Class 12 Ch 10: Vector Algebra - Miscellaneous Exercise

Q1

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Write down a unit vector in XY-plane, making an angle of $30^\circ$ with the positive direction of x-axis.

A unit vector in XY-plane is given by $\vec{r} = \cos \theta \hat{i} + \sin \theta \hat{j}$.

Here $\theta = 30^\circ$.

$\vec{r} = \cos 30^\circ \hat{i} + \sin 30^\circ \hat{j} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$.

$\boxed{\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}}$

Q2

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Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.

Vector $\vec{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.

Scalar components are $(x_2 - x_1), (y_2 - y_1), (z_2 - z_1)$.

Magnitude $|\vec{PQ}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

$\boxed{\text{Components: } x_2-x_1, y_2-y_1, z_2-z_1; \text{ Mag: } \sqrt{\sum(x_2-x_1)^2}}$

Q3

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A girl walks 4 km towards west, then she walks 3 km in a direction $30^\circ$ east of north and stops. Determine the girl’s displacement from her initial point of departure.

Let initial point be O. First displacement $\vec{OA} = -4\hat{i}$.

Second displacement $\vec{AB}$ is 3 km at $30^\circ$ East of North. Angle with x-axis is $90^\circ - 30^\circ = 60^\circ$.

$\vec{AB} = 3(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = 3(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}) = 1.5\hat{i} + 1.5\sqrt{3}\hat{j}$.

Total displacement $\vec{OB} = \vec{OA} + \vec{AB} = -4\hat{i} + 1.5\hat{i} + 1.5\sqrt{3}\hat{j} = -2.5\hat{i} + 1.5\sqrt{3}\hat{j}$.

$\boxed{-\frac{5}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}}$

Q4

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If $\vec{a} = \vec{b} + \vec{c}$, then is it true that $|\vec{a}| = |\vec{b}| + |\vec{c}|$? Justify your answer.

No. By triangle inequality, in a triangle with sides representing vectors $\vec{a}, \vec{b}, \vec{c}$, the sum of lengths of two sides is greater than the third side.

$|\vec{b}| + |\vec{c}| \ge |\vec{b} + \vec{c}| = |\vec{a}|$. Equality holds only if vectors are collinear and in same direction.

$\boxed{\text{No}}$

Q5

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

$|x(\hat{i} + \hat{j} + \hat{k})| = 1 \Rightarrow |x|\sqrt{1^2 + 1^2 + 1^2} = 1$.

$|x|\sqrt{3} = 1 \Rightarrow x = \pm \frac{1}{\sqrt{3}}$.

$\boxed{\pm \frac{1}{\sqrt{3}}}$

Q6

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

Resultant $\vec{c} = \vec{a} + \vec{b} = (2+1)\hat{i} + (3-2)\hat{j} + (-1+1)\hat{k} = 3\hat{i} + \hat{j}$.

$|\vec{c}| = \sqrt{9 + 1} = \sqrt{10}$. Unit vector $\hat{c} = \frac{3\hat{i} + \hat{j}}{\sqrt{10}}$.

Required vector $= 5\hat{c} = \frac{5}{\sqrt{10}}(3\hat{i} + \hat{j}) = \frac{\sqrt{10}}{2}(3\hat{i} + \hat{j})$.

$\boxed{\frac{3\sqrt{10}}{2}\hat{i} + \frac{\sqrt{10}}{2}\hat{j}}$

Q7

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If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} - 2\hat{j} + \hat{k}$, find a unit vector parallel to the vector $2\vec{a} - \vec{b} + 3\vec{c}$.

Let $\vec{v} = 2\vec{a} - \vec{b} + 3\vec{c} = 2(1,1,1) - (2,-1,3) + 3(1,-2,1)$.

$= (2-2+3)\hat{i} + (2+1-6)\hat{j} + (2-3+3)\hat{k} = 3\hat{i} - 3\hat{j} + 2\hat{k}$.

