CBSE Maths › Class 12 PYQs › Three Dimensional Geometry

Three Dimensional Geometry: 5 marks Questions (CBSE Class 12)

11 different 5 marks questions on Three Dimensional Geometry from CBSE Class 12 Maths board exams 2026, newest first.

1 mark (12)2 marks (6)3 marks (2)5 marks (11)

Q355 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/1/1

Check whether the lines given by $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ are parallel or not. If parallel, find the distance between them, otherwise find their point of intersection, if the lines are intersecting.

Show answer & solution

Answer: The lines are not parallel; they intersect at

(- 1, - 1, - 1)

Direction ratios 2, 3, 4 and 5, 2, 1 are not proportional, so the lines are not parallel.
General points: $(2 λ + 1, 3 λ + 2, 4 λ + 3)$ and $(5 μ + 4, 2 μ + 1, μ)$ .
Equate z: $μ = 4 λ + 3$ . Equate y: $3 λ + 2 = 2 (4 λ + 3) + 1 \Rightarrow λ = - 1$ , so $μ = - 1$ .
Check x: $2 (- 1) + 1 = - 1$ and $5 (- 1) + 4 = - 1$ . Satisfied, so the lines intersect.
Point of intersection: $(- 1, - 1, - 1)$ .

Q335 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/1/2

Prove that the line through points A(0, -1, -1) and B(4, 5, 1) intersects the line through points C(3, 9, 4) and D(-4, 4, 4). Hence, write the equation of line passing through the point of intersection of lines AB and CD as well as origin.

Show answer & solution

Answer: The lines intersect at (10, 14, 4); required line

\frac{x}{5} = \frac{y}{7} = \frac{z}{2}

AB: direction $4, 6, 2 \propto 2, 3, 1$ ; general point $(2 λ, 3 λ - 1, λ - 1)$ .
CD: direction $- 7, - 5, 0$ ; general point $(3 - 7 μ, 9 - 5 μ, 4)$ .
Equate z: $λ - 1 = 4 \Rightarrow λ = 5$ . Equate x: $10 = 3 - 7 μ \Rightarrow μ = - 1$ .
Check y: $3 (5) - 1 = 14$ and $9 - 5 (- 1) = 14$ . Consistent, so the lines intersect at (10, 14, 4).
Line through (0, 0, 0) and (10, 14, 4): $\frac{x}{10} = \frac{y}{14} = \frac{z}{4}$ , i.e. $\frac{x}{5} = \frac{y}{7} = \frac{z}{2}$ .

Q325 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/1/3

Show that line AB passing through points A(0, 4, 1), B(2, 3, -1) and the line CD passing through points C(4, 5, 0), D(2, 6, 2) are parallel. Also, find distance between them.

Show answer & solution

Answer: The lines are parallel; distance

= 3

units.

$A B = 2 i^- j^- 2 k^$ and $C D = - 2 i^+ j^+ 2 k^= - A B$ , so AB $∥$ CD.
Take $b = 2 i^- j^- 2 k^$ , $a_{1} = 4 j^+ k^$ (A), $a_{2} = 4 i^+ 5 j^$ (C).
$a_{2} - a_{1} = 4 i^+ j^- k^$ .
$b \times (a_{2} - a_{1}) = (1 + 2) i^- (- 2 + 8) j^+ (2 + 4) k^= 3 i^- 6 j^+ 6 k^$ , magnitude 9.
$∣ b ∣ = 3$ . Distance $= \frac{9}{3} = 3$ units (non-zero, so the lines are distinct).

Q355 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/2/1

A line passing through the points A(1, 2, 3) and B(5, 8, 11) intersects the line $r = 4 i^+ j^+ λ (5 i^+ 2 j^+ k^)$ . Find the co-ordinates of the point of intersection. Hence, write the equation of a line passing through the point of intersection and perpendicular to both the lines.

Show answer & solution

Answer:

(- 1, - 1, - 1)

;

r = - i^- j^- k^+ μ (- 5 i^+ 18 j^- 11 k^)

Line AB: $r = i^+ 2 j^+ 3 k^+ t (4 i^+ 6 j^+ 8 k^)$ ; general point $(1 + 4 t, 2 + 6 t, 3 + 8 t)$ .
Other line: general point $(4 + 5 λ, 1 + 2 λ, λ)$ .
Equating: $3 + 8 t = λ$ and $1 + 4 t = 4 + 5 λ \Rightarrow 1 + 4 t = 19 + 40 t \Rightarrow t = - \frac{1}{2}$ , $λ = - 1$ .
Check: $2 + 6 t = - 1 = 1 + 2 λ$ . Point of intersection $(- 1, - 1, - 1)$ .
Direction perpendicular to both: $(2 i^+ 3 j^+ 4 k^) \times (5 i^+ 2 j^+ k^) = - 5 i^+ 18 j^- 11 k^$ .
Required line: $r = - i^- j^- k^+ μ (- 5 i^+ 18 j^- 11 k^)$ , i.e. $\frac{x + 1}{- 5} = \frac{y + 1}{18} = \frac{z + 1}{- 11}$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q355 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/3/1

