Three Dimensional Geometry: 1 mark Questions (CBSE Class 12)
12 different 1 mark questions on Three Dimensional Geometry from CBSE Class 12 Maths board exams 2026, newest first.
The length of perpendicular drawn from point (2, 5, 7) on line 1 x = 0 y = 0 z is
(A) 2(B) 5(C) 74 (D) 78
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The line has direction ratios 1, 0, 0 and passes through the origin, so it is the x -axis. The foot of the perpendicular from (2, 5, 7) is (2, 0, 0). Length = 5 2 + 7 2 = 74 .
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 ⋅ 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: p x − q = 1 y = r z − s , direction ratios p , 1 , r . Second line: direction ratios p ′ , 1 , r ′ . Perpendicular ⇔ 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.
The length of perpendicular drawn from the point (1, 2, 3) on line 0 x = 1 y = 0 z is
(A) 2(B) 6(C) 10 (D) 14
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The line has direction ratios 0, 1, 0 and passes through the origin, so it is the y -axis. Foot of the perpendicular from (1, 2, 3) is (0, 2, 0). Length = 1 2 + 3 2 = 10 .
The length of perpendicular drawn from the point (3, 4, 2) on the line 0 x = 0 y = 1 z is
(A) 2(B) 9(C) 5(D) 29
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Answer: (C) 5
The line has direction ratios 0, 0, 1 and passes through the origin, so it is the z -axis. Foot of the perpendicular from (3, 4, 2) is (0, 0, 2). Length = 3 2 + 4 2 = 5 .
Direction cosines of the line given by equations : 4 2 x − 1 = 3 1 − y = 6 − z are
(A) 2 , − 3 , − 6 (B) 7 2 , 7 − 3 , 7 − 6 (C) 7 2 , 7 − 3 , 7 6 (D) 61 4 , 61 − 3 , 61 − 6
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Answer: (B) 7 2 , 7 − 3 , 7 − 6
Write in standard form: 2 x − 2 1 = − 3 y − 1 = − 6 z − 0 . Direction ratios 2 , − 3 , − 6 ; 4 + 9 + 36 = 7 . Direction cosines 7 2 , 7 − 3 , 7 − 6 .
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 = 1 ; A is false. cos 0 = 1 ; R is true.
Direction cosines of line x = y = 1 − z are
(A) 1 , 1 , 1 (B) 3 1 , 3 1 , 3 − 1 (C) 0 , 0 , 1 (D) 3 1 , 3 1 , 3 1
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Answer: (B)
3 1 , 3 1 , 3 − 1
Standard form: 1 x = 1 y = − 1 z − 1 ; direction ratios 1 , 1 , − 1 . 1 + 1 + 1 = 3 , so direction cosines are 3 1 , 3 1 , 3 − 1 .
Direction cosines of line 0 1 − x = y = z are
(A) 1 , 1 , 1 (B) 0 , 2 − 1 , 2 − 1 (C) 1 , 0 , 0 (D) 0 , − 1 , − 1
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Answer: (B)
0 , 2 − 1 , 2 − 1
Write the line as 0 x − 1 = 1 y = 1 z (the 0 means x stays equal to 1); direction ratios 0 , 1 , 1 . Direction cosines are ± ( 0 , 2 1 , 2 1 ) . Option (B) 0 , 2 − 1 , 2 − 1 is one valid set (opposite sense). (D) is not a unit vector.
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 = l 2 l 1 + m 2 m 1 + n 2 n 1 (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 ∣ .
Direction ratios of lines l 1 and l 2 are ⟨ 12 , − 3 , 9 ⟩ and ⟨ 4 , q , − p ⟩ respectively. The values of p and q for which l 1 and l 2 are parallel are respectively :
(A) − 1 , 3 (B) 3 , 1 (C) − 3 , − 1 (D) − 1 , − 3
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Answer: (C) − 3 , − 1
Parallel lines have proportional direction ratios: 12 4 = − 3 q = 9 − p . − 3 q = 3 1 gives q = − 1 ; 9 − p = 3 1 gives p = − 3 .p, q = − 3 , − 1 .
Direction ratios of lines l 1 and l 2 respectively are ⟨ 1 , − 2 , 3 ⟩ and ⟨ − 2 , p , − 6 ⟩ . The value of p for which l 1 ∥ l 2 , is :
(A) − 4 (B) 4 (C) − 10 (D) 10
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Answer: (B) 4
Parallel lines have proportional direction ratios: 1 − 2 = − 2 p = 3 − 6 . The common ratio is − 2 , so p = ( − 2 ) ( − 2 ) = 4 .
Direction ratios of lines l 1 and l 2 respectively are ⟨ 1 , 0 , 0 ⟩ and ⟨ 0 , − 1 , 0 ⟩ . The direction ratios of the line perpendicular to both l 1 and l 2 are :
(A) ⟨ 1 , 1 , 0 ⟩ (B) ⟨ 0 , 0 , − 1 ⟩ (C) ⟨ 1 , 1 , 1 ⟩ (D) ⟨ 1 , 0 , − 1 ⟩
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Answer: (B) ⟨ 0 , 0 , − 1 ⟩
A direction perpendicular to both is i ^ × ( − j ^ ) = − k ^ . So the direction ratios are ⟨ 0 , 0 , − 1 ⟩ . Check: ( 0 ) ( 1 ) + ( 0 ) ( 0 ) + ( − 1 ) ( 0 ) = 0 and ( 0 ) ( 0 ) + ( 0 ) ( − 1 ) + ( − 1 ) ( 0 ) = 0 .
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