$|\vec{v}| = \sqrt{9 + 9 + 4} = \sqrt{22}$.

$\boxed{\frac{1}{\sqrt{22}}(3\hat{i} - 3\hat{j} + 2\hat{k})}$

Q8

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Show that the points $A(1, -2, -8)$, $B(5, 0, -2)$ and $C(11, 3, 7)$ are collinear, and find the ratio in which B divides AC.

$\vec{AB} = 4\hat{i} + 2\hat{j} + 6\hat{k}$. $|\vec{AB}| = \sqrt{16+4+36} = \sqrt{56} = 2\sqrt{14}$.

$\vec{BC} = 6\hat{i} + 3\hat{j} + 9\hat{k}$. $|\vec{BC}| = \sqrt{36+9+81} = \sqrt{126} = 3\sqrt{14}$.

$\vec{AC} = 10\hat{i} + 5\hat{j} + 15\hat{k}$. $|\vec{AC}| = \sqrt{100+25+225} = \sqrt{350} = 5\sqrt{14}$.

$|\vec{AB}| + |\vec{BC}| = |\vec{AC}|$, so collinear. Ratio $AB:BC = 2:3$.

$\boxed{2:3}$

Q9

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Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $(2\vec{a} + \vec{b})$ and $(\vec{a} - 3\vec{b})$ externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.

$\vec{R} = \frac{1(\vec{a} - 3\vec{b}) - 2(2\vec{a} + \vec{b})}{1 - 2} = \frac{\vec{a} - 3\vec{b} - 4\vec{a} - 2\vec{b}}{-1} = \frac{-3\vec{a} - 5\vec{b}}{-1} = 3\vec{a} + 5\vec{b}$.

Midpoint of RQ $= \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} - 3\vec{b})}{2} = \frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b} = \vec{P}$.

$\boxed{\vec{R} = 3\vec{a} + 5\vec{b}; \text{ Shown}}$

Q10

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The two adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} + 5\hat{k}$ and $\hat{i} - 2\hat{j} - 3\hat{k}$. Find the unit vector parallel to its diagonal. Also, find its area.

Diagonal $\vec{d} = \vec{a} + \vec{b} = 3\hat{i} - 6\hat{j} + 2\hat{k}$.

Unit vector $\hat{d} = \frac{3\hat{i} - 6\hat{j} + 2\hat{k}}{\sqrt{9+36+4}} = \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k})$.

Area $= |\vec{a} \times \vec{b}| = |\begin{vmatrix} i & j & k \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix}| = |22\hat{i} + 11\hat{j}| = \sqrt{484 + 121} = 11\sqrt{5}$.

$\boxed{\text{Unit vec: } \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}); \text{ Area: } 11\sqrt{5}}$

Q11

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Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$.

Let direction cosines be $l, m, n$. Given $l = m = n$.

$l^2 + m^2 + n^2 = 1 \Rightarrow 3l^2 = 1 \Rightarrow l = \pm \frac{1}{\sqrt{3}}$.

$\boxed{\text{Shown}}$

Q12

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Let $\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$, $\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}$ and $\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$, and $\vec{c} \cdot \vec{d} = 15$.

$\vec{d} \parallel (\vec{a} \times \vec{b})$. $\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = 32\hat{i} - \hat{j} - 14\hat{k}$.

Let $\vec{d} = \lambda(32\hat{i} - \hat{j} - 14\hat{k})$.

$\vec{c} \cdot \vec{d} = 15 \Rightarrow \lambda(2(32) + (-1)(-1) + 4(-14)) = 15 \Rightarrow \lambda(64+1-56) = 15 \Rightarrow 9\lambda = 15 \Rightarrow \lambda = \frac{5}{3}$.

$\boxed{\frac{160}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{70}{3}\hat{k}}$

Q13

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The scalar product of the vector $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of vectors $2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\lambda\hat{i} + 2\hat{j} + 3\hat{k}$ is equal to one. Find the value of $\lambda$.

