Questions and Answers (Board Exam)

Introduction to Board Exam Practice Questions

This section contains 30 board-exam style questions aligned with the Class XII Mathematics pattern. In board exams, presentation is as important as the final answer. Always write the relevant formula first, then substitute the values carefully, and finally simplify the result in a neat step-by-step manner.

For vector questions, common formulas you should clearly state are:

• Magnitude: $|\vec{a}| = \sqrt{x^2+y^2+z^2}$
• Projection of $\vec{a}$ on $\vec{b}$: $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$
• Angle between vectors: $\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$
• Cross product magnitude: $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$
• Internal division formula: $\vec{r} = \dfrac{m\vec{b}+n\vec{a}}{m+n}$
• External division formula: $\vec{r} = \dfrac{m\vec{b}-n\vec{a}}{m-n}$

Question 1

Find the magnitude of the vector $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.

Solution: Step 1: The scalar components of $\vec{a}$ are $x = 2$, $y = -3$, and $z = 6$.

Step 2: Use the formula for the magnitude of a vector: $$|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$$

Step 3: Substitute the values: $$|\vec{a}| = \sqrt{(2)^2 + (-3)^2 + (6)^2}$$

Step 4: Evaluate: $$|\vec{a}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$

Answer: 7

Question 2

Write a unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.

Solution: Step 1: Find the magnitude of $\vec{a}$: $$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$

Step 2: The unit vector in the direction of $\vec{a}$ is $$\hat{a} = \frac{\vec{a}}{|\vec{a}|}$$

Step 3: Therefore, $$\hat{a} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$$

Answer: $\frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$

Question 3

Find the direction cosines of the vector joining the points $A(1, 2, -3)$ and $B(-1, -2, 1)$, directed from $A$ to $B$.

Solution: Step 1: Find the vector $\vec{AB}$: $$\vec{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 4\hat{j} + 4\hat{k}$$

Step 2: Find the magnitude of $\vec{AB}$: $$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$

Step 3: Direction cosines are given by $$l = \frac{x}{r}, \quad m = \frac{y}{r}, \quad n = \frac{z}{r}$$

Step 4: Hence, $$l = \frac{-2}{6} = -\frac{1}{3}, \quad m = \frac{-4}{6} = -\frac{2}{3}, \quad n = \frac{4}{6} = \frac{2}{3}$$

Answer: $-1/3, -2/3, 2/3$

Question 4

Find the projection of the vector $\hat{i} - \hat{j}$ on the vector $\hat{i} + \hat{j}$.

Solution: Step 1: Let $$\vec{a} = \hat{i} - \hat{j}, \qquad \vec{b} = \hat{i} + \hat{j}$$

Step 2: Projection of $\vec{a}$ on $\vec{b}$ is $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$

Step 3: Compute the dot product: $$\vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 1 - 1 = 0$$

Step 4: Compute the magnitude of $\vec{b}$: $$|\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2}$$

Step 5: Therefore, $$\text{Projection} = \frac{0}{\sqrt{2}} = 0$$

Answer: 0

Question 5

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$, respectively, having $\vec{a} \cdot \vec{b} = \sqrt{6}$.

Solution: Step 1: Use the formula $$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$

Step 2: Substitute the values: $$\cos\theta = \frac{\sqrt{6}}{\sqrt{3} \times 2}$$

Step 3: Simplify: $$\cos\theta = \frac{\sqrt{2}\sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{2}}{2}$$

Step 4: Therefore, $$\theta = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$$

So the angle is $45^\circ$ or $\frac{\pi}{4}$.

Answer: $\pi/4$

Question 6

For what value of $\lambda$ are the vectors $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ perpendicular to each other?

Solution: Step 1: Two vectors are perpendicular if their dot product is zero: $$\vec{a} \cdot \vec{b} = 0$$

Step 2: Compute the dot product: $$(2\hat{i} + \lambda\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 0$$ $$(2)(1) + (\lambda)(-2) + (1)(3) = 0$$

Step 3: Simplify: $$2 - 2\lambda + 3 = 0$$ $$5 - 2\lambda = 0$$ $$2\lambda = 5$$ $$\lambda = \frac{5}{2}$$

Answer: $5/2$

Question 7

Find the area of a parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.

