrow{u}\right)^2$

i.e. $\;$ $\left|\overrightarrow{v}\right|^2 + \left|\overrightarrow{w}\right|^2 + 2 \left|\overrightarrow{v}\right| \left|\overrightarrow{w}\right| \cos \beta = \left|u\right|^2$

i.e. $\;$ $\left(4\right)^2 + \left(5\right)^2 + 2 \times 4 \times 5 \times \cos \beta = \left(3\right)^2$

i.e. $\;$ $\cos \beta = \dfrac{9 - 16 - 25}{40} = \dfrac{-32}{40} = \dfrac{-4}{5}$

$\because$ $\;$ $\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} = \overrightarrow{0}$

$\implies$ $\overrightarrow{u} + \overrightarrow{w} = - \overrightarrow{v}$

Then, $\;$ $\left(\overrightarrow{u} + \overrightarrow{w}\right)^2 = \left(- \overrightarrow{v}\right)^2$

i.e. $\;$ $\left|\overrightarrow{u}\right|^2 + \left|\overrightarrow{w}\right|^2 + 2 \left|\overrightarrow{u}\right| \left|\overrightarrow{w}\right| \cos \gamma = \left|v\right|^2$

i.e. $\;$ $\left(3\right)^2 + \left(5\right)^2 + 2 \times 3 \times 5 \times \cos \gamma = \left(4\right)^2$

i.e. $\;$ $\cos \gamma = \dfrac{16 - 9 - 25}{30} = \dfrac{-18}{30} = \dfrac{-3}{5}$

Now, $\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}$

$= \left|\overrightarrow{u}\right| \left|\overrightarrow{v}\right| \cos \alpha + \left|\overrightarrow{v}\right| \left|\overrightarrow{w}\right| \cos \beta + \left|\overrightarrow{w}\right| \left|\overrightarrow{u}\right| \cos \gamma$

$= 3 \times 4 \times 0 + 4 \times 5 \times \left(\dfrac{-4}{5}\right) + 5 \times 3 \times \left(\dfrac{-3}{5}\right) = - 25$

Vector Algebra

Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined with the coordinate axes.

Let $\overrightarrow{p} = \hat{i} + \hat{j} + \hat{k}$

Let $\overrightarrow{p}$ make an angle $\alpha$ with the X axis, $\beta$ with the Y axis and $\gamma$ with the Z axis.

Unit vector along the X axis is $= \hat{x} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k}$

Unit vector along the Y axis is $= \hat{y} = 0 \hat{i} + 1 \hat{j} + 0 \hat{k}$

Unit vector along the Z axis is $= \hat{z} = 0 \hat{i} + 0 \hat{j} + 1 \hat{k}$

Now, $\left|\overrightarrow{p}\right| = \sqrt{\left(1\right)^2 + \left(1\right)^2 + \left(1\right)^2} = \sqrt{3}$

Angle between $\overrightarrow{p}$ and the X axis is the angle between $\overrightarrow{p}$ and $\hat{x}$.

$\overrightarrow{p} \cdot \hat{x} = \left|\overrightarrow{p}\right| \left|\hat{x}\right| \cos \alpha$

$\left|\hat{x}\right| = 1$

$\begin{aligned} \therefore \; \cos \alpha & = \dfrac{\overrightarrow{p} \cdot \hat{x}}{\left|\overrightarrow{p}\right| \left|\hat{x}\right|} \\\\ & = \dfrac{\left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(\hat{i} + 0 \hat{j} + 0 \hat{k}\right)}{\sqrt{3} \times 1} \\\\ & = \dfrac{1}{\sqrt{3}} \;\;\; \cdots \; (1) \end{aligned}$

Angle between $\overrightarrow{p}$ and the Y axis is the angle between $\overrightarrow{p}$ and $\hat{y}$.

$\overrightarrow{p} \cdot \hat{y} = \left|\overrightarrow{p}\right| \left|\hat{y}\right| \cos \beta$

$\left|\hat{y}\right| = 1$

$\begin{aligned} \therefore \; \cos \beta & = \dfrac{\overrightarrow{p} \cdot \hat{y}}{\left|\overrightarrow{p}\right| \left|\hat{y}\right|} \\\\ & = \dfrac{\left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(0 \hat{i} + \hat{j} + 0 \hat{k}\right)}{\sqrt{3} \times 1} \\\\ & = \dfrac{1}{\sqrt{3}} \;\;\; \cdots \; (2) \end{aligned}$

Angle between $\overrightarrow{p}$ and the Z axis is the angle between $\overrightarrow{p}$ and $\hat{z}$.

$\overrightarrow{p} \cdot \hat{z} = \left|\overrightarrow{p}\right| \left|\hat{z}\right| \cos \gamma$

$\left|\hat{z}\right| = 1$

$\begin{aligned} \therefore \; \cos \gamma & = \dfrac{\overrightarrow{p} \cdot \hat{z}}{\left|\overrightarrow{p}\right| \left|\hat{z}\right|} \\\\ & = \dfrac{\left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(0 \hat{i} + 0 \hat{j} + \hat{k}\right)}{\sqrt{3} \times 1} \\\\ & = \dfrac{1}{\sqrt{3}} \;\;\; \cdots \; (3) \end{aligned}$

$\therefore$ $\;$ From equations $(1)$, $(2)$ and $(3)$, $\cos \alpha = \cos \beta = \cos \gamma$

$\implies$ $\alpha = \beta = \gamma$

i.e. $\;$ $\overrightarrow{p}$ is equally inclined with the coordinate axes.

Vector Algebra

Find the value of $m$ for which the vectors $\overrightarrow{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k}$ and $\overrightarrow{b} = \hat{i} + m \hat{j} + 3 \hat{k}$ are

1. perpendicular
2. parallel

If two vectors are perpendicular (angle between the vectors is $\theta = 90^{\circ}$), then $\overrightarrow{P} \cdot \overrightarrow{Q} = \left|\overrightarrow{P}\right| \left|\overrightarrow{Q}\right| \cos \left(90^{\circ}\right) = 0$

If two vectors are parallel (angle between the vectors is $\theta = 0^{\circ}$), then $\overrightarrow{P} \cdot \overrightarrow{Q} = \left|\overrightarrow{P}\right| \left|\overrightarrow{Q}\right| \cos \left(0^{\circ}\right) = \left|\overrightarrow{P}\right| \left|\overrightarrow{Q}\right|$

Now,

$\begin{aligned} \overrightarrow{a} \cdot \overrightarrow{b} & = \left(3 \hat{i} + 2 \hat{j} + 9 \hat{k}\right) \cdot \left(\hat{i} + m \hat{j} + 3 \hat{k}\right) \\\\ & = 3 + 2m + 27 \\\\ & = 30 + 2m \end{aligned}$

and

$\left|\overrightarrow{a}\right| = \sqrt{\left(3\right)^2 + \left(2\right)^2 + \left(9\right)^2} = \sqrt{94}$

$\left|\overrightarrow{b}\right| = \sqrt{\left(1\right)^2 + \left(m\right)^2 + \left(3\right)^2} = \sqrt{10 + m^2}$

1. When $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular, $\overrightarrow{a} \cdot \overrightarrow{b} = 0$
   
   i.e. $\;$ $30 + 2m = 0$ $\implies$ $m = - 15$


2. When $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel, $\overrightarrow{a} \cdot \overrightarrow{b} = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right|$
   
   i.e. $\;$ $30 + 2m = \sqrt{94} \times \sqrt{10 + m^2}$
   
   i.e. $\;$ $900 + 4 m^2 + 120 m = 940 + 94 m^2$
   
   i.e. $\;$ $90 m^2 - 120 m + 40 = 0$
   
   i.e. $\;$ $9m^2 - 12 m + 4 = 0$
   
   i.e. $\;$ $\left(3m - 2\right)^2 = 0$
   
   i.e. $\;$ $3m - 2 = 0$ $\implies$ $m = \dfrac{2}{3}$

Vector Algebra

If $\overrightarrow{a} = \hat{i} + \hat{j} + 2 \hat{k}$ and $\overrightarrow{b} = 3\hat{i} + 2 \hat{j} - \hat{k}$, find $\left(\overrightarrow{a} + 3 \overrightarrow{b}\right) \cdot \left(2 \overrightarrow{a} - \overrightarrow{b}\right)$

