Miscellaneous Exercise Class 12 Maths Vector Algebra

Question 01

Unit vector in XY-plane making angle 30° with positive x-axis.

Let the unit vector be $\vec{r} = \cos\theta \hat{i} + \sin\theta \hat{j}$.
Here $\theta = 30^\circ$.
$\vec{r} = \cos 30^\circ \hat{i} + \sin 30^\circ \hat{j} = \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}$.

Question 02

Scalar components and magnitude of vector joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.

$\vec{PQ} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$.
Scalar Components: $(x_2-x_1), (y_2-y_1), (z_2-z_1)$.
Magnitude: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Question 03

Girl walks 4 km West, then 3 km 30° East of North. Find displacement.

Let origin be O.
$\vec{OA}$ (West) $= -4\hat{i}$.
$\vec{AB}$ (30° East of North) means $90^\circ – 30^\circ = 60^\circ$ with positive X-axis.
$\vec{AB} = 3(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = 3(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}) = \frac{3}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}$.
Total Displacement $\vec{OB} = \vec{OA} + \vec{AB} = -4\hat{i} + 1.5\hat{i} + 1.5\sqrt{3}\hat{j} = -2.5\hat{i} + 1.5\sqrt{3}\hat{j}$.

Question 04

Is $|\vec{a}+\vec{b}| = |\vec{a}| + |\vec{b}|$ true? Justify.

No.

By Triangle Inequality, the sum of lengths of two sides of a triangle is always greater than or equal to the third side.

$|\vec{a}+\vec{b}| \le |\vec{a}| + |\vec{b}|$.

Equality holds only when vectors are collinear and in the same direction.

Question 05

Find $x$ for which $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector.

$|x(\hat{i}+\hat{j}+\hat{k})| = 1 \implies |x| \sqrt{1^2+1^2+1^2} = 1$.
$|x|\sqrt{3} = 1 \implies x = \pm \frac{1}{\sqrt{3}}$.

Question 06

Find vector of magnitude 5 units, parallel to resultant of $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2\hat{j}+\hat{k}$.

Resultant $\vec{c} = \vec{a} + \vec{b} = (2+1)\hat{i} + (3-2)\hat{j} + (-1+1)\hat{k} = 3\hat{i} + \hat{j}$.
$|\vec{c}| = \sqrt{3^2+1^2} = \sqrt{10}$.
Unit Vector $\hat{c} = \frac{3\hat{i}+\hat{j}}{\sqrt{10}}$.
Required Vector $= 5\hat{c} = \frac{15}{\sqrt{10}}\hat{i} + \frac{5}{\sqrt{10}}\hat{j}$.

Question 07

Find unit vector parallel to $2\vec{a}-\vec{b}+3\vec{c}$.

Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2\hat{i}-\hat{j}+3\hat{k}, \vec{c}=\hat{i}-2\hat{j}+\hat{k}$.
$\vec{r} = 2(\hat{i}+\hat{j}+\hat{k}) – (2\hat{i}-\hat{j}+3\hat{k}) + 3(\hat{i}-2\hat{j}+\hat{k})$.
$= (2-2+3)\hat{i} + (2+1-6)\hat{j} + (2-3+3)\hat{k} = 3\hat{i} – 3\hat{j} + 2\hat{k}$.
$|\vec{r}| = \sqrt{9+9+4} = \sqrt{22}$.
Unit Vector $= \frac{3}{\sqrt{22}}\hat{i} – \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}$.

Question 08

Show points A, B, C are collinear and find ratio B divides AC.

A(1, -2, -8), B(5, 0, -2), C(11, 3, 7).
$\vec{AB} = 4\hat{i} + 2\hat{j} + 6\hat{k}$.
$\vec{BC} = 6\hat{i} + 3\hat{j} + 9\hat{k}$.
$\vec{BC} = \frac{3}{2}(4\hat{i} + 2\hat{j} + 6\hat{k}) = \frac{3}{2}\vec{AB}$.
Vectors are parallel and share point B, so A, B, C are collinear.
Ratio = Magnitude Ratio $|\vec{AB}| : |\vec{BC}| = 1 : 1.5 = 2 : 3$.

Question 09

Position vector of R dividing P and Q externally in 1:2. Show P is midpoint of RQ.

$\vec{p} = 2\vec{a}+\vec{b}, \vec{q} = \vec{a}-3\vec{b}$. Ratio 1:2 externally.
$\vec{r} = \frac{1(\vec{a}-3\vec{b}) – 2(2\vec{a}+\vec{b})}{1-2} = \frac{\vec{a}-3\vec{b}-4\vec{a}-2\vec{b}}{-1} = \frac{-3\vec{a}-5\vec{b}}{-1} = 3\vec{a}+5\vec{b}$.
Midpoint of RQ $= \frac{\vec{r}+\vec{q}}{2} = \frac{(3\vec{a}+5\vec{b}) + (\vec{a}-3\vec{b})}{2} = \frac{4\vec{a}+2\vec{b}}{2} = 2\vec{a}+\vec{b} = \vec{p}$.
Hence P is midpoint of RQ.

Question 10

Parallelogram adjacent sides $\vec{a}=2\hat{i}-4\hat{j}+5\hat{k}, \vec{b}=\hat{i}-2\hat{j}-3\hat{k}$. Find unit vector along diagonal and Area.

