Coordinate geometry helps students visualise and analyse objects in space, and Chapter 11: Three Dimensional Geometry is one of the most important and application-oriented chapters in Class 12 Maths. This chapter focuses on understanding the geometry of lines in three-dimensional space, including their equations, angles between lines, and the shortest distance between two lines.
NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry are prepared strictly according to the CBSE syllabus and board exam pattern. These solutions include important CBSE board questions asked between 2018 and 2025 and are explained in a clear, step-by-step manner using simple language and proper mathematical presentation. This helps students gain strong conceptual clarity, practise effectively, and score well in board examinations.
NCERT Solutions For Class 12 Maths Chapter 11 Three Dimensional Geometry
Q.
If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.
Q.
Find the vector and the cartesian equations of the lines that passes through the origin and (5, –2 , 3).
Q.
Findtheshortestdistancebetweenlinesr=6i+2j+2k+λ(i−2j+2k)andr=−4i−k+μ(3i−2j−2k).
Q.
Findtheanglebetweentheplaneswhosevectorequationsarer(2i+2j−k)=5andr(3i−3j+5k)=
Q.
Findtheshortestdistancebetweenthelines7x+1=−6y+1=1z−1and1x−3=−2y−5=1z−7
Q.
Showthatthelines7x−5=−5x+2=1zand1x=2y=3zareperpendiculartoeachother.
Q.
Findthevaluesofpsothatthelines31−x=2p7y−14=2z−3and3p7−7x=1y−5=56−zareatrightangles.
Q.
Findtheanglebetweenthefollowingpairsoflines:(i)r=2i−5j+k+λ(3i+2j+6k)andr=7i−6k+μ(i+2j+2k)(ii)r=3i+j−2k+λ(i−j−2k)andr=2i−j−56k+μ(3i−5j−4k)
Q.
Find the vector and the cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).
Q.
Thecartesianequationofalineis3x−5=7y+4=2z−6.Writeitsvectorform.
Q.
If a line has the direction ratios –18, 12, –4 then what are its direction cosines?
Q.
Findthecartesianequation of the line which passes through the point (−2, 4, −5) and parallel to the line given by 3x+3=5y−4=6z+8.
Q.
Q.
Findtheequationofthelinewhichpassesthroughthepoint (1,2,3) andisparalleltothevector3i^+2j^−2k^.
Q.
Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).
Q.
Show that the line through the points (1,–1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Q.
Showthatthethreelineswithdirectioncosines1312,13−3,13−4;134,1312,133;133,13−4,1312aremutuallyperpendicular.
Q.
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (–1, 1, 2) and (–5, –5, –2).
Q.
Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.
Q.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.
Q. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.
Solution:The direction cosines of a line are given by:
l = cos α, m = cos β, n = cos γ
Here,
Now calculating each cosine:
- l = cos 90° = 0
- m = cos 135° = −1/√2
- n = cos 45° = 1/√2
Therefore, the direction cosines of the line are:
(0, −1/√2, 1/√2)
Q. Find the vector and the Cartesian equations of the line that passes through the origin and the point (5, −2, 3).
Solution: The given line passes through the origin O(0, 0, 0) and the point A(5, −2, 3).The direction ratios of the line are:
⟨5, −2, 3⟩
Vector Equation of the Line:
r = t(5i − 2j + 3k), where t is a scalar.
Cartesian Equation of the Line:
The Cartesian form of the line is given by:
x/5 = y/−2 = z/3
Q. Find the shortest distance between the lines
r = 6i + 2j + 2k + λ(i − 2j + 2k)
and
r = −4i − k + μ(3i − 2j − 2k).
Solution:The given lines are:
Line 1: r = 6i + 2j + 2k + λ(i − 2j + 2k)
Line 2: r = −4i − k + μ(3i − 2j − 2k)
A point on Line 1 is A(6, 2, 2) and its direction ratios are:
d₁ = ⟨1, −2, 2⟩
A point on Line 2 is B(−4, 0, −1) and its direction ratios are:
d₂ = ⟨3, −2, −2⟩
The vector AB is:
AB = ⟨−10, −2, −3⟩
The shortest distance between two skew lines is given by:
|AB · (d₁ × d₂)| / |d₁ × d₂|
First, compute the cross product:
d₁ × d₂ = ⟨8, 8, 4⟩
Now,
|AB · (d₁ × d₂)| = |(−10)(8) + (−2)(8) + (−3)(4)| = 108
|d₁ × d₂| = √(8² + 8² + 4²) = √144 = 12
Shortest distance = 108 / 12 = 9 units
Q. Find the angle between the planes whose vector equations are:
r · (2i + 2j − k) = 5
and
r · (3i − 3j + 5k) = (constant)
Solution: For a plane of the form r · n = d, the vector n is the normal to the plane.The angle between two planes is the angle between their normals.