$\begin{aligned} \overrightarrow{a} + 3 \overrightarrow{b} & = \left(\hat{i} + \hat{j} + 2\hat{k}\right) + \left(3 \hat{i} + 2 \hat{j} - \hat{k}\right) \\\\ & = 10 \hat{i} + 7 \hat{j} - \hat{k} \end{aligned}$

$\begin{aligned} 2 \overrightarrow{a} - \overrightarrow{b} & = 2 \left(\hat{i} + \hat{j} + 2 \hat{k}\right) - \left(3 \hat{i} + 2 \hat{j} - \hat{k}\right) \\\\ & = - \hat{i} + 0 \hat{j} +5 \hat{k} \end{aligned}$

$\begin{aligned} \therefore \; \left(\overrightarrow{a} + 3 \overrightarrow{b}\right) \cdot \left(2 \overrightarrow{a} - \overrightarrow{b}\right) & = \left(10 \hat{i} + 7 \hat{j} - \hat{k}\right) \cdot \left(- \hat{i} + 5 \hat{k}\right) \\\\ & = -10 - 5 = - 15 \end{aligned}$

Vector Algebra

Show that the vectors $\hat{i} - 2 \hat{j} + 3 \hat{k}$, $-2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and $-\hat{j} + 2 \hat{k}$ are coplanar.

Let $\alpha$, $\beta$ and $\gamma$ be three constants.

Let $\overrightarrow{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}$;

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

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

Three vectors are coplanar if $\alpha \overrightarrow{a} + \beta \overrightarrow{b} + \gamma \overrightarrow{c} = \overrightarrow{0}$

i.e. if $\alpha \left(\hat{i} - 2 \hat{j} + 3 \hat{k}\right) + \beta \left(-2 \hat{i} + 3 \hat{j} - 4 \hat{k}\right) + \gamma \left(- \hat{j} + 2 \hat{k}\right) = \overrightarrow{0}$ $\;\;\; \cdots \; (1)$

i.e. if

$\hat{i}$ terms: $\;$ $\alpha - 2 \beta + 0 \gamma = 0$ $\;\;\; \cdots \; (2a)$

$\hat{j}$ terms: $\;$ $- 2 \alpha + 3 \beta - \gamma = 0$ $\;\;\; \cdots \; (2b)$

$\hat{k}$ terms: $\;$ $3 \alpha - 4 \beta + 2 \gamma = 0$ $\;\;\; \cdots \; (2c)$

Solving equations $(2a)$, $(2b)$ and $(2c)$ using the Gauss method, the augmented matrix is

$\begin{bmatrix} 1 & -2 & 0 & | 0 \\ -2 & 3 & -1 & | 0 \\ 3 & -4 & 2 & | 0 \end{bmatrix}$

$R_3 - 3 R_1$ $\implies$

$\begin{bmatrix} 1 & -2 & 0 & | 0 \\ -2 & 3 & -1 & | 0 \\ 0 & 2 & 2 & | 0 \end{bmatrix}$

$R_2 + 2R_1$ $\implies$

$\begin{bmatrix} 1 & -2 & 0 & | 0 \\ 0 & -1 & -1 & | 0 \\ 0 & 2 & 2 & | 0 \end{bmatrix}$

$R_3 + 2 R_2$ $\implies$

$\begin{bmatrix} 1 & -2 & 0 & | 0 \\ 0 & -1 & -1 & | 0 \\ 0 & 0 & 0 & | 0 \end{bmatrix}$

$\implies$ The system has many solutions.

$\therefore$ $\;$ $\exists$ non zero combination of numbers $\alpha$, $\beta$, $\gamma$ such that the linear combination of $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ is equal to $\overrightarrow{0}$.

For example: $\;$ $2 \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} = \overrightarrow{0}$

$\implies$ Vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are linearly dependent.

$\implies$ $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are coplanar.

Vector Algebra

If the position vectors of $P$ and $Q$ are $\hat{i} + 3 \hat{j} - 7 \hat{k}$ and $5 \hat{i} - 2 \hat{j} + 4 \hat{k}$, find $\overrightarrow{PQ}$ and determine its direction cosines.

Position vector of point $P = \hat{i} + 3 \hat{j} - 7 \hat{k}$

Position vector of point $Q = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$

$\begin{aligned} \overrightarrow{PQ} & = \text{position vector of Q} - \text{position vector of P} \\\\ & = \left(5 \hat{i} - 2 \hat{j} + 4 \hat{k}\right) - \left(\hat{i} + 3 \hat{j} - 7 \hat{k}\right) \\\\ & = 4 \hat{i} - 5 \hat{j} + 11 \hat{k} \end{aligned}$

$\left|\overrightarrow{PQ}\right| = \sqrt{\left(4\right)^2 + \left(-5\right)^2 + \left(11\right)^2} = \sqrt{162} = 9 \sqrt{2}$

$\therefore$ $\;$ The direction cosines are $\;$ $\dfrac{4}{9 \sqrt{2}}$, $\dfrac{-5}{9 \sqrt{2}}$, $\dfrac{11}{9 \sqrt{2}}$

Vector Algebra

The vertices of a triangle have position vectors $4 \hat{i} + 5 \hat{j} + 6 \hat{k}$, $5 \hat{i} + 6 \hat{j} + 4 \hat{k}$, $6 \hat{i} + 4 \hat{j} + 5 \hat{k}$. Prove that the triangle is equilateral.

Let $A$, $B$ and $C$ be the vertices of the triangle.

Let

position vector of $A = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$

position vector of $B = 5 \hat{i} + 6 \hat{j} + 4 \hat{k}$

position vector of $C = 6 \hat{i} + 4 \hat{j} + 5 \hat{k}$

Now,

$\begin{aligned} \overrightarrow{AB} & = \text{position vector of B} - \text{position vector of A} \\\\ & = \left(5 \hat{i} + 6 \hat{j} + 4 \hat{k}\right) - \left(4 \hat{i} + 5 \hat{j} + 6 \hat{k}\right) \\\\ & = \hat{i} + \hat{j} - 2 \hat{k} \end{aligned}$

$\begin{aligned} \overrightarrow{BC} & = \text{position vector of C} - \text{position vector of B} \\\\ & = \left(6 \hat{i} + 4 \hat{j} + 5 \hat{k}\right) - \left(5 \hat{i} + 6 \hat{j} + 4 \hat{k}\right) \\\\ & = \hat{i} - 2 \hat{j} + \hat{k} \end{aligned}$

$\begin{aligned} \overrightarrow{CA} & = \text{position vector of A} - \text{position vector of C} \\\\ & = \left(4 \hat{i} + 5 \hat{j} + 6 \hat{k}\right) - \left(6 \hat{i} + 4 \hat{j} + 5 \hat{k}\right) \\\\ & = -2 \hat{i} + \hat{j} + \hat{k} \end{aligned}$

$\left|\overrightarrow{AB}\right| = \sqrt{\left(1\right)^2 + \left(1\right)^2 + \left(-2\right)^2} = \sqrt{6}$

$\left|\overrightarrow{BC}\right| = \sqrt{\left(1\right)^2 + \left(-2\right)^2 + \left(1\right)^2} = \sqrt{6}$

$\left|\overrightarrow{CA}\right| = \sqrt{\left(-2\right)^2 + \left(1\right)^2 + \left(1\right)^2} = \sqrt{6}$

$\therefore$ $\;$ $\left|\overrightarrow{AB}\right| = \left|\overrightarrow{BC}\right| = \left|\overrightarrow{CA}\right|$

$\implies$ $A$, $B$ and $C$ are the vertices of an equilateral triangle.

