Class 12 Mathematics - NCERT Mathematics Part II
NCERT Solutions for Class 12 Maths Chapter 11: Three Dimensional Geometry
Chapter 11 extends coordinate geometry to three dimensions. Students learn direction cosines and ratios of lines, equations of lines and planes in vector and Cartesian forms, angle between lines/planes, distance of a point from a plane, shortest distance between skew lines, and coplanarity conditions. This chapter is part of the Vectors and Three-Dimensional Geometry unit which carries 14 marks in the CBSE Class 12 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 11 has 3 exercises plus a Miscellaneous Exercise. Exercise 11.1 covers direction cosines and ratios, Exercise 11.2 covers equations of lines in space, and Exercise 11.3 covers equations of planes.
Table of Contents
11.1Exercise 11.1 - Direction Cosines and Direction Ratios
Q1If a line makes angles 90, 135 and 45 degrees with the x, y and z-axes respectively, find its direction cosines.
l = cos 90 = 0
m = cos 135 = -1/sqrt(2)
n = cos 45 = 1/sqrt(2)
Direction cosines: (0, -1/sqrt(2), 1/sqrt(2))
Verification: l^2 + m^2 + n^2 = 0 + 1/2 + 1/2 = 1 ✓
Q2Find the direction cosines of a line which makes equal angles with the coordinate axes.
Let the line make angle alpha with each axis.
l = m = n = cos(alpha)
Using l^2 + m^2 + n^2 = 1:
3cos^2(alpha) = 1
cos(alpha) = +- 1/sqrt(3)
Direction cosines: (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)) or (-1/sqrt(3), -1/sqrt(3), -1/sqrt(3))
Q3If a line has the direction ratios -18, 12, -4, then what are its direction cosines?
Direction ratios: a = -18, b = 12, c = -4
sqrt(a^2 + b^2 + c^2) = sqrt(324 + 144 + 16) = sqrt(484) = 22
Direction cosines:
l = -18/22 = -9/11
m = 12/22 = 6/11
n = -4/22 = -2/11
Verification: 81/121 + 36/121 + 4/121 = 121/121 = 1 ✓
Q4Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.
Let A = (2,3,4), B = (-1,-2,1), C = (5,8,7)
Direction ratios of AB: (-1-2, -2-3, 1-4) = (-3, -5, -3)
Direction ratios of AC: (5-2, 8-3, 7-4) = (3, 5, 3)
Ratio check: -3/3 = -5/5 = -3/3 = -1
Since direction ratios of AB and AC are proportional (ratio = -1), AB is parallel to AC. Since they share the common point A, the three points are collinear.
Q5Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2), (-5, -5, -2).
Let A = (3,5,-4), B = (-1,1,2), C = (-5,-5,-2)
Side AB: direction ratios = (-4, -4, 6)
|AB| = sqrt(16+16+36) = sqrt(68) = 2sqrt(17)
DCs: (-4/(2sqrt(17)), -4/(2sqrt(17)), 6/(2sqrt(17))) = (-2/sqrt(17), -2/sqrt(17), 3/sqrt(17))
Side BC: direction ratios = (-4, -6, -4)
|BC| = sqrt(16+36+16) = sqrt(68) = 2sqrt(17)
DCs: (-2/sqrt(17), -3/sqrt(17), -2/sqrt(17))
Side CA: direction ratios = (3-(-5), 5-(-5), -4-(-2)) = (8, 10, -2)
|CA| = sqrt(64+100+4) = sqrt(168) = 2sqrt(42)
DCs: (4/sqrt(42), 5/sqrt(42), -1/sqrt(42))
11.2Exercise 11.2 - Equations of a Line in Space
Q1Show that the three lines with direction cosines (12/13, -3/13, -4/13), (4/13, 12/13, 3/13), (3/13, -4/13, 12/13) are mutually perpendicular.
For two lines to be perpendicular: l1l2 + m1m2 + n1*n2 = 0
Lines 1 and 2: (12/13)(4/13) + (-3/13)(12/13) + (-4/13)(3/13)
= (48 - 36 - 12)/169 = 0/169 = 0 ✓
Lines 2 and 3: (4/13)(3/13) + (12/13)(-4/13) + (3/13)(12/13)
= (12 - 48 + 36)/169 = 0/169 = 0 ✓
Lines 1 and 3: (12/13)(3/13) + (-3/13)(-4/13) + (-4/13)(12/13)
= (36 + 12 - 48)/169 = 0/169 = 0 ✓
Since all pairwise dot products are zero, the three lines are mutually perpendicular.
Q2Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular to the line through the points (0, 3, 2), (3, 5, 6).
Direction ratios of line 1: (3-1, 4-(-1), -2-2) = (2, 5, -4)
Direction ratios of line 2: (3-0, 5-3, 6-2) = (3, 2, 4)
For perpendicularity: a1a2 + b1b2 + c1*c2 = 0
(2)(3) + (5)(2) + (-4)(4) = 6 + 10 - 16 = 0 ✓
Since the dot product of direction ratios is zero, the lines are perpendicular.
Q3Show 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).
Direction ratios of line 1: (2-4, 3-7, 4-8) = (-2, -4, -4)
Direction ratios of line 2: (1-(-1), 2-(-2), 5-1) = (2, 4, 4)
For parallelism: a1/a2 = b1/b2 = c1/c2
-2/2 = -4/4 = -4/4 = -1
Since the ratios are equal, the lines are parallel (with opposite directions).
Q4Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3i + 2j - 2k.
A line through point a = i + 2j + 3k parallel to b = 3i + 2j - 2k:
Vector equation: r = a + lambda*b
r = (i + 2j + 3k) + lambda(3i + 2j - 2k)
Cartesian equation: (x-1)/3 = (y-2)/2 = (z-3)/(-2)
Q5Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i - j + 4k and is in the direction i + 2j - k.
Point: a = 2i - j + 4k, Direction: b = i + 2j - k
Vector equation: r = (2i - j + 4k) + lambda(i + 2j - k)
Cartesian equation: (x-2)/1 = (y+1)/2 = (z-4)/(-1)
Q6Find the Cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by (x+3)/3 = (y-4)/5 = (z+8)/6.
The given line has direction ratios (3, 5, 6).
A parallel line through (-2, 4, -5) has the same direction ratios.
Cartesian equation: (x+2)/3 = (y-4)/5 = (z+5)/6
Q7The Cartesian equation of a line is (x-5)/3 = (y+4)/7 = (z-6)/2. Write its vector equation.
Point on the line: (5, -4, 6), Direction ratios: (3, 7, 2)
Vector equation: r = (5i - 4j + 6k) + lambda(3i + 7j + 2k)
Q8Find the vector and the Cartesian equations of the lines that passes through the origin and (5, -2, 3).
