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.

Three Dimensional Geometry earns its board marks through five askable skills: direction cosines and ratios (conversions and the square-sum identity), writing line equations in both forms, angles between lines with the perpendicular/parallel conditions, shortest distance between skew or parallel lines, and the foot of perpendicular / image of a point — the classic 5-mark finisher.

Presentation rules that earn marks:

1. Standardise first, visibly. If an equation arrives with $3 - x$ or $2z$ in it, rewrite it in standard form as a labelled step — that line of working carries a mark.
2. Name the formula before substituting — "shortest distance $= \dfrac{\left|\left(\vec{b}_1\times\vec{b}_2\right)\cdot\left(\vec{a}_2-\vec{a}_1\right)\right|}{\left|\vec{b}_1\times\vec{b}_2\right|}$" written first survives later arithmetic slips.
3. Show the cross-product determinant in full, top row $\hat{i}, \hat{j}, \hat{k}$.
4. In foot-of-perpendicular problems, verify by dotting the perpendicular with the line's direction and showing zero — one line, one mark, and it catches your own errors.

2-Mark Questions

Q1. Write the direction cosines of a line perpendicular to the $xy$-plane.

Solution:

1. Perpendicular to the $xy$-plane means parallel to the $z$-axis: direction angles $90^\circ, 90^\circ, 0^\circ$ (or $180^\circ$).

Answer: $(0, 0, 1)$ or $(0, 0, -1)$.

Q2. Write the Cartesian equation of the line through $(2, -1, 3)$ with direction ratios $1, -4, 5$.

Solution:

1. Point over ratios: $\dfrac{x - 2}{1} = \dfrac{y + 1}{-4} = \dfrac{z - 3}{5}$.

Answer: $\dfrac{x - 2}{1} = \dfrac{y + 1}{-4} = \dfrac{z - 3}{5}$.

Q3. Show that the lines with direction ratios $(2, 3, -1)$ and $(1, -1, -1)$ are perpendicular.

Solution:

1. Dot the ratios: $2 - 3 + 1 = 0$.

Answer: the dot product vanishes, hence the lines are perpendicular.

Q4. Find the vector equation of the line through $(-1, 2, 3)$ parallel to the line $\vec{r} = 2\hat{i} + \hat{j} + \mu\left(\hat{i} - 2\hat{j} + 3\hat{k}\right)$.

Solution:

1. Borrow the direction, change the anchor: $\vec{r} = \left(-\hat{i} + 2\hat{j} + 3\hat{k}\right) + \lambda\left(\hat{i} - 2\hat{j} + 3\hat{k}\right)$.

Answer: as above — parallel lines differ only in the anchor point.

3-Mark Questions

Q5. Find the angle between the lines $\vec{r} = \left(\hat{i} + \hat{j} - \hat{k}\right) + \lambda\left(\hat{i} - \hat{j} + \hat{k}\right)$ and $\vec{r} = 2\hat{i} + \mu\left(\hat{i} + \hat{j} + 2\hat{k}\right)$.

Solution:

1. Directions: $(1, -1, 1)$ and $(1, 1, 2)$.
2. Dot: $1 - 1 + 2 = 2$; norms $\sqrt{3}$ and $\sqrt{6}$, product $3\sqrt{2}$.
3. Angle: $\cos\theta = \dfrac{2}{3\sqrt{2}} = \dfrac{\sqrt{2}}{3}$.

Answer: $\theta = \cos^{-1}\dfrac{\sqrt{2}}{3}$.

Q6. Find $\lambda$ if the lines with direction ratios $(\lambda, 2, 1)$ and $(1, -\lambda, 3)$ are perpendicular.

Solution:

1. Zero dot: $\lambda - 2\lambda + 3 = 0$.
2. Solve: $-\lambda + 3 = 0$.

Answer: $\lambda = 3$.

Q7. Find the equation of the line through $A(1, 2, 3)$ and the midpoint of the segment joining $(2, 4, 6)$ and $(4, 2, 0)$.

Solution:

1. Midpoint: $M = (3, 3, 3)$.
2. Direction: $M - A = (2, 1, 0)$.
3. Equation: $\dfrac{x - 1}{2} = \dfrac{y - 2}{1}$ with $z = 3$ (the zero third ratio means $z$ is constant).