Vector Algebra

Find the unit vectors parallel to $3 \overrightarrow{a} - 2\overrightarrow{b} + 4 \overrightarrow{c}$ where $\overrightarrow{a} = 3 \hat{i} - \hat{j} - 4 \hat{k}$, $\overrightarrow{b} = - 2 \hat{i} + 4 \hat{j} - 3 \hat{k}$ and $\overrightarrow{c} = \hat{i} + 2 \hat{j} - \hat{k}$

$\begin{aligned} \text{Let } \overrightarrow{P} & = 3 \overrightarrow{a} - 2 \overrightarrow{b} + 4 \overrightarrow{c} \\\\ & = 3 \left(3 \hat{i} - \hat{j} - 4 \hat{k}\right) - 2 \left(-2 \hat{i} + 4 \hat{j} - 3 \hat{k}\right) + 4 \left(\hat{i} + 2 \hat{j} - \hat{k}\right) \\\\ & = 9 \hat{i} - 3 \hat{j} - 12 \hat{k} + 4 \hat{i} - 8 \hat{j} + 6 \hat{k} + 4 \hat{i} + 8 \hat{j} - 4 \hat{k} \\\\ & = 17 \hat{i} - 3 \hat{j} - 10 \hat{k} \end{aligned}$

Let $\hat{p}$ be the unit vector parallel to $\overrightarrow{P}$.

$\begin{aligned} \text{Then, } \hat{p} & = \pm \left(\dfrac{\overrightarrow{P}}{\left|\overrightarrow{P}\right|}\right) \\\\ & = \pm \left(\dfrac{17 \hat{i} - 3 \hat{j} - 10 \hat{k}}{\sqrt{\left(17\right)^2 + \left(-3\right)^2 + \left(-10\right)^2}}\right) \\\\ & = \pm \left(\dfrac{17 \hat{i} - 3 \hat{j} - 10 \hat{k}}{\sqrt{398}}\right) \end{aligned}$

Vector Algebra

Show that the points whose position vectors given by $-2 \hat{i} + 3 \hat{j} + 5 \hat{k}$, $\hat{i} + 2 \hat{j} + 3 \hat{k}$, $7 \hat{i} - \hat{k}$ are collinear.

Let the points be $A$, $B$ and $C$.

Let $O$ be the origin.

Then, $\overrightarrow{OA} = -2 \hat{i} + 3 \hat{j} + 5 \hat{k}$; $\overrightarrow{OB} = \hat{i} + 2 \hat{j} + 5 \hat{k}$ and $\overrightarrow{OC} = 7 \hat{i} - \hat{k}$

$\begin{aligned} \text{Now, } \overrightarrow{AB} & = \overrightarrow{OB} - \overrightarrow{OA} \\\\ & = \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right) - \left(-2 \hat{i} + 3 \hat{j} + 5 \hat{k}\right) \\\\ & = 3 \hat{i} - \hat{j} - 2 \hat{k} \end{aligned}$

$\begin{aligned} \text{and } \overrightarrow{AC} & = \overrightarrow{OC} - \overrightarrow{OA} \\\\ & = \left(7 \hat{i} - \hat{k}\right) - \left(-2 \hat{i} + 3 \hat{j} + 5 \hat{k}\right) \\\\ & = 9 \hat{i} - 3 \hat{j} - 6 \hat{k} \\\\ & = 3 \left(3 \hat{i} - \hat{j} - 2 \hat{k}\right) \\\\ & = 3 \overrightarrow{AB} \end{aligned}$

$\implies$ Vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.

Also, point $A$ is a common point to both the vectors.

$\implies$ The points $A$, $B$ and $C$ are collinear.

Vector Algebra

Prove using vectors, the mid-points of two opposite sides of a quadrilateral and the mid-points of the diagonals are the vertices of a parallelogram.

Let $ABCD$ be a quadrilateral.

$E$ and $F$ are the midpoints of sides $AB$ and $CD$ respectively.

$AC$ and $BD$ are the diagonals of the quadrilateral $ABCD$.

$G$ and $H$ are the mid-points of the diagonals $AC$ and $BD$ respectively.

Position vectors of points $A$, $B$, $C$ and $D$ are $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ and $\overrightarrow{d}$ respectively.

Position vectors of points $E$, $H$, $F$ and $G$ are $\dfrac{\overrightarrow{a} + \overrightarrow{b}}{2}$, $\dfrac{\overrightarrow{b} + \overrightarrow{d}}{2}$, $\dfrac{\overrightarrow{c} + \overrightarrow{d}}{2}$ and $\dfrac{\overrightarrow{a} + \overrightarrow{c}}{2}$ respectively.

To prove that $EHFG$ is a parallelogram, it is sufficient to show that $\overrightarrow{GE} = \overrightarrow{FH}$ and $\overrightarrow{EH} = \overrightarrow{GF}$.

Now,

$\begin{aligned} \overrightarrow{GE} & = \text{position vector of E} - \text{position vector of G} \\\\ & = \left(\dfrac{\overrightarrow{a} + \overrightarrow{b}}{2}\right) - \left(\dfrac{\overrightarrow{a} + \overrightarrow{c}}{2}\right) \\\\ & = \dfrac{\overrightarrow{b} - \overrightarrow{c}}{2} \end{aligned}$

and

$\begin{aligned} \overrightarrow{FH} & = \text{position vector of H} - \text{position vector of F} \\\\ & = \left(\dfrac{\overrightarrow{b} + \overrightarrow{d}}{2}\right) - \left(\dfrac{\overrightarrow{c} + \overrightarrow{d}}{2}\right) \\\\ & = \dfrac{\overrightarrow{b} - \overrightarrow{c}}{2} \end{aligned}$

$\therefore$ $\;$ $\overrightarrow{GE} = \overrightarrow{FH}$

$\implies$ $GE \parallel FH$ and $GE = FH$ $\;\;\; \cdots \; (1)$

Now,

$\begin{aligned} \overrightarrow{EH} & = \text{position vector of H} - \text{position vector of E} \\\\ & = \left(\dfrac{\overrightarrow{b} + \overrightarrow{d}}{2}\right) - \left(\dfrac{\overrightarrow{a} + \overrightarrow{b}}{2}\right) \\\\ & = \dfrac{\overrightarrow{d} - \overrightarrow{a}}{2} \end{aligned}$

and

$\begin{aligned} \overrightarrow{GF} & = \text{position vector of F} - \text{position vector of G} \\\\ & = \left(\dfrac{\overrightarrow{c} + \overrightarrow{d}}{2}\right) - \left(\dfrac{\overrightarrow{a} + \overrightarrow{c}}{2}\right) \\\\ & = \dfrac{\overrightarrow{d} - \overrightarrow{a}}{2} \end{aligned}$

$\therefore$ $\;$ $\overrightarrow{EH} = \overrightarrow{GF}$

$\implies$ $EH \parallel GF$ and $EH = GF$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ From $(1)$ and $(2)$, $EHFG$ is a parallelogram.

Hence proved.

Vector Algebra

Prove that the sum of the vectors directed from the vertices to the mid-points of opposite sides of a triangle is zero.

Let $O$ be the origin.

Let the position vectors of points $A$, $B$ and $C$ be

$\overrightarrow{OA} = \overrightarrow{a}$, $\overrightarrow{OB} = \overrightarrow{b}$ and $\overrightarrow{OC} = \overrightarrow{c}$ respectively.

Let $D$, $E$ and $F$ be the midpoints of $AB$, $AC$ and $BC$ respectively.

Then,

position vector of point $D = \overrightarrow{OD} = \dfrac{\overrightarrow{a} + \overrightarrow{b}}{2}$

position vector of point $E = \overrightarrow{OE} = \dfrac{\overrightarrow{a} + \overrightarrow{c}}{2}$

position vector of point $F = \overrightarrow{OF} = \dfrac{\overrightarrow{b} + \overrightarrow{c}}{2}$

To prove that: $\;$ $\overrightarrow{CD} + \overrightarrow{BE} + \overrightarrow{AF} = \overrightarrow{O}$

Now,

$\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = \dfrac{\overrightarrow{a} + \overrightarrow{b}}{2} - \overrightarrow{c}$

$\overrightarrow{BE} = \overrightarrow{OE} - \overrightarrow{OB} = \dfrac{\overrightarrow{a} + \overrightarrow{c}}{2} - \overrightarrow{b}$

$\overrightarrow{AF} = \overrightarrow{OF} - \overrightarrow{OA} = \dfrac{\overrightarrow{b} + \overrightarrow{c}}{2} - \overrightarrow{a}$

$\begin{aligned} \therefore \; \overrightarrow{CD} + \overrightarrow{BE} + \overrightarrow{AF} & = \dfrac{\overrightarrow{a} + \overrightarrow{b}}{2} - \overrightarrow{c} + \dfrac{\overrightarrow{a} + \overrightarrow{c}}{2} - \overrightarrow{b} + \dfrac{\overrightarrow{b} + \overrightarrow{c}}{2} - \overrightarrow{a} \\\\ & = \dfrac{2 \overrightarrow{a} + 2 \overrightarrow{b} + 2 \overrightarrow{c}}{2} - \left(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}\right) \\\\ & = \left(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}\right) - \left(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}\right) \\\\ & = 0 \\\\ & = \overrightarrow{O} \end{aligned}$

Hence proved.