Solution: Step 1: Area of the parallelogram is $$|\vec{a} \times \vec{b}|$$

Step 2: Compute the cross product: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$$

Expanding, $$= \hat{i}[(-1)(1) - 3(-7)] - \hat{j}[(1)(1) - 3(2)] + \hat{k}[(1)(-7) - (-1)(2)]$$ $$= \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2)$$ $$= 20\hat{i} + 5\hat{j} - 5\hat{k}$$

Step 3: Find the magnitude: $$|\vec{a} \times \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450}$$

Step 4: Simplify: $$\sqrt{450} = \sqrt{225\cdot 2} = 15\sqrt{2}$$

Answer: $15\sqrt{2}$ square units.

Question 8

Find the area of a triangle having the points $A(1, 1, 1)$, $B(1, 2, 3)$, and $C(2, 3, 1)$ as its vertices.

Solution: Step 1: Find two adjacent side vectors: $$\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = \hat{j} + 2\hat{k}$$ $$\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = \hat{i} + 2\hat{j}$$

Step 2: Area of the triangle is $$\frac{1}{2}|\vec{AB} \times \vec{AC}|$$

Step 3: Compute the cross product: $$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}$$ $$= \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) = -4\hat{i} + 2\hat{j} - \hat{k}$$

Step 4: Find the magnitude: $$|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$$

Step 5: Hence, $$\text{Area} = \frac{\sqrt{21}}{2}$$

Answer: $\sqrt{21}/2$ square units.

Question 9

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

Solution: Step 1: Use $$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$$

Step 2: Find the magnitude of the given cross product: $$|\vec{a} \times \vec{b}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$

Step 3: Substitute: $$7 = (2)(7)\sin\theta$$ $$7 = 14\sin\theta$$ $$\sin\theta = \frac{1}{2}$$

Step 4: Therefore the angle between the vectors may be $$\theta = 30^\circ \text{ or } 150^\circ$$ that is, $$\theta = \frac{\pi}{6} \text{ or } \frac{5\pi}{6}$$

Since the sine alone is known, both angles are possible.

Answer: $\pi/6$ or $5\pi/6$

Question 10

Find a unit vector perpendicular to both the vectors $\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$.

Solution: Step 1: A vector perpendicular to both is given by the cross product: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & 3 & -1 \end{vmatrix}$$

Step 2: Expand: $$= \hat{i}[1(-1) - (-2)(3)] - \hat{j}[3(-1) - (-2)(2)] + \hat{k}[3(3) - 1(2)]$$ $$= \hat{i}(-1 + 6) - \hat{j}(-3 + 4) + \hat{k}(9 - 2)$$ $$= 5\hat{i} - \hat{j} + 7\hat{k}$$

Step 3: Find its magnitude: $$|\vec{a} \times \vec{b}| = \sqrt{5^2 + (-1)^2 + 7^2} = \sqrt{25 + 1 + 49} = \sqrt{75} = 5\sqrt{3}$$

Step 4: Therefore, a unit vector perpendicular to both is $$\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \pm \frac{5\hat{i} - \hat{j} + 7\hat{k}}{5\sqrt{3}}$$

Answer: $\pm \frac{1}{5\sqrt{3}}(5\hat{i} - \hat{j} + 7\hat{k})$

Question 11

Show that the points with position vectors $2\hat{i} - \hat{j} + \hat{k}$, $\hat{i} - 3\hat{j} - 5\hat{k}$, and $3\hat{i} - 4\hat{j} - 4\hat{k}$ are the vertices of a right-angled triangle.

Solution: Step 1: Let the points be $$A(2,-1,1), \quad B(1,-3,-5), \quad C(3,-4,-4)$$

Step 2: Find two side vectors meeting at $B$: $$\vec{BA} = (2-1)\hat{i} + (-1+3)\hat{j} + (1+5)\hat{k} = \hat{i} + 2\hat{j} + 6\hat{k}$$ $$\vec{BC} = (3-1)\hat{i} + (-4+3)\hat{j} + (-4+5)\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$$

Step 3: Take their dot product: $$\vec{BA} \cdot \vec{BC} = (1)(2) + (2)(-1) + (6)(1) = 2 - 2 + 6 = 6$$ This is not zero, so angle at $B$ is not a right angle.