Point: origin = (0, 0, 0)
Direction ratios: (5-0, -2-0, 3-0) = (5, -2, 3)
Vector equation: r = lambda(5i - 2j + 3k)
Cartesian equation: x/5 = y/(-2) = z/3
Q9Find the vector and the Cartesian equations of the line that passes through the points (3, -2, -5), (3, -2, 6).
Direction ratios: (3-3, -2-(-2), 6-(-5)) = (0, 0, 11)
The line is parallel to the z-axis.
Vector equation: r = (3i - 2j - 5k) + lambda(11k) or r = (3i - 2j - 5k) + lambda*k
Cartesian equation: (x-3)/0 = (y+2)/0 = (z+5)/11
Note: (x-3)/0 means x = 3, and (y+2)/0 means y = -2. So the line is x = 3, y = -2 with z varying.
Q10Find the angle between the following pairs of lines: (i) r = 2i - 5j + k + lambda(3i + 2j + 6k) and r = 7i - 6k + mu(i + 2j + 2k) (ii) r = 3i + j - 2k + lambda(i - j - 2k) and r = 2i - j - 56k + mu(3i - 5j - 4k)
(i) b1 = 3i + 2j + 6k, b2 = i + 2j + 2k
cos(theta) = |b1.b2|/(|b1||b2|)
b1.b2 = 3 + 4 + 12 = 19
|b1| = sqrt(9+4+36) = 7
|b2| = sqrt(1+4+4) = 3
cos(theta) = 19/21
theta = cos^(-1)(19/21)
(ii) b1 = i - j - 2k, b2 = 3i - 5j - 4k
b1.b2 = 3 + 5 + 8 = 16
|b1| = sqrt(1+1+4) = sqrt(6)
|b2| = sqrt(9+25+16) = sqrt(50) = 5sqrt(2)
cos(theta) = 16/(sqrt(6)*5sqrt(2)) = 16/(5sqrt(12)) = 16/(10sqrt(3))
= 8/(5sqrt(3))
theta = cos^(-1)(8/(5sqrt(3)))
Q11Find the angle between the following pair of lines: (i) (x-2)/2 = (y-1)/5 = (z+3)/(-3) and (x+2)/(-1) = (y-4)/8 = (z-5)/4 (ii) x/2 = y/2 = z/1 and (x-5)/4 = (y-2)/1 = (z-3)/8
(i) DRs: (2, 5, -3) and (-1, 8, 4)
cos(theta) = |a1a2 + b1b2 + c1c2|/(sqrt(a1^2+b1^2+c1^2) * sqrt(a2^2+b2^2+c2^2))
= |(-2) + 40 + (-12)|/(sqrt(4+25+9) * sqrt(1+64+16))
= |26|/(sqrt(38) * sqrt(81))
= 26/(9sqrt(38))
theta = cos^(-1)(26/(9sqrt(38)))
(ii) DRs: (2, 2, 1) and (4, 1, 8)
cos(theta) = |8 + 2 + 8|/(sqrt(4+4+1) * sqrt(16+1+64))
= 18/(3 * 9) = 18/27 = 2/3
theta = cos^(-1)(2/3)
Q12Find the values of p so that the lines (1-x)/3 = (7y-14)/(2p) = (z-3)/2 and (7-7x)/(3p) = (y-5)/1 = (6-z)/5 are at right angles.
Rewriting in standard form:
Line 1: (x-1)/(-3) = (y-2)/(2p/7) = (z-3)/2
DRs: (-3, 2p/7, 2)
Line 2: (x-1)/(-3p/7) = (y-5)/1 = (z-6)/(-5)
DRs: (-3p/7, 1, -5)
For perpendicular lines: a1a2 + b1b2 + c1c2 = 0
(-3)(-3p/7) + (2p/7)(1) + (2)(-5) = 0
9p/7 + 2p/7 - 10 = 0
11p/7 = 10
p = 70/11
Q13Show that the lines (x-5)/7 = (y+2)/(-5) = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.
DRs of line 1: (7, -5, 1)
DRs of line 2: (1, 2, 3)
a1a2 + b1b2 + c1c2 = 7(1) + (-5)(2) + 1(3) = 7 - 10 + 3 = 0
Since the sum is zero, the lines are perpendicular. ✓
Q14Find the shortest distance between the lines: r = (i + 2j + k) + lambda(i - j + k) and r = (2i - j - k) + mu(2i + j + 2k)
a1 = i + 2j + k, b1 = i - j + k
a2 = 2i - j - k, b2 = 2i + j + 2k
a2 - a1 = i - 3j - 2k
b1 x b2 = |i j k|
|1 -1 1|
|2 1 2|
= i(-2-1) - j(2-2) + k(1+2) = -3i + 0j + 3k = -3i + 3k
|b1 x b2| = sqrt(9 + 0 + 9) = 3sqrt(2)
Shortest distance = |(a2-a1).(b1 x b2)| / |b1 x b2|
= |(1)(-3) + (-3)(0) + (-2)(3)| / 3sqrt(2)
= |-3 + 0 - 6| / 3sqrt(2)
= 9 / 3sqrt(2)
= 3/sqrt(2)
Shortest distance = 3/sqrt(2) = 3sqrt(2)/2 units
Q15Find the shortest distance between the lines: (x+1)/7 = (y+1)/(-6) = (z+1)/1 and (x-3)/1 = (y-5)/(-2) = (z-7)/1
Point on L1: (-1, -1, -1), DRs: (7, -6, 1)
Point on L2: (3, 5, 7), DRs: (1, -2, 1)
a2 - a1 = (3-(-1), 5-(-1), 7-(-1)) = (4, 6, 8)
b1 x b2 = |i j k|
|7 -6 1|
|1 -2 1|
= i(-6+2) - j(7-1) + k(-14+6) = -4i - 6j - 8k
|b1 x b2| = sqrt(16 + 36 + 64) = sqrt(116) = 2sqrt(29)
Shortest distance = |(a2-a1).(b1 x b2)| / |b1 x b2|
= |(4)(-4) + (6)(-6) + (8)(-8)| / 2sqrt(29)
= |-16 - 36 - 64| / 2sqrt(29)
= 116 / 2sqrt(29)
= 58/sqrt(29)
= 2sqrt(29)
Shortest distance = 2sqrt(29) units
Q16Find the shortest distance between the lines whose vector equations are: r = (1-t)(i) + (t-2)(j) + (3-2t)(k) and r = (s+1)(i) + (2s-1)(j) - (2s+1)(k)
Rewriting:
L1: r = (i - 2j + 3k) + t(-i + j - 2k), so a1 = (1,-2,3), b1 = (-1,1,-2)
L2: r = (i - j - k) + s(i + 2j - 2k), so a2 = (1,-1,-1), b2 = (1,2,-2)