Vector Algebra

If $ABC$ and $A'B'C'$ are two triangles and $G$ and $G'$ be their corresponding centroids, prove that $\overrightarrow{AA'} + \overrightarrow{BB'} + \overrightarrow{CC'} = 3 \overrightarrow{GG'}$

Let $O$ be the origin.

Let the position vectors of points $A$, $B$ and $C$ be

$\overrightarrow{OA} = \overrightarrow{a}$, $\overrightarrow{OB} = \overrightarrow{b}$ and $\overrightarrow{OC} = \overrightarrow{c}$ respectively.

$G$ is the centroid of $\triangle ABC$.

$\therefore$ $\;$ Position vector of $G = \overrightarrow{OG} = \dfrac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{3}$

Let the position vectors of points $A'$, $B'$ and $C'$ be

$\overrightarrow{OA'} = \overrightarrow{a'}$, $\overrightarrow{OB'} = \overrightarrow{b'}$ and $\overrightarrow{OC'} = \overrightarrow{c'}$ respectively.

$G'$ is the centroid of $\triangle A'B'C'$.

$\therefore$ $\;$ Position vector of $G' = \overrightarrow{OG'} = \dfrac{\overrightarrow{a'} + \overrightarrow{b'} + \overrightarrow{c'}}{3}$

Now,

$\overrightarrow{AA'} = \overrightarrow{OA'} - \overrightarrow{OA} = \overrightarrow{a'} - \overrightarrow{a}$

$\overrightarrow{BB'} = \overrightarrow{OB'} - \overrightarrow{OB} = \overrightarrow{b'} - \overrightarrow{b}$

$\overrightarrow{CC'} = \overrightarrow{OC'} - \overrightarrow{OC} = \overrightarrow{c'} - \overrightarrow{c}$

$\begin{aligned} \therefore \; \overrightarrow{AA'} + \overrightarrow{BB'} + \overrightarrow{CC'} & = \overrightarrow{a'} - \overrightarrow{a} + \overrightarrow{b'} - \overrightarrow{b} + \overrightarrow{c'} - \overrightarrow{c} \\\\ & = \left(\overrightarrow{a'} + \overrightarrow{b'} + \overrightarrow{c'}\right) - \left(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}\right) \\\\ & = 3 \left(\dfrac{\overrightarrow{a'} + \overrightarrow{b'} + \overrightarrow{c'}}{3}\right) - 3 \left(\dfrac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{3}\right) \\\\ & = 3 \left(\overrightarrow{OG'} - \overrightarrow{OG}\right) \\\\ & = 3 \overrightarrow{GG'} \end{aligned}$

Hence proved.

Vector Algebra

Show that the points $A$, $B$, $C$ with position vectors $\;$ $-2 \overrightarrow{a} + 3 \overrightarrow{b} + 5 \overrightarrow{c}$, $\;$ $\overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}$ $\;$ and $\;$ $7 \overrightarrow{a}- \overrightarrow{c}$ $\;$ respectively, are collinear.

Let $O$ be the origin.

The position vectors of points $A$, $B$ and $C$ are $\;$ $-2 \overrightarrow{a} + 3 \overrightarrow{b} + 5 \overrightarrow{c}$, $\;$ $\overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}$ $\;$ and $\;$ $7 \overrightarrow{a} - \overrightarrow{c}$ $\;$ respectively.

i.e. $\;$ $\overrightarrow{OA} = -2 \overrightarrow{a} + 3 \overrightarrow{b} + 5 \overrightarrow{c}$

$\overrightarrow{OB} = \overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}$ $\;$ and

$\overrightarrow{OC} = 7 \overrightarrow{a} - \overrightarrow{c}$

Now, $\;$ $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$

i.e. $\;$ $\overrightarrow{AB} = \left(\overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}\right) - \left(- 2 \overrightarrow{a} + 3 \overrightarrow{b} + 5 \overrightarrow{c}\right) = 3 \overrightarrow{a} - \overrightarrow{b} - 2 \overrightarrow{c}$

and $\;$ $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}$

i.e. $\;$ $\overrightarrow{BC} = \left(7 \overrightarrow{a} - \overrightarrow{c}\right) - \left(\overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}\right) = 6 \overrightarrow{a} - 2 \overrightarrow{b} - 4 \overrightarrow{c}$

i.e. $\;$ $\overrightarrow{BC} = 2 \left(3 \overrightarrow{a} - \overrightarrow{b} - 2 \overrightarrow{c}\right) = 2 \overrightarrow{AB}$

$\implies$ $\overrightarrow{AB}$ $\;$ and $\;$ $\overrightarrow{BC}$ $\;$ are parallel vectors and point $B$ is common to both of them.

$\implies$ $\overrightarrow{AB}$ $\;$ and $\;$ $\overrightarrow{BC}$ $\;$ are collinear vectors.

$\implies$ $A$, $B$ and $C$ are collinear points.

Vector Algebra

If $\overrightarrow{a}$ and $\overrightarrow{b}$ represent two adjacent sides $\overrightarrow{AB}$ and $\overrightarrow{BC}$ respectively of a parallelogram $ABCD$, find the diagonals $\overrightarrow{AC}$ and $\overrightarrow{BD}$.

$ABCD$ is a parallelogram.

$\therefore$ $\;$ $AD = BC$ and $AD \parallel BC$.

Given: $\overrightarrow{AB} = \overrightarrow{a}$; $\;$ $\overrightarrow{BC} = \overrightarrow{b}$

Diagonal $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{a} + \overrightarrow{b}$

$\because$ $\;$ $AD = BC$ and $AD \parallel BC$,

$\therefore$ $\;$ $\overrightarrow{AD} = \overrightarrow{BC} = \overrightarrow{b}$

Now, $\overrightarrow{BA} = - \overrightarrow{AB} = - \overrightarrow{a}$

Diagonal $\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = - \overrightarrow{a} + \overrightarrow{b}$

Analytical Geometry - Conics - Rectangular Hyperbola

Prove that the tangent at any point to the rectangular hyperbola forms with the asymptotes a triangle of constant area.

Let the rectangular hyperbola be: $\;$ $xy = c^2$ $\;\;\; \cdots \; (1)$

The asymptotes of the rectangular hyperbola given by equation $(1)$ are the $X$ and $Y$ coordinate axes.

Any point $P \left(t\right)$ on equation $(1)$ is $\;$ $P \left(t\right) = P \left(ct, \dfrac{c}{t}\right)$

Equation of tangent at point $P \left(t\right)$ is

$x + yt^2 = 2ct$ $\;\;\; \cdots \; (2)$

Putting $y = 0$ in equation $(2)$, we get the coordinates of point $A$ as $A \left(2ct, 0\right)$.

Putting $x = 0$ in equation $(2)$, we get the coordinates of point $B$ as $B \left(0, \dfrac{2c}{t}\right)$

$\begin{aligned} \therefore \; \text{Area of } \triangle AOB & = \dfrac{1}{2} \times OA \times OB \\\\ & = \dfrac{1}{2} \times 2 ct \times \dfrac{2c}{t} \\\\ & = 2c^2 = \text{constant} \end{aligned}$

$\therefore$ $\;$ The tangent at any point to the rectangular hyperbola forms with the asymptotes a triangle of constant area.

Hence proved.

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equations of the asymptotes for the rectangular hyperbola $6x^2 + 5xy - 6y^2 + 12x + 5y + 3 = 0$

Equation of the rectangular hyperbola is

$6x^2 + 5xy - 6y^2 + 12 x + 5y + 3 = 0$ $\;\;\; \cdots \; (1)$

The combined equation of the asymptotes differs from the hyperbola by a constant.