Step 4: Now check angle at $C$ using $$\vec{CB} = (1-3)\hat{i} + (-3+4)\hat{j} + (-5+4)\hat{k} = -2\hat{i} + \hat{j} - \hat{k}$$ $$\vec{CA} = (2-3)\hat{i} + (-1+4)\hat{j} + (1+4)\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$$

Then, $$\vec{CB} \cdot \vec{CA} = (-2)(-1) + (1)(3) + (-1)(5) = 2 + 3 - 5 = 0$$

Step 5: Since the dot product is zero, the sides through $C$ are perpendicular. Hence the triangle is right-angled at $C$.

Answer: Proved.

Question 12

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $\vec{a} + 2\vec{b}$ and $\vec{a} - \vec{b}$ respectively, internally in the ratio 2:1.

Solution: Step 1: Let $$\vec{p} = \vec{a} + 2\vec{b}, \qquad \vec{q} = \vec{a} - \vec{b}$$

Step 2: Since $R$ divides $PQ$ internally in the ratio $2:1$, use $$\vec{r} = \frac{2\vec{q} + 1\vec{p}}{2+1}$$

Step 3: Substitute: $$\vec{r} = \frac{2(\vec{a}-\vec{b}) + (\vec{a}+2\vec{b})}{3}$$

Step 4: Simplify: $$\vec{r} = \frac{2\vec{a}-2\vec{b}+\vec{a}+2\vec{b}}{3} = \frac{3\vec{a}}{3} = \vec{a}$$

Answer: $\vec{a}$

Question 13

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $\vec{a} + 2\vec{b}$ and $\vec{a} - \vec{b}$ respectively, externally in the ratio 2:1.

Solution: Step 1: Use the external division formula: $$\vec{r} = \frac{2\vec{q} - 1\vec{p}}{2-1}$$

Step 2: Substitute $$\vec{p} = \vec{a} + 2\vec{b}, \qquad \vec{q} = \vec{a} - \vec{b}$$

Step 3: Then $$\vec{r} = \frac{2(\vec{a}-\vec{b}) - (\vec{a}+2\vec{b})}{1}$$

Step 4: Simplify: $$\vec{r} = 2\vec{a} - 2\vec{b} - \vec{a} - 2\vec{b} = \vec{a} - 4\vec{b}$$

Answer: $\vec{a} - 4\vec{b}$

Question 14

Find the position vector of the midpoint of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

Solution: Step 1: The position vectors are $$\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}, \qquad \vec{q} = 4\hat{i} + \hat{j} - 2\hat{k}$$

Step 2: Midpoint formula: $$\vec{r} = \frac{\vec{p}+\vec{q}}{2}$$

Step 3: Add: $$\vec{p}+\vec{q} = (2+4)\hat{i} + (3+1)\hat{j} + (4-2)\hat{k} = 6\hat{i} + 4\hat{j} + 2\hat{k}$$

Step 4: Divide by 2: $$\vec{r} = 3\hat{i} + 2\hat{j} + \hat{k}$$

Answer: $3\hat{i} + 2\hat{j} + \hat{k}$

Question 15

If the vectors $\vec{a} = 2\hat{i} + 3\hat{j} - 6\hat{k}$ and $\vec{b} = p\hat{i} - \hat{j} + 2\hat{k}$ are parallel, find the value of $p$.

Solution: Step 1: If two vectors are parallel, then their corresponding components are proportional: $$\frac{2}{p} = \frac{3}{-1} = \frac{-6}{2}$$

Step 2: Simplify the known ratios: $$\frac{3}{-1} = -3, \qquad \frac{-6}{2} = -3$$ So the common ratio is $-3$.

Step 3: Therefore, $$\frac{2}{p} = -3$$ $$2 = -3p$$ $$p = -\frac{2}{3}$$

Answer: $-2/3$

Question 16

Prove that $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$.

Solution: Step 1: Expand using distributive law: $$(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}$$

Step 2: Use the facts $$\vec{a} \times \vec{a} = \vec{0}, \qquad \vec{b} \times \vec{b} = \vec{0}, \qquad \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$$

Step 3: Therefore, $$= \vec{0} + \vec{a} \times \vec{b} - [-(\vec{a} \times \vec{b})] - \vec{0}$$ $$= \vec{a} \times \vec{b} + \vec{a} \times \vec{b}$$ $$= 2(\vec{a} \times \vec{b})$$

Hence proved.