Sum $\vec{s} = (2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k}$. Unit vector $\hat{n} = \frac{\vec{s}}{|\vec{s}|}$.

$(\hat{i} + \hat{j} + \hat{k}) \cdot \hat{n} = 1 \Rightarrow \frac{(2+\lambda) + 6 - 2}{\sqrt{(2+\lambda)^2 + 36 + 4}} = 1$.

$\lambda + 6 = \sqrt{\lambda^2 + 4\lambda + 44} \Rightarrow (\lambda+6)^2 = \lambda^2 + 4\lambda + 44$.

$\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44 \Rightarrow 8\lambda = 8 \Rightarrow \lambda = 1$.

$\boxed{1}$

Q14

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If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.

Let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. Let $\vec{d} = \vec{a} + \vec{b} + \vec{c}$.

$\vec{d} \cdot \vec{a} = |\vec{a}|^2 = k^2$. Also $|\vec{d}| = \sqrt{k^2+k^2+k^2} = k\sqrt{3}$.

$\cos \alpha = \frac{\vec{d} \cdot \vec{a}}{|\vec{d}||\vec{a}|} = \frac{k^2}{k\sqrt{3} \cdot k} = \frac{1}{\sqrt{3}}$. Similarly $\cos \beta = \cos \gamma = \frac{1}{\sqrt{3}}$.

$\boxed{\text{Shown}}$

Q15

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Prove that $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$, if and only if $\vec{a}, \vec{b}$ are perpendicular, given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$.

LHS $= |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$.

Given LHS $= |\vec{a}|^2 + |\vec{b}|^2$, implies $2\vec{a} \cdot \vec{b} = 0$.

Since vectors are non-zero, $\vec{a} \perp \vec{b}$.

$\boxed{\text{Proved}}$

Q16

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If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then $\vec{a} \cdot \vec{b} \ge 0$ only when
(A) $0 \theta \frac{\pi}{2}$
(B) $0 \le \theta \le \frac{\pi}{2}$
(C) $0 \theta \pi$
(D) $0 \le \theta \le \pi$

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos \theta \ge 0$.

Since magnitudes are positive, $\cos \theta \ge 0$.

In $[0, \pi]$, $\cos \theta \ge 0$ for $[0, \frac{\pi}{2}]$.

$\boxed{\text{(B)}}$

Q17

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Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a} + \vec{b}$ is a unit vector if
(A) $\theta = \frac{\pi}{4}$
(B) $\theta = \frac{\pi}{3}$
(C) $\theta = \frac{\pi}{2}$
(D) $\theta = \frac{2\pi}{3}$

$|\vec{a} + \vec{b}|^2 = 1 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 1$.

$1 + 1 + 2(1)(1)\cos \theta = 1 \Rightarrow 2\cos \theta = -1 \Rightarrow \cos \theta = -\frac{1}{2}$.

$\theta = \frac{2\pi}{3}$.

$\boxed{\text{(D)}}$

Q18

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The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is
(A) 0
(B) -1
(C) 1
(D) 3

$\hat{j} \times \hat{k} = \hat{i} \Rightarrow \hat{i} \cdot \hat{i} = 1$.

$\hat{i} \times \hat{k} = -\hat{j} \Rightarrow \hat{j} \cdot (-\hat{j}) = -1$.

$\hat{i} \times \hat{j} = \hat{k} \Rightarrow \hat{k} \cdot \hat{k} = 1$.

Sum $= 1 - 1 + 1 = 1$.

$\boxed{\text{(C)}}$

Q19

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If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$, then $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ when $\theta$ is equal to
(A) 0
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{2}$
(D) $\pi$

$|\vec{a}||\vec{b}||\cos \theta| = |\vec{a}||\vec{b}|\sin \theta$.

$|\cos \theta| = \sin \theta \Rightarrow \tan \theta = 1$ (since $\sin \theta \ge 0$).

$\theta = \frac{\pi}{4}$.

$\boxed{\text{(B)}}$

Miscellaneous Exercise Practice