Normal to plane 1:
n₁ = ⟨2, 2, −1⟩
Normal to plane 2:
n₂ = ⟨3, −3, 5⟩
Angle θ between planes:
cos θ = |n₁ · n₂| / (|n₁| |n₂|)
n₁ · n₂ = (2)(3) + (2)(−3) + (−1)(5) = 6 − 6 − 5 = −5
|n₁ · n₂| = 5
|n₁| = √(2² + 2² + (−1)²) = √(4 + 4 + 1) = √9 = 3
|n₂| = √(3² + (−3)² + 5²) = √(9 + 9 + 25) = √43
Therefore,
cos θ = 5 / (3√43)
Angle between the planes:
θ = cos⁻¹( 5 / (3√43) )
Q. Find the shortest distance between the lines
(x + 1)/7 = (y + 1)/(-6) = (z - 1)/1and
(x - 3)/1 = (y - 5)/(-2) = (z - 7)/1
Solution: Write each line in parametric form.For Line 1, let (x + 1)/7 = (y + 1)/(-6) = (z - 1)/1 = t.
So a point on Line 1 is A(−1, −1, 1) and its direction ratios are:
d₁ = ⟨7, −6, 1⟩
For Line 2, let (x − 3)/1 = (y − 5)/(-2) = (z − 7)/1 = s.
So a point on Line 2 is B(3, 5, 7) and its direction ratios are:
d₂ = ⟨1, −2, 1⟩
Vector AB = B − A = ⟨3 − (−1), 5 − (−1), 7 − 1⟩ = ⟨4, 6, 6⟩
Shortest distance between two skew lines:
|AB · (d₁ × d₂)| / |d₁ × d₂|
Compute d₁ × d₂:
d₁ × d₂ = ⟨−4, −6, −8⟩
|d₁ × d₂| = √( (−4)² + (−6)² + (−8)² ) = √(16 + 36 + 64) = √116 = 2√29
AB · (d₁ × d₂) = ⟨4, 6, 6⟩ · ⟨−4, −6, −8⟩
= (4)(−4) + (6)(−6) + (6)(−8)
= −16 − 36 − 48 = −100
|AB · (d₁ × d₂)| = 100
Therefore, shortest distance =
100 / (2√29) = 50/√29
Shortest distance = 50/√29 units
(or 50√29/29 units)
Q. Show that the lines
(x − 5)/7 = (y + 2)/(−5) = z/1and
x/1 = y/2 = z/3
are perpendicular to each other.
Solution: For a line in symmetric form, the denominators give the direction ratios (d.r.).For Line 1:
(x − 5)/7 = (y + 2)/(−5) = z/1
So direction ratios are d₁ = ⟨7, −5, 1⟩
For Line 2:
x/1 = y/2 = z/3
So direction ratios are d₂ = ⟨1, 2, 3⟩
Two lines are perpendicular if their direction vectors are perpendicular, i.e.
d₁ · d₂ = 0
d₁ · d₂ = (7)(1) + (−5)(2) + (1)(3)
= 7 − 10 + 3
= 0
Since d₁ · d₂ = 0, the given lines are perpendicular to each other.
Q. Find the values of p so that the lines
(1 − x)/3 = (7y − 14)/(2p) = (z − 3)/2and
(7 − 7x)/(3p) = (y − 5)/1 = (6 − z)/5
are at right angles.
Solution: Two lines are at right angles if their direction ratios are perpendicular, i.e. their dot product is 0.Line 1:
(1 − x)/3 = (7y − 14)/(2p) = (z − 3)/2 = t
1 − x = 3t ⇒ x = 1 − 3t
7y − 14 = 2pt ⇒ y = 2 + (2p/7)t
z − 3 = 2t ⇒ z = 3 + 2t
Direction ratios of Line 1: d₁ = ⟨−3, 2p/7, 2⟩
Multiply by 7 to avoid fraction: d₁ = ⟨−21, 2p, 14⟩
Line 2:
(7 − 7x)/(3p) = (y − 5)/1 = (6 − z)/5 = s
7 − 7x = 3ps ⇒ x = 1 − (3p/7)s
y − 5 = s ⇒ y = 5 + s
6 − z = 5s ⇒ z = 6 − 5s
Direction ratios of Line 2: d₂ = ⟨−3p/7, 1, −5⟩
Multiply by 7: d₂ = ⟨−3p, 7, −35⟩
For right angles:
d₁ · d₂ = 0
⟨−21, 2p, 14⟩ · ⟨−3p, 7, −35⟩ = 0
(−21)(−3p) + (2p)(7) + (14)(−35) = 0
63p + 14p − 490 = 0
77p = 490
p = 490/77 = 70/11
Hence, p = 70/11.
FAQs: NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry
Q1. Why is Three Dimensional Geometry important for Class 12 Maths?
This chapter builds spatial understanding and has strong applications in mathematics and physics. It also carries good weightage in CBSE board exams.
Q2. How many marks are usually asked from Chapter 11 in board exams?
Generally, 5 to 8 marks worth of questions are asked, mostly from equations of lines and shortest distance between two lines.
Q3. Are NCERT Solutions enough to score well in this chapter?
Yes. NCERT textbook exercises and NCERT Solutions are fully sufficient for CBSE board exams, as most questions are directly based on NCERT problems.
Q4. Which topics are most important in Three Dimensional Geometry?
The most important topics include:
Q5. Why do students find this chapter difficult?
Students often struggle with:
Regular practice and clear understanding of formulas help overcome these difficulties.
Q6. Is Three Dimensional Geometry useful for competitive exams?
Yes. This chapter is very important for JEE and other competitive exams, especially for questions related to vector-based geometry and spatial reasoning.