Answer: Proved.

Question 17

Prove the parallelogram identity: $|\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2)$.

Solution: Step 1: Expand the first term: $$|\vec{a} + \vec{b}|^2 = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}$$

Step 2: Expand the second term: $$|\vec{a} - \vec{b}|^2 = (\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b}$$

Step 3: Add the two equations: $$|\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = 2|\vec{a}|^2 + 2|\vec{b}|^2$$

Step 4: Factor out 2: $$= 2(|\vec{a}|^2 + |\vec{b}|^2)$$

Hence proved.

Answer: Proved.

Question 18

If $|\vec{a}| = 2$, $|\vec{b}| = 3$, and $\vec{a} \cdot \vec{b} = 4$, find $|\vec{a} - \vec{b}|$.

Solution: Step 1: Use the identity $$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$$

Step 2: Substitute the values: $$|\vec{a} - \vec{b}|^2 = 2^2 + 3^2 - 2(4) = 4 + 9 - 8 = 5$$

Step 3: Take positive square root: $$|\vec{a} - \vec{b}| = \sqrt{5}$$

Answer: $\sqrt{5}$

Question 19

Find a vector in the direction of vector $\vec{a} = \hat{i} - 2\hat{j}$ that has magnitude 7 units.

Solution: Step 1: Find the magnitude of $\vec{a}$: $$|\vec{a}| = \sqrt{1^2 + (-2)^2} = \sqrt{5}$$

Step 2: Unit vector in the direction of $\vec{a}$ is $$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}$$

Step 3: Multiply by the required magnitude 7: $$\vec{v} = 7\hat{a} = 7\left(\frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}\right) = \frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j}$$

Answer: $\frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j}$

Question 20

If $|\vec{a}| = 10$, $|\vec{b}| = 2$, and $\vec{a} \cdot \vec{b} = 12$, find $|\vec{a} \times \vec{b}|$.

Solution: Step 1: Use Lagrange's identity: $$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$

Step 2: Substitute the values: $$|\vec{a} \times \vec{b}|^2 + 12^2 = 10^2 \cdot 2^2$$ $$|\vec{a} \times \vec{b}|^2 + 144 = 400$$

Step 3: Solve: $$|\vec{a} \times \vec{b}|^2 = 256$$ $$|\vec{a} \times \vec{b}| = 16$$

Answer: 16

Question 21

Write the value of $(\hat{i} \times \hat{j}) \cdot \hat{k} + \hat{i} \cdot \hat{j}$.

Solution: Step 1: Evaluate the cross product: $$\hat{i} \times \hat{j} = \hat{k}$$

Step 2: Substitute: $$(\hat{k}) \cdot \hat{k} + \hat{i} \cdot \hat{j}$$

Step 3: Now, $$\hat{k} \cdot \hat{k} = 1, \qquad \hat{i} \cdot \hat{j} = 0$$

Step 4: Therefore, $$1 + 0 = 1$$

Answer: 1

Question 22

Find a vector $\vec{x}$ which is perpendicular to both $\vec{a} = 4\hat{i} + 5\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 4\hat{j} + 5\hat{k}$ such that $\vec{x} \cdot \vec{c} = 21$, where $\vec{c} = 3\hat{i} + \hat{j} - \hat{k}$.

Solution: Step 1: A vector perpendicular to both $\vec{a}$ and $\vec{b}$ must be parallel to their cross product. Compute: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & -1 \\ 1 & -4 & 5 \end{vmatrix}$$

Expanding, $$= \hat{i}[5(5) - (-1)(-4)] - \hat{j}[4(5) - (-1)(1)] + \hat{k}[4(-4) - 5(1)]$$ $$= 21\hat{i} - 21\hat{j} - 21\hat{k}$$

Step 2: So let $$\vec{x} = \lambda(21\hat{i} - 21\hat{j} - 21\hat{k}) = 21\lambda(\hat{i} - \hat{j} - \hat{k})$$