$\therefore$ $\;$ The combined equation of the asymptotes is

$6x^2 + 5xy - 6y^2 + 12x + 5y + k = 0$

Consider

$\begin{aligned} 6x^2 + 5xy - 6y^2 & = 6x^2 + 9xy - 4xy - 6y^2 \\\\ & = 3x \left(2x + 3y\right) - 2y \left(2x + 3y\right) \\\\ & = \left(3x - 2y\right) \left(2x + 3y\right) \end{aligned}$

$\therefore$ $\;$ The separate equations of asymptotes are

$3x - 2y + \ell = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $2x + 3y + m = 0$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $\left(3x - 2y + \ell\right) \left(2x + 3y + m\right) = 6x^2 + 5xy -6y^2 + 12x + 5y + k$

Equating the coefficients of $x$ terms we get: $\;$ $3m + 2 \ell = 12$ $\;\;\; \cdots \; (3a)$

Equating the coefficients of $y$ terms we get: $\;$ $-2m + 3 \ell = 5$ $\;\;\; \cdots \; (3b)$

Equating the constant terms we get: $\;$ $\ell \times m = k$ $\;\;\; \cdots \; (3c)$

Solving equations $(3a)$ and $(3b)$ simultaneously we get,

$\ell = 3$ $\;$ and $\;$ $m = 2$

Substituting the values of $\ell$ and $m$ in equations $(2a)$ and $(2b)$ respectively, we get the separate equations of asymptotes as

$3x - 2y + 3 = 0$ $\;$ and $\;$ $2x + 3y + 2 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equations of the asymptotes for the rectangular hyperbola $2xy + 3x + 4y + 1 = 0$

Equation of given rectangular hyperbola is

$2xy + 3x + 4y + 1 = 0$

i.e. $\;$ $xy + \dfrac{3}{2}x + 2y + \dfrac{1}{2} = 0$

i.e. $\;$ $xy + \dfrac{3}{2}x + 2y + \dfrac{1}{2} + 3 = 3$

i.e. $\;$ $xy + \dfrac{3}{2}x + 2y + \dfrac{7}{2} = 3$

i.e. $\;$ $y \left(x + 2\right) + \dfrac{3}{2} \left(x + 2\right) = 3$

i.e. $\;$ $\left(x + 2\right) \left(y + \dfrac{3}{2}\right) - 3 = 0$

The equation of a rectangular hyperbola differs from the combined equation of asymptotes by a constant.

$\therefore$ $\;$ The combined equation of asymptotes is

$\left(x + 2\right) \left(y + \dfrac{3}{2}\right) = 0$

$\therefore$ $\;$ The separate equations of asymptotes are

$x+ 2 = 0$ $\;$ and $\;$ $y + \dfrac{3}{2} = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the rectangular hyperbola which has its center at $\left(2,1\right)$, one of its asymptotes $3x - y - 5 = 0$ and which passes through the point $\left(1, -1\right)$.

Let the equation of the required rectangular hyperbola be

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$

One of the asymptotes of the required rectangular hyperbola is

$3x - y - 5 = 0$ $\;\;\; \cdots \; (2)$

Slope of the given asymptote is $= m_1 = 3$

The asymptotes of a rectangular hyperbola are perpendicular to each other.

$\therefore$ $\;$ Slope of the second asymptote $= m_2 = \dfrac{-1}{m_1} = \dfrac{-1}{3}$

Let the equation of the second asymptote be

$y = \dfrac{-1}{3}x + p$ $\;$ where $p$ is a constant.

i.e. $\;$ $x + 3y = 3p$ $\;\;\; \cdots \; (3)$

Solving equations $(2)$ and $(3)$ simultaneously we have

$x = \dfrac{15 + 3p}{10}$

The asymptotes intersect at the center.

Given: Center $= \left(h, k\right) = \left(2, 1\right)$

$\implies$ $\dfrac{15 + 3p}{10} = 2$ $\implies$ $p = \dfrac{5}{3}$

Substituting the value of $p$ in equation $(3)$, the equation of the second asymptote is

$x + 3y - 5 = 0$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ The combined equation of the asymptotes is

$\left(3x - y - 5\right) \left(x + 3y - 5\right) = 0$ $\;\;\; \cdots \; (5)$

The equation of rectangular hyperbola differs from the combined equation of asymptotes by a constant.

$\therefore$ $\;$ The equation of the rectangular hyperbola is

$\left(3x - y - 5\right) \left(x + 3y - 5\right) + \ell = 0$ $\;\;\; \cdots \; (6)$

Equation $(6)$ passes through the point $\left(1, -1\right)$.

$\therefore$ $\;$ We have

$\left(3 + 1 - 5\right) \left(1 - 3 - 5\right) + \ell = 0$ $\implies$ $\ell = -7$

$\therefore$ $\;$ The required equation of rectangular hyperbola is

$\left(3x - y - 5\right) \left(x + 3y - 5\right) - 7 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

A standard rectangular hyperbola has its vertices at $\left(5, 7\right)$ and $\left(-3, -1\right)$. Find its equation and asymptotes.

Let the required equation of the general rectangular hyperbola be

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$

The vertices of the rectangular hyperbola are $\left(5, 7\right)$ and $\left(-3, -1\right)$.

Let the center of the rectangular hyperbola be $\left(h, k\right)$.

The midpoint of the vertices of a rectangular hyperbola is its center.

$\therefore$ $\;$ We have, $\;$ $h = \dfrac{5 - 3}{2} = 1$ $\;$ and $\;$ $k = \dfrac{7 - 1}{2} = 3$

$\therefore$ $\;$ The center of the rectangular hyperbola is $\left(h, k\right) = \left(1, 3\right)$

Substituting the values of $h$ and $k$ in equation $(1)$ we have,

$\left(x -1\right) \left(y - 3\right) = c^2$ $\;\;\; \cdots \; (2)$

The vertices $\left(5, 7\right)$ and $\left(-3, -1\right)$ lie on the hyperbola.

$\therefore$ $\;$ We have from equation $(2)$,

$\left(5 - 1\right) \left(7 - 3\right) = c^2$ $\implies$ $c^2 = 16$

Substituting the value of $c^2$ in equation $(2)$, the required equation of rectangular hyperbola is

$\left(x -1\right) \left(y - 3\right) = 16$

The equation of rectangular hyperbola differs from the combined equation of asymptotes by a constant.

$\therefore$ $\;$ The combined equation of the asymptotes is $\;$ $\left(x -1\right) \left(y - 3\right) = 0$

$\therefore$ $\;$ The equations of the asymptotes are

$x -1 = 0$ $\;$ and $\;$ $y - 3 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the rectangular hyperbola which which has for one of its asymptotes the line $x + 2y - 5 = 0$ and passes through the points $\left(6,0\right)$ and $\left(-3,0\right)$.

Given: One of the asymptotes of the rectangular hyperbola is

$x + 2y - 5 = 0$ $\;\;\; \cdots \; (1)$

$\therefore$ $\;$ Slope of the asymptote $= m_1 = \dfrac{-1}{2}$

The asymptotes of a rectangular hyperbola are at right angles to one another.

$\therefore$ $\;$ Slope of the second asymptote $= m_2 = \dfrac{-1}{m_1} = 2$

$\therefore$ $\;$ Let the equation of the second asymptote be

$y = 2x + k$ $\;$ where $k$ is a constant.

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

$\therefore$ $\;$ The combined equation of the asymptotes is

$\left(x + 2y - 5\right) \left(2x - y + k\right) = 0$ $\;\;\; \cdots \; (3)$

The equation of rectangular hyperbola differs from the combined equation of asymptotes by a constant.

$\therefore$ $\;$ The equation of hyperbola is

$\left(x + 2y - 5\right) \left(2x - y + k\right) + \ell = 0$ $\;\;\; \cdots \; (4)$

Equation $(4)$ passes through the points $\left(6, 0\right)$ and $\left(-3, 0\right)$.