Represent the equations of lines $l_{1}$ and $l_{2}$ in vector form and check whether they are intersecting or not.
$l_{1} : \frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}$
$l_{2} : \frac{x + 1}{- 1} = \frac{2 - y}{- 2} = \frac{z - 5}{5}$

Show answer & solution

Answer:

l_{1} : r = - 3 i^+ j^+ 5 k^+ λ (- 3 i^+ j^+ 5 k^)

l_{2} : r = - i^+ 2 j^+ 5 k^+ μ (- i^+ 2 j^+ 5 k^)

; the lines intersect (at the origin).

$l_{1}$ passes through $(- 3, 1, 5)$ with d.r. $⟨ - 3, 1, 5 ⟩$ : $r = - 3 i^+ j^+ 5 k^+ λ (- 3 i^+ j^+ 5 k^)$ .
$l_{2}$ is $\frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ : through $(- 1, 2, 5)$ , d.r. $⟨ - 1, 2, 5 ⟩$ : $r = - i^+ 2 j^+ 5 k^+ μ (- i^+ 2 j^+ 5 k^)$ .
The direction ratios are not proportional, so the lines are not parallel.
$b_{1} \times b_{2} = (- 3 i^+ j^+ 5 k^) \times (- i^+ 2 j^+ 5 k^) = - 5 i^+ 10 j^- 5 k^$ .
$a_{2} - a_{1} = 2 i^+ j^$ , and $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = - 10 + 10 + 0 = 0$ .
Shortest distance is 0, so the lines intersect (λ = −1 and μ = −1 both give the origin).

Also asked in: 2026 65/3/2, 2026 65/3/3

Q35 (OR) (OR)5 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/3/1

Opposite sides of a square are along the lines :
$r = i^+ 2 j^- 4 k^+ λ (2 i^+ 3 j^+ 6 k^)$
$r = 3 i^+ 3 j^- 5 k^+ μ (2 i^+ 3 j^+ 6 k^)$
Find the area of the square if direction ratios of other pair of opposite sides of the square are given by $⟨ - 3, 6, p ⟩$ . Also, find the value of p.

Show answer & solution

Answer: Area

= \frac{293}{49}

sq units;

p = - 2

The lines are parallel with $b = 2 i^+ 3 j^+ 6 k^$ , $∣ b ∣ = 7$ . The side of the square is the distance between them.
$a_{2} - a_{1} = 2 i^+ j^- k^$ .
$b \times (a_{2} - a_{1}) = (2 i^+ 3 j^+ 6 k^) \times (2 i^+ j^- k^) = - 9 i^+ 14 j^- 4 k^$ , of magnitude $81 + 196 + 16 = 293$ .
Side $d = \frac{293}{7}$ , so area $= d^{2} = \frac{293}{49}$ sq units.
The other sides are perpendicular to these: $2 (- 3) + 3 (6) + 6 p = 0$ , so $p = - 2$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q355 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/4/1

Find the equation of a line (in vector and cartesian form) that passes through the point of intersection of lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ and is parallel to the vector $3 i^+ 2 j^- 8 k^$ .

Show answer & solution

Answer:

r = (- i^- j^- k^) + t (3 i^+ 2 j^- 8 k^)

;

\frac{x + 1}{3} = \frac{y + 1}{2} = \frac{z + 1}{- 8}

General points: $(2 λ + 1, 3 λ + 2, 4 λ + 3)$ and $(5 μ + 4, 2 μ + 1, μ)$ .
Equating: $4 λ + 3 = μ$ and $3 λ + 2 = 2 μ + 1 = 8 λ + 7 \Rightarrow λ = - 1$ , $μ = - 1$ .
Check x: $2 (- 1) + 1 = - 1 = 5 (- 1) + 4$ . Point of intersection $(- 1, - 1, - 1)$ .
Vector form: $r = (- i^- j^- k^) + t (3 i^+ 2 j^- 8 k^)$ .
Cartesian form: $\frac{x + 1}{3} = \frac{y + 1}{2} = \frac{z + 1}{- 8}$ .

