Board Exam-Pattern Questions

How This Bank Works

Note: These are original practice questions written in the CBSE/State Board pattern — matched to the style, difficulty and mark-distribution of recent board papers, not copied from any specific year's paper. Attempt each question on paper before reading the solution.

Vector Algebra questions on the board paper are short and formula-driven — the chapter typically contributes one or two 2-mark questions, a 3-mark computation, and occasionally shares a 5-mark slot with Three Dimensional Geometry. The five askable skills: unit vectors and direction cosines, collinearity and the section formula, dot product (angles, projections, perpendicularity conditions), cross product (perpendicular vectors, areas), and short vector proofs (rhombus diagonals, angle in a semicircle, algebraic identities).

Presentation rules that earn marks:

1. State the formula before substituting. "Projection of $\vec{a}$ on $\vec{b}$ $= \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$" written first earns its mark even if arithmetic later slips.
2. Show the determinant for every cross product — examiners look for the $3 \times 3$ array with $\hat{i}, \hat{j}, \hat{k}$ in the top row.
3. Keep vectors and scalars visually distinct: arrows on vectors, none on magnitudes. Writing $\vec{a} = 3$ costs presentation marks; $|\vec{a}| = 3$ does not.
4. In proofs, end with the statement proved — "hence the diagonals are perpendicular" — mirroring the question's wording.

2-Mark Questions

Q1. Find the unit vector in the direction of $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.

Solution:

1. Magnitude: $|\vec{a}| = \sqrt{4 + 9 + 36} = 7$.
2. Divide: $\hat{a} = \dfrac{1}{7}\left(2\hat{i} - 3\hat{j} + 6\hat{k}\right)$.

Answer: $\dfrac{2}{7}\hat{i} - \dfrac{3}{7}\hat{j} + \dfrac{6}{7}\hat{k}$.

Q2. Find $\lambda$ if $2\hat{i} + \lambda\hat{j} + \hat{k}$ is perpendicular to $\hat{i} - 2\hat{j} + 3\hat{k}$.

Solution:

1. Perpendicular means zero dot: $2 - 2\lambda + 3 = 0$.
2. Solve: $\lambda = \dfrac{5}{2}$.

Answer: $\lambda = \dfrac{5}{2}$.

Q3. If $|\vec{a}| = 10$, $|\vec{b}| = 2$ and $\vec{a}\cdot\vec{b} = 12$, find $\left|\vec{a} \times \vec{b}\right|$.

Solution:

1. Lagrange's identity: $\left|\vec{a}\times\vec{b}\right|^2 = |\vec{a}|^2|\vec{b}|^2 - \left(\vec{a}\cdot\vec{b}\right)^2 = 400 - 144 = 256$.

Answer: $\left|\vec{a}\times\vec{b}\right| = 16$.

Q4. Write the direction cosines of the vector $-2\hat{i} + \hat{j} - 5\hat{k}$.

Solution:

1. Magnitude: $\sqrt{4 + 1 + 25} = \sqrt{30}$.
2. Divide each component: $\left(-\dfrac{2}{\sqrt{30}}, \dfrac{1}{\sqrt{30}}, -\dfrac{5}{\sqrt{30}}\right)$.

Answer: $l = -\dfrac{2}{\sqrt{30}}$, $m = \dfrac{1}{\sqrt{30}}$, $n = -\dfrac{5}{\sqrt{30}}$ — signs stay with their components.

3-Mark Questions

Q5. Find the vector $\vec{c}$ such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a}\cdot\vec{c} = 3$, where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$.

Solution:

1. Set $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ and expand the cross: $\vec{a}\times\vec{c} = (z - y)\hat{i} + (x - z)\hat{j} + (y - x)\hat{k} = 0\hat{i} + \hat{j} - \hat{k}$.
2. Match components: $z = y$, $x - z = 1$, $y - x = -1$ — consistent, giving $x = z + 1$, $y = z$.
3. Use the dot condition: $x + y + z = 3 \Rightarrow (z + 1) + z + z = 3 \Rightarrow z = \dfrac{2}{3}$.

Answer: $\vec{c} = \dfrac{1}{3}\left(5\hat{i} + 2\hat{j} + 2\hat{k}\right)$. Check: $\vec{a}\cdot\vec{c} = \dfrac{5+2+2}{3} = 3$. ✓

Q6. Find the area of the 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:

1. Diagonal formula: area $= \dfrac{1}{2}\left|\vec{d}_1 \times \vec{d}_2\right|$ (from Example 20 of the Solved Examples: the side-parallelogram has half the area of the diagonal-parallelogram).
2. Cross: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} = (4 - 6)\hat{i} - (12 + 2)\hat{j} + (-9 - 1)\hat{k} = -2\hat{i} - 14\hat{j} - 10\hat{k}$.
3. Halve the magnitude: $\dfrac{1}{2}\sqrt{4 + 196 + 100} = \dfrac{1}{2}\sqrt{300} = 5\sqrt{3}$.

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

Q7. If $\hat{a}$ and $\hat{b}$ are unit vectors and $\theta$ is the angle between them, show that $\sin\dfrac{\theta}{2} = \dfrac{1}{2}\left|\hat{a} - \hat{b}\right|$.