$\therefore$ $\;$ We have

$\left(6 + 0 - 5\right) \left(12 - 0 + k\right) + \ell = 0$ $\implies$ $k + \ell = - 12$ $\;\;\; \cdots \; (5a)$

and $\left(-3 + 0 - 5\right) \left(-6 - 0 + k\right) + \ell = 0$ $\implies$ $-8k + \ell = -48$ $\;\;\; \cdots \; (5b)$

Solving equations $(5a)$ and $(5b)$ simultaneously we have,

$k = 4$ $\;$ and $\;$ $\ell = -16$

Substituting the values of $k$ and $\ell$ in equation $(4)$, the equation of the required rectangular hyperbola is

$\left(x + 2y - 5\right) \left(2x - y + 4\right) - 16 = 0$

i.e. $\left(x + 2y - 5\right) \left(2x - y + 4\right) = 16$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the tangent and normal at $\left(-2, \dfrac{1}{4}\right)$ to the rectangular hyperbola $2xy - 2x - 8y - 1 = 0$

Equation of given rectangular hyperbola is

$2xy - 2x - 8y -1 = 0$

i.e. $\;$ $2xy - 2x - 8y = 1$

i.e. $\;$ $2xy - 2x - 8y + 8 = 9$

i.e. $\;$ $xy - x - 4y + 4 = \dfrac{9}{2}$

i.e. $\;$ $x \left(y - 1\right) - 4 \left(y - 1\right) = \dfrac{9}{2}$

i.e. $\;$ $\left(x - 4\right) \left(y - 1\right) = \dfrac{9}{2}$ $\;\;\; \cdots \; (1)$

The standard equation of rectangular hyperbola with center at $\left(h, k\right)$ is:

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (2)$

Comparing equations $(1)$ and $(2)$ we have

Center $= \left(h, k\right) = \left(4, 1\right)$ $\;$ and $\;$ $c^2 = \dfrac{9}{2}$

Equation of tangent to the rectangular hyperbola at the point $\left(x_1, y_1\right)$ is

$x \left(y_1 - k\right) + y \left(x_1 - h\right) = c^2 - hk + x_1y_1$

Given: Point $\left(x_1, y_1\right) = \left(-2, \dfrac{1}{4}\right)$

$\therefore$ $\;$ The required equation of tangent is

$x \left(\dfrac{1}{4} - 1\right) + y \left(-2 - 4\right) = \dfrac{9}{2} - \left(4 \times 1\right) + \left(-2 \times \dfrac{1}{4}\right)$

i.e. $\;$ $\dfrac{-3}{4}x - 6y = 0$

i.e. $\;$ $\dfrac{-x}{4}- 2y = 0$

i.e. $\;$ $x + 8y = 0$

Equation of normal to the rectangular hyperbola at the point $\left(x_1, y_1\right)$ is

$x \left(x_1 - h\right) - y \left(y_1 - k\right) = x_1 \left(x_1 - h\right) - y_1 \left(y_1 - k\right)$

$\therefore$ $\;$ The required equation of normal is

$x \left(-2 - 4\right) - y \left(\dfrac{1}{4} - 1\right) = -2 \left(-2 - 4\right) - \dfrac{1}{4} \left(\dfrac{1}{4} - 1\right)$

i.e. $\;$ $- 6x + \dfrac{3}{4}y = \dfrac{195}{16}$

i.e. $\;$ $- 2x + \dfrac{y}{4} = \dfrac{65}{16}$

i.e. $\;$ $32x - 4y + 65 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the standard rectangular hyperbola whose center is $\left(\dfrac{-1}{2}, \dfrac{-1}{2}\right)$ and which passes through the point $\left(1, \dfrac{1}{4}\right)$.

The equation of standard rectangular hyperbola with center at $\left(h, k\right)$ is:

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$

Given: Center $= \left(h, k\right) = \left(\dfrac{-1}{2}, \dfrac{-1}{2}\right)$

Putting the values of $h$ and $k$ in equation $(1)$, the required equation of rectangular hyperbola becomes

$\left(x + \dfrac{1}{2}\right) \left(y + \dfrac{1}{2}\right) = c^2$ $\;\;\; \cdots \; (2)$

Equation $(2)$ passes through the point $\left(1, \dfrac{1}{4}\right)$.

$\therefore$ $\;$ We have

$\left(1 + \dfrac{1}{2}\right) \left(\dfrac{1}{4} + \dfrac{1}{2}\right) = c^2$ $\implies$ $c^2 = \dfrac{9}{8}$

$\therefore$ $\;$ The required equation of the rectangular hyperbola [equation $(2)$] is

$\left(x + \dfrac{1}{2}\right) \left(y + \dfrac{1}{2}\right) = \dfrac{9}{8}$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the angle between the asymptotes of the hyperbola $\;$ $24x^2 - 8y^2 = 27$

Equation of hyperbola is $\;$ $24x^2 - 8y^2 = 27$

i.e. $\;$ $\dfrac{x^2}{27 / 24} - \dfrac{y^2}{27 / 8} = 1$

i.e. $\;$ $\dfrac{x^2}{9 / 8} - \dfrac{y^2}{27 / 8} = 1$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = \dfrac{9}{8}$ $\implies$ $a = \dfrac{3}{2 \sqrt{2}}$

and $b^2 = \dfrac{27}{8}$ $\implies$ $b = \dfrac{3 \sqrt{3}}{2 \sqrt{2}}$

The angle between the asymptotes is

$\begin{aligned} 2 \alpha & = 2 \tan^{-1} \left(\dfrac{b}{a}\right) \\\\ & = 2 \tan^{-1} \left(\dfrac{3 \sqrt{3} / 2 \sqrt{2}}{3 / 2 \sqrt{2}}\right) \\\\ & = 2 \tan^{-1} \left(\sqrt{3}\right) \end{aligned}$

i.e. $\;$ $2 \alpha = \dfrac{2 \pi}{3}$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the equation of the hyperbola if its asymptotes are parallel to $\;$ $x + 2y - 12 = 0$ $\;$ and $\;$ $x - 2y+ 8 = 0$, $\;$ $\left(2,4\right)$ is the center of the hyperbola and it passes through $\left(2,0\right)$.

Equations of the given straight lines are

$x + 2y - 12 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x - 2y + 8 = 0$ $\;\;\; \cdots \; (2)$

Slope of equation $(1)$ is $= m_1 = \dfrac{-1}{2}$

Slope of equation $(2)$ is $= m_2 = \dfrac{1}{2}$

$\because$ $\;$ the asymptotes are parallel to the given lines [equations $(1)$ and $(2)$],

$\therefore$ $\;$ the slopes of the asymptotes are $\dfrac{1}{2}$ and $\dfrac{-1}{2}$

Let the equations of the asymptotes be

$y = \dfrac{1}{2}x + p$ $\;\;\; \cdots \; (3a)$ $\;$ and $\;$ $y = -\dfrac{1}{2}x + q$ $\;\;\; \cdots \; (3b)$

where $p$ and $q$ are the Y intercepts.

Solving equations $(3a)$ and $(3b)$ simultaneously gives the point of intersection of the asymptotes.

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

$\dfrac{1}{2}x + p = \dfrac{-1}{2}x + q$ $\implies$ $x = q - p$

Substituting the value of $x$ in equation $(3a)$ gives

$y = \dfrac{1}{2} \left(q - p\right) + p$ $\implies$ $y = \dfrac{1}{2} \left(p + q\right)$

The asymptotes intersect at the center.

Given: Center of hyperbola $= \left(2, 4\right)$

$\therefore$ $\;$ We have $\;$ $q - p = 2$ $\;\;\; \cdots \; (4a)$

and $\;$ $\dfrac{1}{2} \left(p + q\right) = 4$ $\implies$ $p + q = 8$ $\;\;\; \cdots \; (4b)$

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

$2q = 10$ $\implies$ $q = 5$

Substituting the value of $q$ in equation $(4a)$ gives

$p = q - 2 = 3$

$\therefore$ $\;$ The equations of the asymptotes are

$y = \dfrac{1}{2}x + 3$ $\;\;$ i.e. $\;$ $x - 2y + 6 = 0$

and $\;$ $y = \dfrac{-1}{2}x + 5$ $\;\;$ i.e. $\;$ $x + 2y - 10 = 0$

$\therefore$ $\;$ The combined equations of the asymptotes is

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) = 0$ $\;\;\; \cdots \; (5)$

The equation of the hyperbola differs from the combined equation of the asymptotes by a constant.