Diagonal $\vec{d} = \vec{a}+\vec{b} = 3\hat{i} – 6\hat{j} + 2\hat{k}$.
$|\vec{d}| = \sqrt{9+36+4} = 7$. Unit vector $= \frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})$.
Area $= |\vec{a} \times \vec{b}| = |\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix}|$.
$= |22\hat{i} + 11\hat{j} + 0\hat{k}| = 11|2\hat{i}+\hat{j}| = 11\sqrt{5}$.

Question 11

Show direction cosines of vector equally inclined to axes are $\pm \frac{1}{\sqrt{3}}$.

Let angles be $\alpha, \beta, \gamma$. Equally inclined $\implies \alpha = \beta = \gamma$.
$l=m=n$.
$l^2+m^2+n^2=1 \implies 3l^2=1 \implies l = \pm \frac{1}{\sqrt{3}}$.
DCs are $(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}})$.

Question 12

$\vec{a}=\hat{i}+4\hat{j}+2\hat{k}, \vec{b}=3\hat{i}-2\hat{j}+7\hat{k}, \vec{c}=2\hat{i}-\hat{j}+4\hat{k}$. Find $\vec{d} \perp \vec{a}, \vec{b}$ and $\vec{c}\cdot\vec{d}=15$.

$\vec{d} \perp \vec{a}, \vec{b} \implies \vec{d} \parallel (\vec{a} \times \vec{b})$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = 32\hat{i} – \hat{j} – 14\hat{k}$.
Let $\vec{d} = \lambda(32\hat{i} – \hat{j} – 14\hat{k})$.
$\vec{c} \cdot \vec{d} = 15 \implies (2\hat{i}-\hat{j}+4\hat{k}) \cdot \lambda(32\hat{i}-\hat{j}-14\hat{k}) = 15$.
$\lambda(64 + 1 – 56) = 15 \implies 9\lambda = 15 \implies \lambda = 5/3$.
$\vec{d} = \frac{5}{3}(32\hat{i} – \hat{j} – 14\hat{k})$.

Question 13

Find $\lambda$ if $(\hat{i}+\hat{j}+\hat{k}) \cdot \text{Unit Vector}(\vec{a}+\vec{b}) = 1$.

$\vec{a} = 2\hat{i}+4\hat{j}-5\hat{k}, \quad \vec{b} = \lambda\hat{i}+2\hat{j}+3\hat{k}$.
Sum $\vec{S} = (2+\lambda)\hat{i} + 6\hat{j} – 2\hat{k}$.
Unit Vector $\hat{S} = \frac{\vec{S}}{|\vec{S}|}$.
$(\hat{i}+\hat{j}+\hat{k}) \cdot \frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda)^2+36+4}} = 1$.
$(2+\lambda) + 6 – 2 = \sqrt{(2+\lambda)^2+40}$.
$\lambda+6 = \sqrt{\lambda^2+4\lambda+44}$. Square both sides.
$\lambda^2+12\lambda+36 = \lambda^2+4\lambda+44 \implies 8\lambda = 8 \implies \lambda = 1$.

Question 14

If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular equal magnitude vectors, show $\vec{a}+\vec{b}+\vec{c}$ is equally inclined.

Let $|\vec{a}|=|\vec{b}|=|\vec{c}|=k$. Also $\vec{a}\cdot\vec{b} = \dots = 0$.
Let $\vec{r} = \vec{a}+\vec{b}+\vec{c}$.
$|\vec{r}|^2 = \vec{r}\cdot\vec{r} = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 0 = 3k^2 \implies |\vec{r}| = k\sqrt{3}$.
$\cos\alpha = \frac{\vec{r}\cdot\vec{a}}{|\vec{r}||\vec{a}|} = \frac{|\vec{a}|^2 + 0 + 0}{k\sqrt{3} \cdot k} = \frac{k^2}{k^2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Similarly $\cos\beta = \frac{1}{\sqrt{3}}$ and $\cos\gamma = \frac{1}{\sqrt{3}}$. Equally inclined.

Question 15

Prove $(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = |\vec{a}|^2+|\vec{b}|^2 \iff \vec{a} \perp \vec{b}$.

LHS $= |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}$.
Equation holds iff $2\vec{a}\cdot\vec{b} = 0 \implies \vec{a}\cdot\vec{b} = 0 \implies \vec{a} \perp \vec{b}$.

Questions 16 — 19

Multiple Choice Questions.

16. $\vec{a}\cdot\vec{b} \ge 0$ if…

$|\vec{a}||\vec{b}|\cos\theta \ge 0$. Magnitudes are positive.
$\cos\theta \ge 0 \implies 0 \le \theta \le \pi/2$.

Correct Option: (B)

17. $|\vec{a}+\vec{b}|=1$ for unit vectors.

$|\vec{a}+\vec{b}|^2 = 1 \implies 1 + 1 + 2\cos\theta = 1 \implies \cos\theta = -1/2$.
$\theta = 2\pi/3$.

Correct Option: (D)

18. Value of $\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$.

$\hat{i}\cdot\hat{i} + \hat{j}\cdot(-\hat{j}) + \hat{k}\cdot\hat{k} = 1 – 1 + 1 = 1$.

Correct Option: (C)

19. $|\vec{a}\cdot\vec{b}| = |\vec{a}\times\vec{b}|$.

Correct Option: (B)

Vector Algebra Review

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