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CBSE · Class 12 · Mathematics

Three-dimensional Geometry

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The direction cosines of a line with direction ratios 3, 2, −6 are:
  1. 3/7, 2/7, −6/7
  2. 3/49, 2/49, −6/49
  3. 3/11, 2/11, −6/11
  4. 2/7, 3/7, −6/7
2.
The direction ratios of the line through (−2, −5, 4) and (−4, −4, −1) are:
  1. −2, −5, 4
  2. 1, −2, −5
  3. −6, −9, 3
  4. −2, 1, −5
3.
The Cartesian equation of the line through (5, 2, 5) with direction ratios −3, −3, 3 is:
  1. (x − 6)/(−3) = (y − 2)/(−3) = (z − 5)/3
  2. (x + 5)/(−3) = (y + 2)/(−3) = (z + 5)/3
  3. (x + 3)/5 = (y + 3)/2 = (z − 3)/5
  4. (x − 5)/(−3) = (y − 2)/(−3) = (z − 5)/3
4.
The angle between lines with direction ratios −6, 3, 2 and −9, −8, −12 is:
  1. sin−1(6/119)
  2. cos−1(1/21)
  3. cos−1(6/119)
  4. π/2
5.
The direction cosines of the x-axis are:
  1. 1, 0, 0
  2. 1, 1, 1
  3. 0, 1, 0
  4. 0, 0, 1
6.
A line has direction cosines l = 12/13, m = 4/13 and n. Then n is:
  1. ±3/169
  2. ±−3/13
  3. ±9/169
  4. ±3/13
7.
The point on the line r = (2î − 3ĵ + 2k̂) + λ(2î + ĵ + 4k̂) for λ = 2 is:
  1. (6, −1, 10)
  2. (6, −5, 8)
  3. (8, 0, 14)
  4. (−2, −5, −6)
8.
Assertion (A): The lines with direction ratios 6, 5, 5 and −15, −11, 29 are perpendicular.
Reason (R): The direction cosines l, m, n of a line satisfy l2 + m2 + n2 = 1.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Find the vector and Cartesian equations of the line through (−2, −4, 4) and (−5, 3, −1).
10.
Find the angle between the lines (x − 3)/(−12) = (y + 1)/12 = (z + 4)/1 and (x − 1)/(−11) = (y − 5)/(−2) = (z + 3)/10.
11.
Find the distance between the parallel lines r = (î − 5ĵ − 4k̂) + λ(−8î − 4ĵ + k̂) and r = (î − 4ĵ − 5k̂) + μ(−8î − 4ĵ + k̂).
12.
Find the shortest distance between the lines r = (3î − ĵ + 2k̂) + λ(î + 3ĵ − 4k̂) and r = (−î + 3ĵ − 3k̂) + μ(−î + 4ĵ − 2k̂).
13.
Find k so that the lines (x − 5)/k = (y + 3)/(−5) = (z + 5)/(−5) and (x − 5)/(−4) = (y − 3)/(−5) = (z + 5)/(−6) are perpendicular.
14.
Find the foot of the perpendicular from P(−6, 0, 1) to the line r = (2î + 5ĵ − 2k̂) + λ(4î − ĵ − 5k̂).
15.
Find the direction cosines of the line joining (−3, −1, 2) and (1, −13, −1).
16.
Write the line (x − 2)/(−4) = (y − 1)/(−5) = (z + 1)/1 in vector form.
Section C — Word problems4 × 4 = 16 marks
17.
Find the shortest distance between the lines r = (−3ĵ − k̂) + λ(4î + 4ĵ − 6k̂) and r = (−î − 4ĵ − 3k̂) + μ(−î − 2ĵ + 6k̂).
18.
Find the foot of the perpendicular from P(25, 8, −6) to the line (x − 3)/5 = (y + 2)/5 = (z + 4)/(−6), the length of the perpendicular, and the image of P in the line.
19.
Show that the lines (x + 6)/1 = (y − 2)/(−2) = (z + 4)/1 and (x + 15)/6 = (y + 16)/6 = (z − 9)/(−5) intersect, and find the point of intersection.
20.
Find the vector and Cartesian equations of the line through (−5, −4, −2) parallel to r = (−î + ĵ − 5k̂) + λ(−4î + 8ĵ + k̂), and the distance between the two lines.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Mathematics

Answer key

Three-dimensional Geometry · AW-12M-TDG-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) 3/7, 2/7, −6/7divide each by √(32 + 22 + (−6)2) = 7
2(d) −2, 1, −5subtract the coordinates
3(d) (x − 5)/(−3) = (y − 2)/(−3) = (z − 5)/3(x − x1)/a = (y − y1)/b = (z − z1)/c
4(c) cos−1(6/119)cos θ = |6|/(7 × 17)
5(a) 1, 0, 0—
6(d) ±3/13n2 = 1 − l2 − m2 = 9/169
7(a) (6, −1, 10)(2, −3, 2) + 2(2, 1, 4)
8(b) Both A and R are true, but R is not the correct explanation of A.A is true: a1a2 + b1b2 + c1c2 = 0. R is true but does not explain A.
9r = (−2î − 4ĵ + 4k̂) + λ(−3î + 7ĵ − 5k̂); (x + 2)/(−3) = (y + 4)/7 = (z − 4)/(−5)direction (−3, 7, −5)
10θ = cos−1(118/255)cos θ = |118|/(17 × 15)
11(1/9)√137|b × (a2 − a1)|/|b| = √137/9
12(51/185)√185b1 × b2 = 10î + 6ĵ + 7k̂; |(a2 − a1) · (b1 × b2)| = 51; distance = 51/√185
13k = 55/4(−4)k + (−5)(−5) + (−5)(−6) = 0
14(−2, 6, 3)a general point is (2, 5, −2) + λ(4, −1, −5); (point − P) · b = 0 gives λ = −1
154/13, −12/13, −3/13direction ratios 4, −12, −3; length 13
16r = (2î + ĵ − k̂) + λ(−4î − 5ĵ + k̂)read off the point and the direction ratios
177/11a2 − a1 = −î − ĵ − 2k̂; b1 × b2 = 12î − 18ĵ − 4k̂; distance = |(a2 − a1) · (b1 × b2)|/|b1 × b2| = 14/√484
18foot (13, 8, −16); length 2√61; image (1, 8, −26)λ = 2; the image is 2 × foot − P
19(−3, −4, −1)general points (−6, 2, −4) + λ(1, −2, 1) and (−15, −16, 9) + μ(6, 6, −5) agree for λ = 3, μ = 2
20r = (−5î − 4ĵ − 2k̂) + μ(−4î + 8ĵ + k̂); (x + 5)/(−4) = (y + 4)/8 = (z + 2)/1; distance (1/3)√401|b × (P − a)|/|b| = √3609/9