$\therefore$ $\;$ The equation of the hyperbola is of the form

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + k = 0$ $\;\;\; \cdots \; (6)$

The required hyperbola passes through the point $\left(2, 0\right)$.

$\therefore$ $\;$ We have from equation $(6)$,

$\left(2 - 0 + 6\right) \left(2 + 0 - 10\right) + k = 0$

$\implies$ $k = 64$

Substituting the value of $k$ in equation $(6)$, the equation of the required hyperbola is

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + 64 = 0$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the equation of the asymptotes to the hyperbola $8x^2 + 10xy - 3y^2 - 2x + 4y -2 = 0$

Equation of given hyperbola is $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y - 2 = 0$ $\;\;\; \cdots \; (1)$

The combined equation of the asymptotes differs from the hyperbola by a constant only.

$\therefore$ $\;$ Let the combined equation of the asymptotes be: $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y + k = 0$

Now,

$\begin{aligned} 8x^2 + 10xy - 3y^2 & = 8x^2 + 12xy - 2xy - 3y^2 \\\\ & = 4x \left(2x + 3y\right) - y \left(2x + 3y\right) \\\\ & = \left(4x - y\right) \left(2x + 3y\right) \end{aligned}$

$\therefore$ $\;$ The separate equations of the asymptotes are

$4x - y + \ell = 0$ $\;\;\; \cdots \; (2a)$ and

$2x + 3y + m = 0$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $\left(4x - y + \ell\right) \left(2x + 3y + m\right) = 8x^2 + 10xy - 3y^2 - 2x + 4y + k$ $\;\;\; \cdots \; (3)$

Equating the coefficient of the $x$ terms we have

$4 m + 2 \ell = -2$

i.e. $\;$ $2 m + \ell = -1$ $\;\;\; \cdots \; (4a)$

Equating the coefficient of the $y$ terms we have

$- m + 3 \ell = 4$ $\;\;\; \cdots \; (4b)$

Equating the constant term we have

$\ell m = k$ $\;\;\; \cdots \; (4c)$

Solving equations $(4a)$ and $(4b)$ simultaneously we get

$\ell = 1$ $\;$ and $m = -1$

$\therefore$ $\;$ We have from equation $(4c)$, $\;$ $k = -1$

$\therefore$ $\;$ The separate equations of the asymptotes are

$4x - y + 1 = 0$ $\;$ and $\;$ $2x + 3y - 1 = 0$

The combined equation of the asymptotes is

$8x^2 + 10xy - 3y^2 - 2x + 4y - 1 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the chord of contact of tangents from the point $\left(5,3\right)$ to the hyperbola $4x^2 - 6y^2 = 24$.

Given equation of hyperbola is $\;$ $4x^2 - 6y^2 = 24$

i.e. $\;$ $\dfrac{x^2}{24 / 4} - \dfrac{y^2}{24 / 6} = 1$

i.e. $\;$ $\dfrac{x^2}{6} - \dfrac{y^2}{4} = 1$

The transverse axis of the given hyperbola is along the X axis.

Comparing with the standard equation of the hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 6$ $\;$ and $\;$ $b^2 = 4$

Equation of chord of contact of tangents from the point $\left(x_1, y_1\right)$ to the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is: $\;$ $\dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1$

Here $\;$ $\left(x_1, y_1\right) = \left(5,3\right)$

$\therefore$ $\;$ The required equation is

$\dfrac{5x}{6} - \dfrac{3y}{4} = 1$

i.e. $\;$ $10x - 9y - 12 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the chord of contact of tangents from the point $\left(-3,1\right)$ to the parabola $y^2 = 8x$.

Equation of parabola is $\;$ $y^2 = 8x$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 8 \implies a = 2$

Equation of the chord of contact of tangents from the point $\left(x_1, y_1\right)$ to the parabola $y^2 = 4ax$ is

$yy_1 = 2a \left(x + x_1\right)$

Here, $\left(x_1, y_1\right) = \left(-3,1\right)$

$\therefore$ $\;$ Required equation is

$1 \times y = 2 \times 2 \times \left(x - 3\right)$

i.e. $\;$ $4x - y - 12 = 0$

Analytical Geometry - Conics - Tangents and Normals

Show that the line $x - y + 4 = 0$ is a tangent to the ellipse $x^2 + 3y^2 = 12$. Find the coordinates of the point of contact.

Equation of given line is $\;$ $x - y + 4 = 0$

i.e. $\;$ $y = x + 4$ $\;\;\; \cdots \; (1)$

Slope of given line $= m = 1$

Intercept of given line $= c = 4$

Equation of ellipse is $\;$ $x^2 + 3y^2 = 12$

i.e. $\;$ $\dfrac{x^2}{12} + \dfrac{y^2}{12 / 3} = 1$

i.e. $\;$ $\dfrac{x^2}{12} + \dfrac{y^2}{4} = 1$ $\;\;\; \cdots \; (2)$

The major axis of the ellipse given by equation $(2)$ is along the X axis.

$\therefore$ $\;$ Comparing equation $(2)$ with the tandard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 12$ $\;$ and $\;$ $b^2 = 4$

Condition that the given line is a tangent to the ellipse is $\;$ $c^2 = a^2 m^2 + b^2$ $\;\;\; \cdots \; (3)$

Substituting the values of $c$, $a^2$, $m$ and $b^2$ in equation $(3)$ we have,

$\left(4\right)^2 = 12 \times \left(1\right)^2 + 4$

i.e. $\;$ $16 = 16$ $\;$ which is true.

$\therefore$ $\;$ The line given by equation $(1)$ is a tangent to the ellipse given by equation $(2)$.

The point of contact of the tangent with the ellipse is $\;$ $\left(\dfrac{-a^2 m}{c}, \dfrac{b^2}{c}\right)$

i.e. $\;$ $\left(\dfrac{-12 \times 1}{4}, \dfrac{4}{4}\right)$ $\;\;$ i.e. $\;$ $\left(-3, 1\right)$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the two tangents that can be drawn from the point $\left(1, 2\right)$ to the hyperbola $2x^2 - 3y^2 = 6$.

Equation of hyperbola is $\;$ $2x^2 - 3y^2 = 6$

i.e. $\;$ $\dfrac{x^2}{6 / 2} - \dfrac{y^2}{6 / 3} = 1$

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

The transverse axis of the hyperbola given by equation $(1)$ is along the X axis.

$\therefore$ $\;$ Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 3$ $\;$ and $b^2 = 2$

Let the equation of the tangent be $\;$ $y = mx + \sqrt{a^2 m^2 - b^2}$ $\;\;\; \cdots \; (2)$

Substituting the values of $a^2$ and $b^2$ in equation $(2)$, we have,

$y = mx + \sqrt{3m^2 - 2}$ $\;\;\; \cdots \; (3)$

Equation $(3)$ passes through the point $\left(1, 2\right)$.

$\therefore$ $\;$ We have from equation $(3)$,

$2 = m + \sqrt{3m^2 - 2}$

i.e. $\;$ $2 - m = \sqrt{3m^2 - 2}$ $\;\;\; \cdots \; (4)$

Squaring both sides of equation $(4)$ we get,

$4 + m^2 - 4m = 3m^2 - 2$

i.e. $\;$ $2m^2 + 4m - 6 = 0$

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

i.e. $\;$ $\left(m + 3\right) \left(m - 1\right) = 0$

i.e. $\;$ $m = -3$ $\;$ or $\;$ $m = 1$

Substituting $m = -3$ in equation $(3)$ gives

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

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

Substituting $m = 1$ in equation $(3)$ gives

$y = x + \sqrt{3 \times \left(1\right)^2 - 2}$

i.e. $\;$ $x - y + 1 = 0$ $\;\;\; \cdots \; (5b)$

Equations $(5a)$ and $(5b)$ are the required equations of tangents.

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the two tangents that can be drawn from the point $\left(2,-3\right)$ to the parabola $y^2 = 4x$.