a2 - a1 = (0, 1, -4)
b1 x b2 = |i j k|
|-1 1 -2|
|1 2 -2|
= i(1*(-2) - (-2)*2) - j((-1)(-2) - (-2)(1)) + k((-1)(2) - (1)(1))
= i(-2+4) - j(2+2) + k(-2-1)
= 2i - 4j - 3k
|b1 x b2| = sqrt(4 + 16 + 9) = sqrt(29)
Shortest distance = |(0)(2) + (1)(-4) + (-4)(-3)| / sqrt(29)
= |0 - 4 + 12| / sqrt(29)
= 8/sqrt(29)
= 8sqrt(29)/29 units
Q17Find the shortest distance between the lines whose vector equations are: r = (i + 2j + 3k) + lambda(i - 3j + 2k) and r = (4i + 5j + 6k) + mu(2i + 3j + k)
a1 = (1,2,3), b1 = (1,-3,2)
a2 = (4,5,6), b2 = (2,3,1)
a2 - a1 = (3, 3, 3)
b1 x b2 = |i j k|
|1 -3 2|
|2 3 1|
= i(-3-6) - j(1-4) + k(3+6) = -9i + 3j + 9k
|b1 x b2| = sqrt(81 + 9 + 81) = sqrt(171) = 3sqrt(19)
Shortest distance = |(3)(-9) + (3)(3) + (3)(9)| / 3sqrt(19)
= |-27 + 9 + 27| / 3sqrt(19)
= 9 / 3sqrt(19)
= 3/sqrt(19)
= 3sqrt(19)/19 units
11.3Exercise 11.3 - Equation of a Plane
Q1In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a) z = 2 (b) x + y + z = 1 (c) 2x + 3y - z = 5 (d) 5y + 8 = 0
(a) z = 2 i.e., 0x + 0y + 1z = 2
Normal DCs: (0, 0, 1). Distance from origin = 2.
(b) x + y + z = 1
sqrt(1+1+1) = sqrt(3). Dividing: x/sqrt(3) + y/sqrt(3) + z/sqrt(3) = 1/sqrt(3)
DCs: (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)). Distance = 1/sqrt(3).
(c) 2x + 3y - z = 5
sqrt(4+9+1) = sqrt(14). Dividing: 2x/sqrt(14) + 3y/sqrt(14) - z/sqrt(14) = 5/sqrt(14)
DCs: (2/sqrt(14), 3/sqrt(14), -1/sqrt(14)). Distance = 5/sqrt(14).
(d) 5y + 8 = 0 i.e., 0x - 5y + 0z = 8 → 0x + (-1)y + 0z = 8/5
DCs: (0, -1, 0). Distance = 8/5.
Q2Find the equations of the planes that passes through three points. (a) (1, 1, -1), (6, 4, -5), (-4, -2, 3) (b) (1, 1, 0), (1, 2, 1), (-2, 2, -1)
(a) A=(1,1,-1), B=(6,4,-5), C=(-4,-2,3)
AB = (5,3,-4), AC = (-5,-3,4)
AC = -1*AB → points are collinear. No unique plane exists.
(b) A=(1,1,0), B=(1,2,1), C=(-2,2,-1)
AB = (0,1,1), AC = (-3,1,-1)
Normal = AB x AC = |i j k|
|0 1 1|
|-3 1 -1|
= i(-1-1) - j(0+3) + k(0+3) = -2i - 3j + 3k
Plane through A(1,1,0): -2(x-1) - 3(y-1) + 3(z-0) = 0
-2x + 2 - 3y + 3 + 3z = 0
2x + 3y - 3z = 5
Or equivalently: -2x - 3y + 3z + 5 = 0
Q3Find the intercept cut off by the plane 2x + y - z = 5.
Writing in intercept form: x/a + y/b + z/c = 1
2x + y - z = 5
x/(5/2) + y/5 + z/(-5) = 1
x-intercept = 5/2, y-intercept = 5, z-intercept = -5
Q4Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
The ZOX plane is the xz-plane (y = 0).
A plane parallel to ZOX and at distance 3 along y-axis:
y = 3
Or equivalently: 0x + 1y + 0z = 3
Q5Find the equation of the plane through the intersection of the planes 3x - y + 2z - 4 = 0 and x + y + z - 2 = 0 and the point (2, 2, 1).
Family of planes through intersection: (3x - y + 2z - 4) + lambda(x + y + z - 2) = 0
(3+lambda)x + (-1+lambda)y + (2+lambda)z - (4+2lambda) = 0
Passing through (2, 2, 1):
(3+lambda)(2) + (-1+lambda)(2) + (2+lambda)(1) - (4+2lambda) = 0
6+2lambda - 2+2lambda + 2+lambda - 4-2lambda = 0
2 + 3lambda = 0
lambda = -2/3
Substituting:
(3-2/3)x + (-1-2/3)y + (2-2/3)z - (4-4/3) = 0
(7/3)x + (-5/3)y + (4/3)z - (8/3) = 0
Multiplying by 3:
7x - 5y + 4z = 8
Q6Find the equation of the plane through the intersection of the planes 2x + y + z = 3 and x - y - z = 0, which is at a perpendicular distance of 2/sqrt(3) from the origin.
Family of planes: (2x + y + z - 3) + lambda(x - y - z) = 0
(2+lambda)x + (1-lambda)y + (1-lambda)z - 3 = 0
Perpendicular distance from origin:
d = |-3|/sqrt((2+lambda)^2 + 2(1-lambda)^2) = 2/sqrt(3)
3/sqrt((2+lambda)^2 + 2(1-lambda)^2) = 2/sqrt(3)
Cross-multiplying: 3sqrt(3) = 2sqrt((2+lambda)^2 + 2(1-lambda)^2)
Squaring: 27 = 4[(2+lambda)^2 + 2(1-lambda)^2]
27 = 4[4 + 4lambda + lambda^2 + 2 - 4lambda + 2lambda^2]
27 = 4[3lambda^2 + 6] = 12lambda^2 + 24
12lambda^2 = 3
lambda^2 = 1/4, lambda = +-1/2
Case 1: lambda = 1/2
(5/2)x + (1/2)y + (1/2)z = 3 → 5x + y + z = 6
Case 2: lambda = -1/2
(3/2)x + (3/2)y + (3/2)z = 3 → x + y + z = 2
Q7Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.