Step 3: Use the condition $\vec{x} \cdot \vec{c} = 21$: $$21\lambda(\hat{i} - \hat{j} - \hat{k}) \cdot (3\hat{i} + \hat{j} - \hat{k}) = 21$$

Step 4: Compute the dot product inside: $$(1)(3) + (-1)(1) + (-1)(-1) = 3 - 1 + 1 = 3$$ So, $$21\lambda \cdot 3 = 21$$ $$63\lambda = 21$$ $$\lambda = \frac{1}{3}$$

Step 5: Therefore, $$\vec{x} = 21\left(\frac{1}{3}\right)(\hat{i} - \hat{j} - \hat{k}) = 7\hat{i} - 7\hat{j} - 7\hat{k}$$

Answer: $7\hat{i} - 7\hat{j} - 7\hat{k}$

Question 23

Find the projection of the vector $\vec{a} + \vec{b}$ on the vector $\vec{a} - \vec{b}$, where $\vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$.

Solution: Step 1: Find $$\vec{u} = \vec{a} + \vec{b}, \qquad \vec{v} = \vec{a} - \vec{b}$$

Step 2: Compute them: $$\vec{u} = (2+1)\hat{i} + (-2+2)\hat{j} + (1-2)\hat{k} = 3\hat{i} - \hat{k}$$ $$\vec{v} = (2-1)\hat{i} + (-2-2)\hat{j} + (1-(-2))\hat{k} = \hat{i} - 4\hat{j} + 3\hat{k}$$

Step 3: Projection of $\vec{u}$ on $\vec{v}$ is $$\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$$

Step 4: Compute the dot product: $$\vec{u} \cdot \vec{v} = (3)(1) + (0)(-4) + (-1)(3) = 3 - 3 = 0$$

Step 5: Hence the projection is zero.

Answer: 0

Question 24

Write the vector component of $\vec{a} = 2\hat{i} + 3\hat{j}$ along the direction of $\vec{b} = \hat{i} + \hat{j}$.

Solution: Step 1: The vector component of $\vec{a}$ along $\vec{b}$ is the vector projection: $$\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$$

Step 2: Compute the dot product: $$\vec{a} \cdot \vec{b} = (2)(1) + (3)(1) = 5$$

Step 3: Compute $|\vec{b}|^2$: $$|\vec{b}|^2 = 1^2 + 1^2 = 2$$

Step 4: Therefore, $$\text{Vector component} = \frac{5}{2}(\hat{i} + \hat{j}) = \frac{5}{2}\hat{i} + \frac{5}{2}\hat{j}$$

Answer: $\frac{5}{2}\hat{i} + \frac{5}{2}\hat{j}$

Question 25

Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes $OX, OY$, and $OZ$.

Solution: Step 1: Let $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$ Its magnitude is $$|\vec{a}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$$

Step 2: Find its direction cosines: $$l = \frac{1}{\sqrt{3}}, \quad m = \frac{1}{\sqrt{3}}, \quad n = \frac{1}{\sqrt{3}}$$

Step 3: Since direction cosines are the cosines of the angles with the coordinate axes, $$\cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt{3}}$$

Step 4: Therefore, $$\alpha = \beta = \gamma$$ So the vector is equally inclined to all three axes.

Answer: Proved.

Question 26

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 $\vec{c}$.

Solution: Step 1: Let $$|\vec{a}| = |\vec{b}| = |\vec{c}| = \lambda$$ and since they are mutually perpendicular, $$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$$

Step 2: Let $$\vec{r} = \vec{a} + \vec{b} + \vec{c}$$ Then $$|\vec{r}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = 3\lambda^2$$ Hence, $$|\vec{r}| = \lambda\sqrt{3}$$

Step 3: Find the angle $\theta_1$ between $\vec{r}$ and $\vec{a}$: $$\cos\theta_1 = \frac{\vec{r} \cdot \vec{a}}{|\vec{r}||\vec{a}|} = \frac{(\vec{a}+\vec{b}+\vec{c})\cdot\vec{a}}{\lambda\sqrt{3}\cdot\lambda} = \frac{\lambda^2}{\lambda^2\sqrt{3}} = \frac{1}{\sqrt{3}}$$

Step 4: Similarly, $$\cos\theta_2 = \frac{1}{\sqrt{3}}, \qquad \cos\theta_3 = \frac{1}{\sqrt{3}}$$ Thus all three angles are equal.