Equation of parabola is $\;$ $y^2 = 4x$ $\;\;\; \cdots \; (1)$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives $\;$ $a = 1$

Let the equation of tangent be $\;$ $y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (2)$

Substituting the value of $a$ in equation $(2)$ gives

$y = mx + \dfrac{1}{m}$ $\;\;\; \cdots \; (3)$

Equation $(3)$ passes through the point $\left(2, -3\right)$.

$\therefore$ $\;$ We have $\;$ $-3 = 2m + \dfrac{1}{m}$

i.e. $\;$ $2m^2 + 3m + 1 = 0$

i.e. $\;$ $\left(2m + 1\right) \left(m + 1\right) = 0$

i.e. $\;$ $m = -1$ $\;$ or $\;$ $m = \dfrac{-1}{2}$

When $m = -1$, we have from equation $(3)$, $\;$ $y = -x -1$

i.e. $\;$ $x + y + 1 = 0$ $\;\;\; \cdots \; (4a)$

When $m = \dfrac{-1}{2}$, we have from equation $(3)$, $\;$ $y = \dfrac{-x}{2} - \dfrac{1}{1 / 2}$

i.e. $\;$ $y = \dfrac{-x}{2} - 2$

i.e. $\;$ $x + 2y + 4 = 0$ $\;\;\; \cdots \; (4b)$

Equations $(4a)$ and $(4b)$ are the required equations of tangents.

Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangents to the hyperbola $4x^2 - y^2 = 64$, which are parallel to $10x-3y+9=0$.

Equation of hyperbola is $\;$ $4x^2 - y^2 = 64$

i.e. $\;$ $\dfrac{x^2}{64 / 4} - \dfrac{y^2}{64} = 1$

i.e. $\;$ $\dfrac{x^2}{16} - \dfrac{y^2}{64} = 1$ $\;\;\; \cdots \; (1)$

The transverse axis of the hyperbola given by equation $(1)$ is along the X axis.

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 16 \implies a = 4$ $\;$ and $\;$ $b^2 = 64 \implies b = 8$

Equation of given line is $\;$ $10x - 3y + 9 = 0$

i.e. $\;$ $y = \dfrac{10}{3} x + 3$

Slope of given line $= m = \dfrac{10}{3}$

$\because$ $\;$ the required tangents are parallel to the given line,

$\therefore$ $\;$ slope of required tangents $= m = \dfrac{10}{3}$

Equations of tangents (with slope $m$) to the hyperbola are $\;$ $y = mx \pm \sqrt{a^2 m^2 - b^2}$

$\therefore$ $\;$ Required equations of tangents to the hyperbola are

$y = \dfrac{10}{3} x \pm \sqrt{16 \times \dfrac{100}{9} - 64}$

i.e. $\;$ $y = \dfrac{10}{3} x \pm \sqrt{\dfrac{1024}{9}}$

i.e. $\;$ $y = \dfrac{10}{3} x \pm \dfrac{32}{3}$

i.e. $\;$ $10 x - 3y \pm 32 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangents to the ellipse $\dfrac{x^2}{20}+ \dfrac{y^2}{5} = 1$, which are perpendicular to $x+y+2=0$.

Equation of ellipse is $\;$ $\dfrac{x^2}{20} + \dfrac{y^2}{5} = 1$ $\;\;\; \cdots \; (1)$

The major axis of the ellipse given by equation $(1)$ is along the X axis.

Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 20$ $\;$ and $\;$ $b^2 = 5$

Equation of given line is $\;$ $x + y + 2 = 0$

i.e. $\;$ $y = -x - 2$

Slope of given line $= m_1 = -1$

$\because$ $\;$ the required tangents are perpendicular to the given line,

slope of tangent $= m = \dfrac{-1}{m_1} = 1$

Equations of tangents (with slope $m$) to the ellipse are $\;$ $y = mx \pm \sqrt{a^2 m^2 + b^2}$

$\therefore$ $\;$ Required equations of tangents to the ellipse are

$y = 1 \times x \pm \sqrt{20 \times 1^2 + 5}$

i.e. $\;$ $y = x \pm 5$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangent to the parabola $y^2 = 6x$, parallel to $3x - 2y + 5 = 0$.

Equation of parabola is $\;$ $y^2 = 6x$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 6 \implies a = \dfrac{3}{2}$

Equation of given line is $\;$ $3x - 2y + 5 = 0$

i.e. $\;$ $y = \dfrac{3}{2}x + \dfrac{5}{2}$

$\therefore$ $\;$ Slope of the given line $= m = \dfrac{3}{2}$

$\because$ $\;$ the tangent to the parabola is is parallel to the given line, slope of tangent to the parabola $= m = \dfrac{3}{2}$

The equation of tangent (with slope m) to the parabola $y^2 = 4ax$ is $\;$ $y = mx + \dfrac{a}{m}$

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

$y = \dfrac{3}{2} x + \dfrac{3 / 2}{3 / 2}$

i.e. $\;$ $y = \dfrac{3}{2}x + 1$

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

Analytical Geometry - Conics - Tangents and Normals

Find the equations of tangent and normal to the ellipse $x^2 + 4y^2 = 32$ at $\theta = \dfrac{\pi}{4}$

Equation of ellipse is $\;$ $x^2 + 4y^2 = 32$

i.e. $\;$ $\dfrac{x^2}{32} + \dfrac{y^2}{32 / 4} = 1$

i.e. $\;$ $\dfrac{x^2}{32} + \dfrac{y^2}{8} = 1$ $\;\;\; \cdots \; (1)$

The major axis of the ellipse given by equation $(1)$ is along the X axis.

Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 32 \implies a = 4 \sqrt{2}$ $\;$ and $\;$ $b^2 = 8 \implies b = 2 \sqrt{2}$

$\theta = \dfrac{\pi}{4}$ represents the point $\left(a \cos \theta, b \sin \theta\right)$

i.e. the point $\left(4 \sqrt{2} \cos \left(\dfrac{\pi}{4}\right), 2 \sqrt{2} \sin \left(\dfrac{\pi}{4}\right)\right)$ $\;$ i.e. $\left(4,2\right)$

Equation of tangent to the ellipse at the point $\left(x_1, y_1\right)$ is $\;$ $\dfrac{xx_1}{a^2} + \dfrac{yy_1}{b^2} = 1$

Equation of normal to the ellipse at the point $\left(x_1, y_1\right)$ is $\;$ $\dfrac{a^2 x}{x_1} - \dfrac{b^2 y}{y_1} = a^2 - b^2$

Here $\left(x_1, y_1\right) = \left(4,2\right)$

$\therefore$ $\;$ The required equation of tangent to the ellipse is

$\dfrac{4x}{32} + \dfrac{2y}{8} = 1$

i.e. $\;$ $\dfrac{x}{8} + \dfrac{y}{4} = 1$

i.e. $\;$ $x + 2y - 8 = 0$

The required equation of normal to the ellipse is

$\dfrac{32x}{4} - \dfrac{8y}{2} = 32 - 8$

i.e. $\;$ $8x - 4y = 24$

i.e. $\;$ $2x - y - 6 = 0$

ALTERNATELY

Equation of tangent to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ in parametric form is $\;$ $\dfrac{x \cos \theta}{a} + \dfrac{y \sin \theta}{b} = 1$

Equation of normal to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ in parametric form is $\;$ $\dfrac{ax}{\cos \theta} - \dfrac{b y}{\sin \theta} = a^2 - b^2$

$\therefore$ $\;$ Required equation of tangent is

$\dfrac{x \cos \left(\dfrac{\pi}{4}\right)}{4 \sqrt{2}} + \dfrac{y \sin \left(\dfrac{\pi}{4}\right)}{2 \sqrt{2}} = 1$

i.e. $\;$ $\dfrac{x}{8} + \dfrac{y}{4} = 1$

i.e. $\;$ $x + 2y - 8 = 0$

Required equation of normal is

$\dfrac{4 \sqrt{2} x}{\cos \left(\dfrac{\pi}{4}\right)} - \dfrac{2 \sqrt{2} y}{\sin \left(\dfrac{\pi}{4}\right)} = 32 - 8$

i.e. $\;$ $8x - 4y = 24$

i.e. $\;$ $2x - y - 6 = 0$