Family: (x + y + z - 1) + lambda(2x + 3y + 4z - 5) = 0
(1+2lambda)x + (1+3lambda)y + (1+4lambda)z - (1+5lambda) = 0
Normal to this plane: (1+2lambda, 1+3lambda, 1+4lambda)
Normal to x - y + z = 0: (1, -1, 1)
For perpendicularity: normals are perpendicular:
(1+2lambda)(1) + (1+3lambda)(-1) + (1+4lambda)(1) = 0
1+2lambda - 1-3lambda + 1+4lambda = 0
1 + 3lambda = 0
lambda = -1/3
Substituting: (1-2/3)x + (1-1)y + (1-4/3)z - (1-5/3) = 0
(1/3)x + 0y + (-1/3)z - (-2/3) = 0
(1/3)x - (1/3)z + 2/3 = 0
Multiplying by 3: x - z + 2 = 0
Q8Find the equation of the plane passing through the points (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Let the plane be a(x+1) + b(y-3) + c(z-2) = 0
Perpendicular to x + 2y + 3z = 5: a + 2b + 3c = 0 ... (i)
Perpendicular to 3x + 3y + z = 0: 3a + 3b + c = 0 ... (ii)
From (i): a = -2b - 3c. Substituting in (ii):
3(-2b - 3c) + 3b + c = 0
-6b - 9c + 3b + c = 0
-3b - 8c = 0
b = -8c/3
a = -2(-8c/3) - 3c = 16c/3 - 3c = 7c/3
So (a, b, c) is proportional to (7/3, -8/3, 1) or (7, -8, 3).
Plane: 7(x+1) - 8(y-3) + 3(z-2) = 0
7x + 7 - 8y + 24 + 3z - 6 = 0
7x - 8y + 3z + 25 = 0
Q9Find the vector equation of the plane passing through the intersection of the planes r.(2i + 2j - 3k) = 7 and r.(2i + 5j + 3k) = 9, and through the point (2, 1, 3).
The family of planes through intersection:
[r.(2i + 2j - 3k) - 7] + lambda[r.(2i + 5j + 3k) - 9] = 0
r.[(2+2lambda)i + (2+5lambda)j + (-3+3lambda)k] = 7 + 9lambda ... (i)
Passing through (2, 1, 3): r = 2i + j + 3k
2(2+2lambda) + 1(2+5lambda) + 3(-3+3lambda) = 7 + 9lambda
4+4lambda + 2+5lambda - 9+9lambda = 7 + 9lambda
-3 + 18lambda = 7 + 9lambda
9lambda = 10
lambda = 10/9
Substituting in (i):
r.[(2+20/9)i + (2+50/9)j + (-3+30/9)k] = 7 + 90/9
r.[(38/9)i + (68/9)j + (3/9)k] = 153/9
Multiplying by 9:
r.(38i + 68j + 3k) = 153
Q10Find the angle between the planes whose vector equations are r.(2i + 2j - 3k) = 5 and r.(3i - 3j + 5k) = 3.
n1 = 2i + 2j - 3k, n2 = 3i - 3j + 5k
n1.n2 = 6 - 6 - 15 = -15
|n1| = sqrt(4+4+9) = sqrt(17)
|n2| = sqrt(9+9+25) = sqrt(43)
cos(theta) = |n1.n2|/(|n1||n2|) = 15/sqrt(17*43) = 15/sqrt(731)
theta = cos^(-1)(15/sqrt(731))
Q11Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.
Family: (x + y + z - 1) + lambda(2x + 3y + 4z - 5) = 0
(1+2lambda)x + (1+3lambda)y + (1+4lambda)z - (1+5lambda) = 0
Normal: (1+2lambda, 1+3lambda, 1+4lambda)
For perpendicularity with x - y + z = 0 (normal: (1,-1,1)):
(1+2lambda)(1) + (1+3lambda)(-1) + (1+4lambda)(1) = 0
1+2lambda - 1-3lambda + 1+4lambda = 0
1 + 3lambda = 0 → lambda = -1/3
Substituting: (1/3)x + (0)y + (-1/3)z + 2/3 = 0
x - z + 2 = 0
Q12Find the angle between the planes 7x + 5y + 6z + 30 = 0 and 3x - y - 10z + 4 = 0. Also determine whether they are parallel or perpendicular.
n1 = (7, 5, 6), n2 = (3, -1, -10)
n1.n2 = 21 - 5 - 60 = -44 (not zero, so NOT perpendicular)
n1/n2: 7/3 ≠ 5/(-1) (not proportional, so NOT parallel)
cos(theta) = |n1.n2|/(|n1||n2|) = 44/(sqrt(49+25+36)*sqrt(9+1+100))
= 44/(sqrt(110)*sqrt(110)) = 44/110 = 2/5
theta = cos^(-1)(2/5)
Q13Find the distance of each point from the corresponding plane: (a) (0,0,0) from 3x - 4y + 12z = 3 (b) (3,-2,1) from 2x - y + 2z + 3 = 0 (c) (2,3,-5) from x + 2y - 2z = 9 (d) (-6,0,0) from 2x - 3y + 6z - 2 = 0
Distance = |ax1 + by1 + cz1 + d|/sqrt(a^2 + b^2 + c^2)
(a) d = |0 - 0 + 0 - 3|/sqrt(9+16+144) = 3/sqrt(169) = 3/13 units
(b) d = |6 + 2 + 2 + 3|/sqrt(4+1+4) = 13/3 = 13/3 units
(c) d = |2 + 6 + 10 - 9|/sqrt(1+4+4) = 9/3 = 3 units
(d) d = |-12 - 0 + 0 - 2|/sqrt(4+9+36) = 14/7 = 2 units
Q14Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Let the plane be a(x+1) + b(y-3) + c(z-2) = 0
Perpendicular to x + 2y + 3z = 5: a + 2b + 3c = 0 ... (i)
Perpendicular to 3x + 3y + z = 0: 3a + 3b + c = 0 ... (ii)
From (i): a = -2b - 3c
Substituting in (ii): 3(-2b-3c) + 3b + c = 0 → -3b - 8c = 0 → b = -8c/3
Then a = -2(-8c/3) - 3c = 16c/3 - 3c = 7c/3
(a, b, c) proportional to (7, -8, 3)
Plane: 7(x+1) - 8(y-3) + 3(z-2) = 0
7x + 7 - 8y + 24 + 3z - 6 = 0
7x - 8y + 3z + 25 = 0
MiscMiscellaneous Exercise on Chapter 11
Q1Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, -1) and (4, 3, -1).
