Three Dimensional Geometry: CBSE Class 12 Previous Year Questions

31 different questions from Three Dimensional Geometry (NCERT Chapter 11) asked in CBSE Class 12 Maths board exams 2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions12 questions 2 marks questions6 questions 3 marks questions2 questions 5 marks questions11 questions

Most asked Three Dimensional Geometry questions

Q201 markAssertion–ReasonThree Dimensional Geometry CBSE 2026 · 65/1/1

Assertion (A): Lines given by $x = p y + q, z = r y + s$ and $x = p^{'} y + q^{'}, z = r^{'} y + s^{'}$ are perpendicular to each other when $p p^{'} + r r^{'} = 1$ .
Reason (R): Two lines $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ are perpendicular to each other if $b_{1} \cdot b_{2} = 0$ .

(A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true and Reason (R) is false.
(D)Assertion (A) is false and Reason (R) is true.

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Answer: (D) Assertion (A) is false and Reason (R) is true.

First line: $\frac{x - q}{p} = \frac{y}{1} = \frac{z - s}{r}$ , direction ratios $p, 1, r$ .
Second line: direction ratios $p^{'}, 1, r^{'}$ .
Perpendicular $\Leftrightarrow p p^{'} + 1 + r r^{'} = 0$ , i.e. $p p^{'} + r r^{'} = - 1$ . So A is false.
R is the standard condition for perpendicular lines, so R is true.

Also asked in: 2026 65/1/2, 2026 65/1/3

Q201 markAssertion–ReasonThree Dimensional Geometry CBSE 2026 · 65/2/1

Assertion (A) : A line can have direction cosines $< 1, 1, 1 >$
Reason (R) : $cos θ = 1$ is possible for $θ = 0$ .

(A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

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Answer: (D) Assertion (A) is false, but Reason (R) is true.

For direction cosines $l^{2} + m^{2} + n^{2} = 1$ , but $1^{2} + 1^{2} + 1^{2} = 3 \neq = 1$ ; A is false.
$cos 0 = 1$ ; R is true.

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

Q161 markMCQThree Dimensional Geometry CBSE 2026 · 65/3/1

If $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are direction cosines of lines $L_{1}$ and $L_{2}$ respectively and $θ$ is the acute angle between them, then :

(A) $cos θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$
(B) $sin θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$
(C) $tan θ = \frac{l _{1}}{l _{2}} + \frac{m _{1}}{m _{2}} + \frac{n _{1}}{n _{2}}$
(D) $cos θ = ∣ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} ∣$

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Answer: (D)

cos θ = ∣ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} ∣

The angle between the direction vectors satisfies $cos ϕ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$ , which may be negative.
For the acute angle we take the absolute value: $cos θ = ∣ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} ∣$ .

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

Q252 marksVery Short AnswerThree Dimensional Geometry CBSE 2026 · 65/3/1

If the lines $\frac{x - 3}{1} = \frac{1 - y}{1} = \frac{z + 2}{p}$ and $\frac{2 - x}{3} = \frac{y + 1}{5} = \frac{z + 56}{2 p}$ are perpendicular to each other, then find the value(s) of p.

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Answer:

p = \pm 2

In standard form the first line is $\frac{x - 3}{1} = \frac{y - 1}{- 1} = \frac{z + 2}{p}$ : d.r. $⟨ 1, - 1, p ⟩$ .
The second is $\frac{x - 2}{- 3} = \frac{y + 1}{5} = \frac{z + 56}{2 p}$ : d.r. $⟨ - 3, 5, 2 p ⟩$ .
Perpendicular: $(1) (- 3) + (- 1) (5) + p (2 p) = 0$ , so $2 p^{2} = 8$ .
$p = \pm 2$ .

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

Q25 (OR) (OR)2 marksVery Short AnswerThree Dimensional Geometry CBSE 2026 · 65/3/1

Find the vector equation of a line passing through the origin and perpendicular to both the lines $r = 2 i^- j^+ 2 k^+ λ (3 i^+ 4 j^+ 2 k^)$ and $r = μ (i^- j^+ k^)$ .

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Answer:

r = λ (6 i^- j^- 7 k^)

A direction perpendicular to both is $(3 i^+ 4 j^+ 2 k^) \times (i^- j^+ k^)$ .
$= i^(4 + 2) - j^(3 - 2) + k^(- 3 - 4) = 6 i^- j^- 7 k^$ .
Line through the origin: $r = λ (6 i^- j^- 7 k^)$ .

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

Q252 marksVery Short AnswerThree Dimensional Geometry CBSE 2026 · 65/5/1

Find the angle between the following pair of lines :
$\frac{x - 2}{3} = \frac{y + 5}{2} = \frac{1 - z}{- 6}$ and
$\frac{x - 7}{1} = \frac{y}{2} = \frac{6 - z}{- 2}$

Show answer & solution

Answer:

cos^{- 1} (\frac{19}{21})

Write the lines as $\frac{x - 2}{3} = \frac{y + 5}{2} = \frac{z - 1}{6}$ and $\frac{x - 7}{1} = \frac{y}{2} = \frac{z - 6}{2}$ .
Direction ratios: (3, 2, 6) and (1, 2, 2).
$cos θ = \frac{3 + 4 + 12}{49 9} = \frac{19}{21}$ .
$θ = cos^{- 1} (\frac{19}{21})$ .

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

Q303 marksShort AnswerThree Dimensional Geometry CBSE 2026 · 65/4/1

Find a point on the line $\frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2}$ at a distance of $2$ units from the point (1, 2, 3).

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Answer: (2, 1, 3) or

(\frac{4}{17}, \frac{37}{17}, \frac{31}{17})

Write the line as $\frac{x - 2}{3} = \frac{y - 1}{- 2} = \frac{z - 3}{2} = λ$ ; a general point is $(3 λ + 2, 1 - 2 λ, 2 λ + 3)$ .
Distance from (1, 2, 3): $(3 λ + 1)^{2} + (- 1 - 2 λ)^{2} + (2 λ)^{2} = 2$ .
$17 λ^{2} + 10 λ = 0 \Rightarrow λ = 0$ or $λ = - \frac{10}{17}$ .
Points: $(2, 1, 3)$ and $(\frac{4}{17}, \frac{37}{17}, \frac{31}{17})$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q30 (OR) (OR)3 marksShort AnswerThree Dimensional Geometry CBSE 2026 · 65/4/1

Find the shortest distance between the lines
$r = (4 + λ) i^+ (2 λ - 1) j^- 3 λ k^$
$r = (1 + 2 μ) i^+ (2 - 5 μ) k^+ (4 μ - 1) j^$ .

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Answer:

\frac{6}{5}

units

Line 1: $a_{1} = 4 i^- j^$ , $b_{1} = i^+ 2 j^- 3 k^$ .
Line 2: $a_{2} = i^- j^+ 2 k^$ , $b_{2} = 2 i^+ 4 j^- 5 k^$ .
$b_{1} \times b_{2} = 2 i^- j^$ , $∣ b_{1} \times b_{2} ∣ = 5$ .
$a_{2} - a_{1} = - 3 i^+ 2 k^$ ; $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = - 6$ .
SD $= \frac{∣ - 6∣}{5} = \frac{6}{5}$ units.

Also asked in: 2026 65/4/2, 2026 65/4/3

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.

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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.
$l_{1} : \frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}$
$l_{2} : \frac{x + 1}{- 1} = \frac{2 - y}{- 2} = \frac{z - 5}{5}$

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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

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