Solution:

1. Square the difference: $\left|\hat{a}-\hat{b}\right|^2 = 1 + 1 - 2\cos\theta = 2\left(1 - \cos\theta\right)$.
2. Half-angle: $1 - \cos\theta = 2\sin^2\dfrac{\theta}{2}$, so $\left|\hat{a}-\hat{b}\right|^2 = 4\sin^2\dfrac{\theta}{2}$.
3. Take roots: $\left|\hat{a}-\hat{b}\right| = 2\sin\dfrac{\theta}{2}$ (both sides non-negative for $0 \leq \theta \leq \pi$).

Answer: $\sin\dfrac{\theta}{2} = \dfrac{1}{2}\left|\hat{a}-\hat{b}\right|$, as required.

Q8. Find the projection of $\vec{b} + \vec{c}$ on $\vec{a}$, where $\vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}$.

Solution:

1. Sum first: $\vec{b} + \vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}$.
2. Dot and divide: $\left(\vec{b}+\vec{c}\right)\cdot\vec{a} = 6 - 2 + 2 = 6$; $|\vec{a}| = \sqrt{4+4+1} = 3$.

Answer: projection $= \dfrac{6}{3} = 2$.

5-Mark Questions

Q9. Find the vector $\vec{d}$ perpendicular to both $\vec{a} = 4\hat{i} + 5\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 4\hat{j} + 5\hat{k}$, satisfying $\vec{d}\cdot\vec{c} = 21$ where $\vec{c} = 3\hat{i} + \hat{j} - \hat{k}$.

Solution:

1. Perpendicular to both: $\vec{d} = \lambda\left(\vec{a}\times\vec{b}\right)$.
2. Cross: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & -1 \\ 1 & -4 & 5 \end{vmatrix} = (25 - 4)\hat{i} - (20 + 1)\hat{j} + (-16 - 5)\hat{k} = 21\hat{i} - 21\hat{j} - 21\hat{k}$.
3. Dot condition: $\vec{c}\cdot\left(\vec{a}\times\vec{b}\right) = 63 - 21 + 21 = 63$, so $63\lambda = 21$, $\lambda = \dfrac{1}{3}$.

Answer: $\vec{d} = 7\hat{i} - 7\hat{j} - 7\hat{k}$.

Q10. Find the area of the triangle with vertices $A(1, -1, 2)$, $B(2, 1, -1)$ and $C(3, -1, 1)$.

Solution:

1. Side vectors from $A$: $\overrightarrow{AB} = \hat{i} + 2\hat{j} - 3\hat{k}$, $\overrightarrow{AC} = 2\hat{i} - \hat{k}$.
2. Cross: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 0 & -1 \end{vmatrix} = (-2 - 0)\hat{i} - (-1 + 6)\hat{j} + (0 - 4)\hat{k} = -2\hat{i} - 5\hat{j} - 4\hat{k}$.
3. Halve the magnitude: $\left|\overrightarrow{AB}\times\overrightarrow{AC}\right| = \sqrt{4 + 25 + 16} = \sqrt{45} = 3\sqrt{5}$.

Answer: area $= \dfrac{3\sqrt{5}}{2}$ square units.

Q11. For any two nonzero vectors $\vec{a}$ and $\vec{b}$, prove that $|\vec{a}|\vec{b} + |\vec{b}|\vec{a}$ is perpendicular to $|\vec{a}|\vec{b} - |\vec{b}|\vec{a}$.

Solution:

1. Dot the two combinations: $\left(|\vec{a}|\vec{b} + |\vec{b}|\vec{a}\right)\cdot\left(|\vec{a}|\vec{b} - |\vec{b}|\vec{a}\right)$.
2. Expand (difference of squares): $|\vec{a}|^2\left(\vec{b}\cdot\vec{b}\right) - |\vec{b}|^2\left(\vec{a}\cdot\vec{a}\right) = |\vec{a}|^2|\vec{b}|^2 - |\vec{b}|^2|\vec{a}|^2 = 0$ (the middle terms $\mp|\vec{a}||\vec{b}|\,\vec{a}\cdot\vec{b}$ cancel by commutativity).

Answer: the dot product vanishes, hence the vectors are perpendicular — geometrically, these are the diagonals of a rhombus built on $|\vec{a}|\vec{b}$ and $|\vec{b}|\vec{a}$, two vectors of equal length $|\vec{a}||\vec{b}|$.

Q12. Using vectors, prove that the angle inscribed in a semicircle is a right angle.

Solution:

1. Set up: let the circle have centre $O$ and radius $r$, with diameter endpoints $A$ and $B$ having position vectors $\vec{a}$ and $-\vec{a}$ (so $|\vec{a}| = r$). Let $P$ be any point on the circle, position vector $\vec{p}$ with $|\vec{p}| = r$.
2. The two chord vectors: $\overrightarrow{PA} = \vec{a} - \vec{p}$ and $\overrightarrow{PB} = -\vec{a} - \vec{p}$.
3. Dot them: $\left(\vec{a} - \vec{p}\right)\cdot\left(-\vec{a} - \vec{p}\right) = -|\vec{a}|^2 + \vec{p}\cdot\vec{p} = |\vec{p}|^2 - |\vec{a}|^2 = r^2 - r^2 = 0$.

Answer: $\overrightarrow{PA} \perp \overrightarrow{PB}$ for every position of $P$ on the circle — the angle in a semicircle is right, proved in three lines with one dot product.