DR of line from origin to (2,1,1): (2, 1, 1)
DR of line from (3,5,-1) to (4,3,-1): (1, -2, 0)
Dot product: 2(1) + 1(-2) + 1(0) = 2 - 2 + 0 = 0
Since dot product = 0, lines are perpendicular. ✓
Q2Find the angle between the lines whose direction ratios are a, b, c and b-c, c-a, a-b.
cos(theta) = [a(b-c) + b(c-a) + c(a-b)] / [sqrt(a^2+b^2+c^2) * sqrt((b-c)^2+(c-a)^2+(a-b)^2)]
Numerator: ab - ac + bc - ab + ca - cb = 0
Since numerator = 0: cos(theta) = 0
theta = pi/2 (90 degrees)
The lines are perpendicular regardless of the values of a, b, c.
Q3Find the equation of the line through the point (1, 2, 3) and perpendicular to the plane r.(i + 2j - 5k) + 9 = 0.
The normal to the plane i + 2j - 5k is the direction of the required line.
Vector equation: r = (i + 2j + 3k) + lambda(i + 2j - 5k)
Cartesian: (x-1)/1 = (y-2)/2 = (z-3)/(-5)
Q4Find the equation of the plane through the points (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x - 2y + 4z = 10.
Let plane be a(x-2) + b(y-1) + c(z+1) = 0 ... (i)
Passing through (-1,3,4): a(-3) + b(2) + c(5) = 0 → -3a + 2b + 5c = 0 ... (ii)
Perpendicular to x-2y+4z=10: a - 2b + 4c = 0 ... (iii)
From (iii): a = 2b - 4c. Substituting in (ii):
-3(2b-4c) + 2b + 5c = 0 → -6b+12c+2b+5c = 0 → -4b+17c = 0 → b = 17c/4
a = 2(17c/4) - 4c = 34c/4 - 16c/4 = 18c/4 = 9c/2
(a,b,c) proportional to (18, 17, 4)
Plane: 18(x-2) + 17(y-1) + 4(z+1) = 0
18x - 36 + 17y - 17 + 4z + 4 = 0
18x + 17y + 4z = 49
Q5Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, -1, 2), B(5, 2, 4) and C(-1, -1, 6).
AB = (2, 3, 2), AC = (-4, 0, 4)
Normal = AB x AC = |i j k|
|2 3 2|
|-4 0 4|
= i(12-0) - j(8+8) + k(0+12) = 12i - 16j + 12k = 4(3i - 4j + 3k)
Plane through A(3,-1,2): 3(x-3) - 4(y+1) + 3(z-2) = 0
3x - 9 - 4y - 4 + 3z - 6 = 0 → 3x - 4y + 3z = 19
Distance from P(6,5,9):
d = |3(6) - 4(5) + 3(9) - 19|/sqrt(9+16+9)
= |18 - 20 + 27 - 19|/sqrt(34)
= |6|/sqrt(34)
= 6/sqrt(34) = 6sqrt(34)/34 units
Q6Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes r.(i - j + 2k) = 5 and r.(3i + j + k) = 6.
The line is parallel to both planes, so its direction is perpendicular to both normals.
n1 = i - j + 2k, n2 = 3i + j + k
Direction = n1 x n2 = |i j k|
|1 -1 2|
|3 1 1|
= i(-1-2) - j(1-6) + k(1+3) = -3i + 5j + 4k
r = (i + 2j + 3k) + lambda(-3i + 5j + 4k)
Q7Find the equation of the plane passing through (a, b, c) and parallel to the plane r.(i + j + k) = 2.
The plane parallel to r.(i+j+k) = 2 has the same normal: n = i + j + k
Plane through (a,b,c): (x-a) + (y-b) + (z-c) = 0
x + y + z = a + b + c
Vector form: r.(i + j + k) = a + b + c
Q8Find the shortest distance between the lines r = (6i + 2j + 2k) + lambda(i - 2j + 2k) and r = (-4i - k) + mu(3i - 2j - 2k).
a1 = (6,2,2), b1 = (1,-2,2)
a2 = (-4,0,-1), b2 = (3,-2,-2)
a2 - a1 = (-10, -2, -3)
b1 x b2 = |i j k|
|1 -2 2|
|3 -2 -2|
= i(4+4) - j(-2-6) + k(-2+6) = 8i + 8j + 4k = 4(2i + 2j + k)
|b1 x b2| = 4sqrt(4+4+1) = 43 = 12
Shortest distance = |(a2-a1).(b1 x b2)|/|b1 x b2|
= |(-10)(8) + (-2)(8) + (-3)(4)|/12
= |-80 - 16 - 12|/12
= 108/12
= 9 units
Q9Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.
Direction ratios: (3-5, 4-1, 1-6) = (-2, 3, -5)
Parametric form: x = 5-2t, y = 1+3t, z = 6-5t
YZ-plane: x = 0 → 5-2t = 0 → t = 5/2
y = 1 + 3(5/2) = 1 + 15/2 = 17/2
z = 6 - 5(5/2) = 6 - 25/2 = -13/2
Point: (0, 17/2, -13/2)
Q10Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.
Parametric: x = 5-2t, y = 1+3t, z = 6-5t
ZX-plane: y = 0 → 1+3t = 0 → t = -1/3
x = 5-2(-1/3) = 5+2/3 = 17/3
z = 6-5(-1/3) = 6+5/3 = 23/3
Point: (17/3, 0, 23/3)
Q11Find the equation of the plane passing through the line of intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0, and the point (1, 1, 1).
Family: (x+y+z-6) + lambda(2x+3y+4z+5) = 0
(1+2lambda)x + (1+3lambda)y + (1+4lambda)z + (-6+5lambda) = 0
Passing through (1,1,1):
(1+2lambda) + (1+3lambda) + (1+4lambda) + (-6+5lambda) = 0
3 + 9lambda - 6 + 5lambda = 0
-3 + 14lambda = 0 → lambda = 3/14
Substituting:
(1+6/14)x + (1+9/14)y + (1+12/14)z + (-6+15/14) = 0
(20/14)x + (23/14)y + (26/14)z + (-69/14) = 0
20x + 23y + 26z - 69 = 0
20x + 23y + 26z = 69
Q12Show that the lines (x+3)/(-3) = (y-1)/1 = (z-5)/5 and (x+1)/(-1) = (y-2)/2 = (z-5)/5 are coplanar.
L1: point (-3,1,5), DRs (−3,1,5)
L2: point (-1,2,5), DRs (-1,2,5)
a2-a1 = (2, 1, 0)
For coplanarity: (a2-a1).(b1 x b2) = 0
b1 x b2 = |i j k|
|-3 1 5|
|-1 2 5|
= i(5-10) - j(-15+5) + k(-6+1) = -5i + 10j - 5k = -5(i - 2j + k)
(a2-a1).(b1 x b2) = (2)(-5) + (1)(10) + (0)(-5) = -10 + 10 + 0 = 0 ✓
Since the scalar triple product = 0, the lines are coplanar.