Answer: $\dfrac{x - 1}{2} = \dfrac{y - 2}{1}, \ z = 3$ — equivalently $\vec{r} = \left(\hat{i} + 2\hat{j} + 3\hat{k}\right) + \lambda\left(2\hat{i} + \hat{j}\right)$.

Q8. Show that the lines $\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4}$ and $\dfrac{x - 4}{5} = \dfrac{y - 1}{2} = \dfrac{z}{1}$ intersect, and find their point of intersection.

Solution:

1. Parametrise both: $(1 + 2t, \ 2 + 3t, \ 3 + 4t)$ and $(4 + 5s, \ 1 + 2s, \ s)$.
2. Equate and solve two equations: from the $z$'s, $s = 3 + 4t$; substituting into the $x$'s: $1 + 2t = 19 + 20t$, so $t = -1$, $s = -1$.
3. Verify with the $y$'s: $2 - 3 = -1$ and $1 - 2 = -1$. ✓

Answer: the lines intersect at $(-1, -1, -1)$ — the third equation must be checked; two-equation agreement alone does not prove intersection.

5-Mark Questions

Q9. Find the shortest distance between the lines $\vec{r} = \left(\hat{i} + 2\hat{j} + 3\hat{k}\right) + \lambda\left(\hat{i} + \hat{k}\right)$ and $\vec{r} = \left(2\hat{i} + 4\hat{j} + 5\hat{k}\right) + \mu\left(2\hat{i} + \hat{j}\right)$.

Solution:

1. Cross: $\left(1, 0, 1\right) \times \left(2, 1, 0\right) = (0 - 1, \ 2 - 0, \ 1 - 0) = (-1, 2, 1)$, magnitude $\sqrt{6}$.
2. Joining vector: $(1, 2, 2)$.
3. Triple product: $-1 + 4 + 2 = 5$.

Answer: $d = \dfrac{5}{\sqrt{6}} = \dfrac{5\sqrt{6}}{6}$.

Q10. Find the vector and Cartesian equations of the line through $(1, 2, 3)$ perpendicular to both lines with direction ratios $(1, 2, -2)$ and $(2, -1, 3)$.

Solution:

1. Direction = cross: $(1, 2, -2) \times (2, -1, 3) = (6 - 2, \ -4 - 3, \ -1 - 4) = (4, -7, -5)$.
2. Vector form: $\vec{r} = \left(\hat{i} + 2\hat{j} + 3\hat{k}\right) + \lambda\left(4\hat{i} - 7\hat{j} - 5\hat{k}\right)$.
3. Cartesian form: $\dfrac{x - 1}{4} = \dfrac{y - 2}{-7} = \dfrac{z - 3}{-5}$.

Answer: as above. Verification: $(4, -7, -5)\cdot(1, 2, -2) = 4 - 14 + 10 = 0$ ✓ and $(4, -7, -5)\cdot(2, -1, 3) = 8 + 7 - 15 = 0$ ✓.

Q11. Find the foot of the perpendicular from $P(1, 6, 3)$ to the line $\dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 2}{3}$, and the length of the perpendicular.

Solution:

1. Parametrise the line: $Q = (t, \ 1 + 2t, \ 2 + 3t)$.
2. Perpendicularity: $\overrightarrow{PQ} = (t - 1, \ 2t - 5, \ 3t - 1)$ must satisfy $\overrightarrow{PQ}\cdot(1, 2, 3) = 0$: $(t - 1) + 2(2t - 5) + 3(3t - 1) = 14t - 14 = 0$, so $t = 1$.
3. Foot and length: $Q = (1, 3, 5)$; $PQ = \sqrt{0 + 9 + 4} = \sqrt{13}$.

Answer: foot $(1, 3, 5)$, perpendicular length $\sqrt{13}$.

Q12. Find the image of the point $P(1, 6, 3)$ in the line $\dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 2}{3}$.

Solution:

1. Foot of the perpendicular (from Q11): $Q = (1, 3, 5)$.
2. Reflect: the image $P'$ satisfies $Q = \dfrac{P + P'}{2}$, so $P' = 2Q - P = (2 - 1, \ 6 - 6, \ 10 - 3)$.

Answer: $P' = (1, 0, 7)$ — the foot is the midpoint of a point and its image; one subtraction finishes the problem.