Answer: Proved.

Question 27

Find the area of a parallelogram whose diagonals are $\vec{d}_1 = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d}_2 = \hat{i} - 3\hat{j} + 4\hat{k}$.

Solution: Step 1: Area in terms of diagonals is $$\text{Area} = \frac{1}{2}|\vec{d}_1 \times \vec{d}_2|$$

Step 2: Compute the cross product: $$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix}$$ $$= \hat{i}(4-6) - \hat{j}(12-(-2)) + \hat{k}(-9-1) = -2\hat{i} - 14\hat{j} - 10\hat{k}$$

Step 3: Find the magnitude: $$|\vec{d}_1 \times \vec{d}_2| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{4+196+100} = \sqrt{300} = 10\sqrt{3}$$

Step 4: Hence, $$\text{Area} = \frac{1}{2}(10\sqrt{3}) = 5\sqrt{3}$$

Answer: $5\sqrt{3}$ square units.

Question 28

Find the sine of the angle between the vectors $\vec{a} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k}$.

Solution: Step 1: Use $$\sin\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$$

Step 2: Compute the cross product: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & -2 & 4 \end{vmatrix}$$ $$= \hat{i}(4-(-4)) - \hat{j}(12-4) + \hat{k}(-6-2) = 8\hat{i} - 8\hat{j} - 8\hat{k}$$

Step 3: Magnitude of cross product: $$|\vec{a} \times \vec{b}| = \sqrt{8^2 + (-8)^2 + (-8)^2} = \sqrt{192} = 8\sqrt{3}$$

Step 4: Magnitudes of the vectors: $$|\vec{a}| = \sqrt{3^2+1^2+2^2} = \sqrt{14}$$ $$|\vec{b}| = \sqrt{2^2+(-2)^2+4^2} = \sqrt{24} = 2\sqrt{6}$$

Step 5: Therefore, $$\sin\theta = \frac{8\sqrt{3}}{\sqrt{14} \cdot 2\sqrt{6}} = \frac{2}{\sqrt{7}}$$

Answer: $2/\sqrt{7}$

Question 29

Find a vector of magnitude 6 which is perpendicular to both the vectors $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = -2\hat{i} + \hat{j} - 2\hat{k}$.

Solution: Step 1: Find a vector perpendicular to both using cross product: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix}$$

Step 2: Expand: $$= \hat{i}[(-1)(-2) - 3(1)] - \hat{j}[4(-2) - 3(-2)] + \hat{k}[4(1) - (-1)(-2)]$$ $$= -\hat{i} + 2\hat{j} + 2\hat{k}$$

Step 3: Magnitude of this perpendicular vector: $$|\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{9} = 3$$

Step 4: Unit vector in that direction is $$\hat{n} = \frac{-\hat{i} + 2\hat{j} + 2\hat{k}}{3}$$

Step 5: Multiply by 6 to get a vector of magnitude 6: $$\vec{v} = 6\hat{n} = 6\left(\frac{-\hat{i} + 2\hat{j} + 2\hat{k}}{3}\right) = -2\hat{i} + 4\hat{j} + 4\hat{k}$$

The opposite vector is also valid.

Answer: $-2\hat{i} + 4\hat{j} + 4\hat{k}$

Question 30

A force $\vec{F} = 2\hat{i} + \hat{j} - \hat{k}$ acts on a particle and displaces it from point $A(1, 2, 3)$ to point $B(5, 4, 1)$. Find the work done by the force.

Solution: Step 1: Find the displacement vector: $$\vec{d} = \vec{AB} = (5-1)\hat{i} + (4-2)\hat{j} + (1-3)\hat{k} = 4\hat{i} + 2\hat{j} - 2\hat{k}$$

Step 2: Work done is $$W = \vec{F} \cdot \vec{d}$$

Step 3: Compute the dot product: $$W = (2\hat{i} + \hat{j} - \hat{k}) \cdot (4\hat{i} + 2\hat{j} - 2\hat{k})$$ $$= (2)(4) + (1)(2) + (-1)(-2) = 8 + 2 + 2 = 12$$

Answer: 12 units.