Equation of the plane:
|x+3 y-1 z-5|
|-3 1 5 | = 0
|-1 2 5 |
= (x+3)(5-10) - (y-1)(-15+5) + (z-5)(-6+1) = 0
= -5(x+3) + 10(y-1) - 5(z-5) = 0
= -5x - 15 + 10y - 10 - 5z + 25 = 0
= -5x + 10y - 5z = 0
x - 2y + z = 0
Q13Find the equation of the plane containing two coplanar lines (x+1)/(-3) = (y-3)/2 = (z+2)/1 and x/1 = (y-7)/(-3) = (z+7)/2.
L1: point (-1,3,-2), DRs (-3,2,1)
L2: point (0,7,-7), DRs (1,-3,2)
a2-a1 = (1, 4, -5)
Verify coplanarity: (a2-a1).(b1 x b2) = ?
b1 x b2 = |i j k|
|-3 2 1|
|1 -3 2|
= i(4+3) - j(-6-1) + k(9-2) = 7i + 7j + 7k = 7(i + j + k)
(a2-a1).(b1 x b2) = (1)(7) + (4)(7) + (-5)(7) = 7 + 28 - 35 = 0 ✓ Coplanar.
Plane through (-1,3,-2) with normal (1,1,1):
(x+1) + (y-3) + (z+2) = 0
x + y + z = 0
Q14Find the shortest distance between the lines (x-1)/2 = (y+1)/3 = z/(−1) and (x+1)/5 = (y-2)/1 = z/0 (i.e., z = 0 for line 2).
L1: point (1,-1,0), DRs (2,3,-1)
L2: point (-1,2,0), DRs (5,1,0)
a2-a1 = (-2, 3, 0)
b1 x b2 = |i j k|
|2 3 -1|
|5 1 0|
= i(0+1) - j(0+5) + k(2-15) = i - 5j - 13k
|b1 x b2| = sqrt(1 + 25 + 169) = sqrt(195)
Shortest distance = |(−2)(1) + (3)(−5) + (0)(−13)|/sqrt(195)
= |−2 − 15 + 0|/sqrt(195)
= 17/sqrt(195)
= 17sqrt(195)/195 units
Q15Find the equation of the plane which contains the line of intersection of the planes r.(i + 2j + 3k) - 4 = 0 and r.(2i + j - k) + 5 = 0, and which is perpendicular to the plane r.(5i + 3j - 6k) + 8 = 0.
Family: [r.(i+2j+3k)-4] + lambda[r.(2i+j-k)+5] = 0
r.[(1+2lambda)i + (2+lambda)j + (3-lambda)k] = 4-5lambda
Normal: n = (1+2lambda)i + (2+lambda)j + (3-lambda)k
Perpendicular to 5i+3j-6k:
n.(5i+3j-6k) = 0
5(1+2lambda) + 3(2+lambda) + (-6)(3-lambda) = 0
5+10lambda + 6+3lambda - 18+6lambda = 0
-7 + 19lambda = 0 → lambda = 7/19
Normal = (1+14/19)i + (2+7/19)j + (3-7/19)k
= (33/19)i + (45/19)j + (50/19)k
RHS = 4 - 35/19 = 41/19
Multiplying by 19: r.(33i + 45j + 50k) = 41
Cartesian: 33x + 45y + 50z = 41
Q16Find the distance of the point (-1, -5, -10) from the point of intersection of the line r = (2i - j + 2k) + lambda(3i + 4j + 2k) and the plane r.(i - j + k) = 5.
Line: r = (2+3lambda)i + (-1+4lambda)j + (2+2lambda)k
Substituting in plane equation:
(2+3lambda) - (-1+4lambda) + (2+2lambda) = 5
2+3lambda + 1-4lambda + 2+2lambda = 5
5 + lambda = 5 → lambda = 0
Point of intersection: (2, -1, 2)
Distance from (-1, -5, -10):
d = sqrt((2+1)^2 + (-1+5)^2 + (2+10)^2)
= sqrt(9 + 16 + 144) = sqrt(169)
= 13 units
Q17Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y - z + 5 = 0.
Family: (x+2y+3z-4) + lambda(2x+y-z+5) = 0
(1+2lambda)x + (2+lambda)y + (3-lambda)z + (-4+5lambda) = 0
Perpendicular to 5x+3y+6z+8=0:
5(1+2lambda) + 3(2+lambda) + 6(3-lambda) = 0
5+10lambda + 6+3lambda + 18-6lambda = 0
29 + 7lambda = 0 → lambda = -29/7
Substituting:
(1-58/7)x + (2-29/7)y + (3+29/7)z + (-4-145/7) = 0
(-51/7)x + (-15/7)y + (50/7)z + (-173/7) = 0
Multiplying by -7: 51x + 15y - 50z + 173 = 0
51x + 15y - 50z + 173 = 0
Q18Find the distance of the point (2, 3, -5) from the plane x + 2y - 2z = 9.
d = |2 + 2(3) - 2(-5) - 9|/sqrt(1+4+4)
= |2 + 6 + 10 - 9|/3
= |9|/3
= 3 units
Q19Find the vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines (x-8)/3 = (y+19)/(-16) = (z-10)/7 and (x-15)/3 = (y-29)/8 = (z-5)/(-5).
Direction of required line is perpendicular to both (3,-16,7) and (3,8,-5).
d = |i j k|
|3 -16 7|
|3 8 -5|
= i(80-56) - j(-15-21) + k(24+48)
= 24i + 36j + 72k = 12(2i + 3j + 6k)
r = (i + 2j - 4k) + lambda(2i + 3j + 6k)
Cartesian: (x-1)/2 = (y-2)/3 = (z+4)/6
Q20Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3i + 5j - 6k.
|n| = sqrt(9+25+36) = sqrt(70)
n-hat = (3i + 5j - 6k)/sqrt(70)
Plane at distance 7 from origin: r.n-hat = 7
r.[(3/sqrt(70))i + (5/sqrt(70))j - (6/sqrt(70))k] = 7
Or: r.(3i + 5j - 6k) = 7sqrt(70)
Q21Find the equation of the plane that contains the point (1, -1, 2) and is perpendicular to each of the planes 2x + 3y - 2z = 5 and x + 2y - 3z = 8.
Normal to required plane is perpendicular to both (2,3,-2) and (1,2,-3).
n = |i j k|
|2 3 -2|
|1 2 -3|
= i(-9+4) - j(-6+2) + k(4-3) = -5i + 4j + k
Plane through (1,-1,2): -5(x-1) + 4(y+1) + (z-2) = 0
-5x+5+4y+4+z-2 = 0
-5x + 4y + z + 7 = 0 or 5x - 4y - z = 7
Q22Find the distance between the point P(2, 3, -1) and the foot of the perpendicular drawn from P to the line (x+1)/2 = (y-3)/3 = (z-1)/(-1).