Q355 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/4/2

Find the vector and cartesian equations of the line passing through the point of intersection of the lines $r = (i^+ j^- k^) + λ (3 i^- j^)$ and $r = (4 i^- k^) + μ (2 i^+ 3 k^)$ and parallel to the line $\frac{x - 1}{- 2} = \frac{7 - y}{- 3} = z$ .

Show answer & solution

Answer:

r = (4 i^- k^) + t (- 2 i^+ 3 j^+ k^)

;

\frac{x - 4}{- 2} = \frac{y}{3} = \frac{z + 1}{1}

General points: $(1 + 3 λ, 1 - λ, - 1)$ and $(4 + 2 μ, 0, - 1 + 3 μ)$ .
Equating y: $1 - λ = 0 \Rightarrow λ = 1$ ; z: $- 1 = - 1 + 3 μ \Rightarrow μ = 0$ ; x: $1 + 3 = 4 = 4 + 0$ . Point $(4, 0, - 1)$ .
The given line is $\frac{x - 1}{- 2} = \frac{y - 7}{3} = \frac{z}{1}$ , so its direction ratios are $- 2, 3, 1$ .
Vector form: $r = (4 i^- k^) + t (- 2 i^+ 3 j^+ k^)$ .
Cartesian form: $\frac{x - 4}{- 2} = \frac{y}{3} = \frac{z + 1}{1}$ .

Q355 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/4/3

Find the length of the perpendicular drawn from the point P(1, 2, 3) to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Also, find the equation of the perpendicular line joining P and the foot of the perpendicular.

Show answer & solution

Answer: Length 7 units; foot (3, 5, 9); line

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}

, i.e.

r = (i^+ 2 j^+ 3 k^) + t (2 i^+ 3 j^+ 6 k^)

A general point on the line is $Q (3 λ + 6, 2 λ + 7, - 2 λ + 7)$ , so $P Q = (3 λ + 5) i^+ (2 λ + 5) j^+ (4 - 2 λ) k^$ .
$P Q ⊥ (3 i^+ 2 j^- 2 k^)$ : $3 (3 λ + 5) + 2 (2 λ + 5) - 2 (4 - 2 λ) = 0 \Rightarrow 17 λ + 17 = 0 \Rightarrow λ = - 1$ .
Foot $Q (3, 5, 9)$ ; $P Q = 2 i^+ 3 j^+ 6 k^$ , length $= 4 + 9 + 36 = 7$ units.
Line PQ: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ , or $r = (i^+ 2 j^+ 3 k^) + t (2 i^+ 3 j^+ 6 k^)$ .

Q355 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/5/1

Find the foot of the perpendicular from the point (0, 2, 3) on the line $\frac{- x - 3}{- 5} = \frac{1 - y}{- 2} = \frac{3 z + 12}{9}$ and hence find the length of the perpendicular.

Show answer & solution

Answer: Foot (2, 3,

- 1

); length

21

units.

Standard form: $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ .
General point $Q (5 t - 3, 2 t + 1, 3 t - 4)$ ; $P Q = (5 t - 3, 2 t - 1, 3 t - 7)$ with P(0, 2, 3).
$P Q ⊥ (5, 2, 3)$ : $25 t - 15 + 4 t - 2 + 9 t - 21 = 0 \Rightarrow 38 t = 38 \Rightarrow t = 1$ .
Foot Q = (2, 3, $- 1$ ).
$P Q = 2^{2} + 1^{2} + (- 4)^{2} = 21$ units.

Also asked in: 2026 65/5/2, 2026 65/5/3

Q35 (OR) (OR)5 marksLong AnswerThree Dimensional Geometry CBSE 2026 · 65/5/1

Find the value of p if the shortest distance between the lines
$r = (i^+ 2 j^+ k^) + λ (i^- j^+ k^)$ and
$r = (p i^- j^- k^) + μ (2 i^+ j^+ 2 k^)$
is $\frac{3}{2}$ units.

Show answer & solution

Answer:

p = 2

p = - 4

$b_{1} \times b_{2} = i^12 j^- 1 1 k^12 = - 3 i^+ 3 k^$ , magnitude $32$ .
$a_{2} - a_{1} = (p - 1) i^- 3 j^- 2 k^$ .
$(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = - 3 (p - 1) - 6 = - 3 p - 3$ .
SD $= \frac{∣ - 3 p - 3∣}{3 2} = \frac{∣ p + 1∣}{2} = \frac{3}{2}$ .
$∣ p + 1∣ = 3 \Rightarrow p = 2$ or $p = - 4$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

← Vector Algebra Linear Programming →

Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on MonoMath CBSE →