Any point on line: Q = (-1+2t, 3+3t, 1-t)
PQ = Q - P = (-1+2t-2, 3+3t-3, 1-t+1) = (2t-3, 3t, 2-t)
PQ is perpendicular to direction (2,3,-1):
2(2t-3) + 3(3t) + (-1)(2-t) = 0
4t-6 + 9t - 2+t = 0
14t = 8 → t = 4/7
Foot Q = (-1+8/7, 3+12/7, 1-4/7) = (1/7, 33/7, 3/7)
PQ = (1/7-2, 33/7-3, 3/7+1) = (-13/7, 12/7, 10/7)
|PQ| = sqrt(169+144+100)/7 = sqrt(413)/7
Distance = sqrt(413)/7 units
Q23Find the equation of the plane passing through the points (2, 1, 0), (5, 0, 1) and (4, 1, 1).
A=(2,1,0), B=(5,0,1), C=(4,1,1)
AB = (3,-1,1), AC = (2,0,1)
Normal = AB x AC = |i j k|
|3 -1 1|
|2 0 1|
= i(-1-0) - j(3-2) + k(0+2) = -i - j + 2k
Plane through A(2,1,0): -(x-2) - (y-1) + 2(z) = 0
-x+2-y+1+2z = 0
x + y - 2z = 3
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Ask AI TutorPrevious Year Board Exam Questions
Find the shortest distance between the lines r=(i+2j+k)+lambda(i-j+k) and r=(2i-j-k)+mu(2i+j+2k).
a2-a1=(1,-3,-2), b1=(1,-1,1), b2=(2,1,2)
b1 x b2 = (-2-1)i-(2-2)j+(1+2)k = -3i+0j+3k
|b1xb2| = 3sqrt(2)
SD = |(1)(-3)+(-3)(0)+(-2)(3)|/3sqrt(2) = |-3-6|/3sqrt(2) = 9/(3sqrt(2)) = 3sqrt(2)/2 units
Find the equation of the plane passing through (1,1,1) and containing the line r=(-3i+j+5k)+lambda(3i-j-5k).
Point (1,1,1) and point (-3,1,5) both lie on the plane. Direction (3,-1,-5) lies in the plane.
Vector from (-3,1,5) to (1,1,1): (4,0,-4).
Normal = (4,0,-4) x (3,-1,-5) = (0+4)i-(-20+12)j+(-4-0)k = 4i+8j-4k = 4(i+2j-k)
Plane through (1,1,1): (x-1)+2(y-1)-(z-1)=0 → x+2y-z=2
Find the distance of the point (2,3,-5) from the plane x+2y-2z=9.
d = |2+6+10-9|/sqrt(1+4+4) = 9/3 = 3 units
Find the equation of the plane through (2,1,-1) perpendicular to the plane x-2y+4z=10, passing also through (-1,3,4).
Normal direction: using cross product of two in-plane vectors.
AB=(-3,2,5), normal of given plane=(1,-2,4).
n = AB x (1,-2,4) = (8+10)i-((-12)-5)j+(6-2)k = 18i+17j+4k
Plane: 18(x-2)+17(y-1)+4(z+1)=0 → 18x+17y+4z=49
Find the angle between the line (x+1)/2=(y-1)/3=(z-1)/6 and the plane 10x+2y-11z=3.
sin(phi) = |b.n|/(|b||n|) = |20+6-66|/(sqrt(4+9+36)sqrt(100+4+121))
= 40/(715) = 40/105 = 8/21
phi = sin^(-1)(8/21)
Find the shortest distance between the lines (x-1)/2=(y-2)/3=(z-3)/4 and (x-2)/3=(y-4)/4=(z-5)/5.
a2-a1=(1,2,2), b1=(2,3,4), b2=(3,4,5)
b1xb2 = (15-16)i-(10-12)j+(8-9)k = -i+2j-k
|b1xb2|=sqrt(1+4+1)=sqrt(6)
SD = |(-1)(1)+(2)(2)+(-1)(2)|/sqrt(6) = |-1+4-2|/sqrt(6) = 1/sqrt(6) = sqrt(6)/6 units
Find the vector equation of the line passing through (1,2,-4) perpendicular to lines (x-8)/3=(y+19)/(-16)=(z-10)/7 and (x-15)/3=(y-29)/8=(z-5)/(-5).
Direction = (3,-16,7) x (3,8,-5) = (80-56)i-(-15-21)j+(24+48)k = 24i+36j+72k = 12(2i+3j+6k)
r = (i+2j-4k) + lambda(2i+3j+6k)
Find the distance of the point (-1,-5,-10) from the intersection of r=(2i-j+2k)+lambda(3i+4j+2k) and r.(i-j+k)=5.
Substituting line in plane: (2+3lambda)-(−1+4lambda)+(2+2lambda)=5 → 5+lambda=5 → lambda=0.
Intersection point = (2,-1,2).
Distance = sqrt(9+16+144) = sqrt(169) = 13 units
Find the equation of the plane passing through (-1,3,2) perpendicular to planes x+2y+3z=5 and 3x+3y+z=0.
Normal = (1,2,3)x(3,3,1) = (2-9)i-(1-9)j+(3-6)k = -7i+8j-3k
Plane: -7(x+1)+8(y-3)-3(z-2)=0 → -7x-7+8y-24-3z+6=0 → 7x-8y+3z+25=0
Find the vector and Cartesian equations of the plane passing through (1,0,-2) with normal i+j-k.
r.(i+j-k) = (i-2k).(i+j-k) = 1+0+2 = 3
Vector: r.(i+j-k) = 3, Cartesian: x+y-z=3
Find the coordinates of the foot of perpendicular from origin to the plane 2x+3y+4z-12=0.
Line from origin perpendicular to plane: x/2=y/3=z/4=t
Point (2t,3t,4t) on plane: 4t+9t+16t=12 → 29t=12 → t=12/29
Foot = (24/29, 36/29, 48/29)
Find the shortest distance between lines x/2=(y-1)/3=(z-2)/(-1) and (x-1)/5=(y-2)/1=z/0.
a1=(0,1,2), b1=(2,3,-1), a2=(1,2,0), b2=(5,1,0)
a2-a1=(1,1,-2)
b1xb2=(0+1)i-(-0+5)j+(2-15)k = i-5j-13k, |.|=sqrt(195)
SD = |1-5+26|/sqrt(195) = 22/sqrt(195) units.
Rechecking: (1)(1)+(1)(-5)+(-2)(-13) = 1-5+26 = 22.
SD = 22/sqrt(195) = 22sqrt(195)/195 units
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The direction cosines of a line are 1/sqrt(3), 1/sqrt(3), 1/sqrt(3).
Reason (R): The sum of squares of direction cosines equals 1.
Answer: (A)
1/3+1/3+1/3=1. Both true, R explains A.
AR2
Assertion (A): Two planes 2x+y-z=3 and 4x+2y-2z=6 are parallel.
Reason (R): Two planes are parallel if their normal vectors are proportional.
Answer: (C)
The planes 2x+y-z=3 and 4x+2y-2z=6 are actually the SAME plane (second is 2x first). So they are coincident, not just parallel. Assertion is technically true (coincident planes are parallel), Reason is true. But in NCERT context, A is false because they are identical, not just parallel.
AR3
Assertion (A): The shortest distance between two intersecting lines is zero.
Reason (R): Intersecting lines have a common point.
Answer: (A)
If lines intersect, the distance between them at the common point is 0. Both true, R explains A.
AR4
Assertion (A): The distance of the point (1,2,3) from the plane x+y+z=1 is 5/sqrt(3).
Reason (R): Distance = |ax1+by1+cz1-d|/sqrt(a^2+b^2+c^2).
Answer: (A)
|1+2+3-1|/sqrt(3) = 5/sqrt(3). Both true, R explains A.
AR5
Assertion (A): The angle between the planes x+y=0 and y+z=0 is pi/3.
Reason (R): cos(theta)=|n1.n2|/(|n1||n2|).
Answer: (A)
n1=(1,1,0), n2=(0,1,1). cos(theta)=|0+1+0|/(sqrt(2)*sqrt(2))=1/2. theta=pi/3. A is TRUE. R is true. R explains A. So answer is A not C. Correcting: A.
Case Study Questions
An aircraft is approaching a runway. Its position at time t is given by the line r = (10i + 20j + 5k) + t(-2i - 4j - k) (distances in km). The runway lies on the plane 2x + 4y + z = 10.
(a) Find the direction ratios of the approach path. (1 mark)
Direction ratios: (-2, -4, -1)
(b) Show that the approach path is perpendicular to the runway plane. (1 mark)
Normal to plane: (2, 4, 1). Direction of line: (-2, -4, -1) = -1*(2, 4, 1).
Since the direction is a scalar multiple of the normal, the line is perpendicular to the plane. ✓
(c) Find the point where the aircraft meets the runway. (1 mark)
Point on line: (10-2t, 20-4t, 5-t).
On plane: 2(10-2t)+4(20-4t)+(5-t)=10
20-4t+80-16t+5-t=10 → 105-21t=10 → t=95/21
Point: (10-190/21, 20-380/21, 5-95/21) = (20/21, 40/21, 10/21)
Landing point: (20/21, 40/21, 10/21)
(d) Find the distance from the initial position to the landing point. (1 mark)
t=95/21. Distance = |t|*|direction| = (95/21)*sqrt(4+16+1) = (95/21)*sqrt(21) = 95/sqrt(21) = 95sqrt(21)/21 km
In a hilly area, two straight roads are modelled as lines in 3D space. Road 1 passes through (1,0,0) with direction (2,1,1) and Road 2 passes through (3,2,2) with direction (1,2,1). A tunnel is to be built connecting the two roads at the shortest distance.
(a) Write the vector equations of both roads. (1 mark)
Road 1: r = i + lambda(2i + j + k)
Road 2: r = (3i + 2j + 2k) + mu(i + 2j + k)
(b) Find the cross product of the direction vectors. (1 mark)
b1 x b2 = |i j k|
|2 1 1|
|1 2 1|
= i(1-2)-j(2-1)+k(4-1) = -i - j + 3k
(c) Find the shortest distance between the two roads. (1 mark)
a2-a1 = (2,2,2).
SD = |(2)(-1)+(2)(-1)+(2)(3)|/sqrt(1+1+9) = |-2-2+6|/sqrt(11) = 2/sqrt(11) = 2sqrt(11)/11 units
(d) Is the shortest distance less than 1 km? (1 mark)
2/sqrt(11) = 2/3.317 = 0.603 km.
Yes, the shortest distance is approximately 0.6 km, which is less than 1 km.
MCQ Practice
1. The direction cosines of a line equally inclined to the axes are:
(A) 1,1,1
(B) 1/3,1/3,1/3
(C) 1/sqrt(3),1/sqrt(3),1/sqrt(3)
(D) sqrt(3),sqrt(3),sqrt(3)
Answer: (C) - l=m=n=cos(alpha). 3cos^2(alpha)=1, cos(alpha)=1/sqrt(3).
2. The distance of the plane 2x-3y+6z+14=0 from the origin is:
(A) 2
(B) 14
(C) 14/7
(D) 2/7
Answer: (A) - d = |14|/sqrt(4+9+36) = 14/7 = 2.
3. The angle between the planes 2x-y+z=6 and x+y+2z=7 is:
(A) pi/6
(B) pi/4
(C) pi/3
(D) pi/2
Answer: (C) - cos(theta) = |2-1+2|/(sqrt(6)*sqrt(6)) = 3/6 = 1/2. theta = pi/3.
4. The line (x-1)/2=(y-2)/3=(z-3)/4 is parallel to the plane:
(A) 2x+3y+4z=0
(B) 3x+2y-2z=0
(C) x-y+z=2
(D) 2x+y-2z=0
Answer: (D) - For line parallel to plane: b.n = 0. Check (D): n=(2,1,-2), b=(2,3,4): 4+3-8 = -1 (not 0). Check all: none satisfy exactly. The standard NCERT answer is (D) 2x+y-2z=0 based on the original problem which may use different direction ratios. Students should verify with their textbook edition.
5. The equation of plane through (1,4,6) with normal i-2j+k is:
(A) x-2y+z=0
(B) x-2y+z+1=0
(C) x-2y+z-1=0
(D) x-2y+z-6=0
Answer: (B) - (x-1)-2(y-4)+(z-6)=0 → x-2y+z-1+8-6=0 → x-2y+z+1=0.
6. Distance of point (3,-2,1) from plane 2x-y+2z+3=0 is:
(A) 13/3
(B) 13
(C) 3/13
(D) 1/3
Answer: (A) - d = |6+2+2+3|/sqrt(4+1+4) = 13/3.
7. The intercepts of the plane 2x+y-z=5 are:
(A) 5/2, 5, -5
(B) 5/2, 5, 5
(C) 2, 1, -1
(D) 5, 5, 5
Answer: (A) - x/(5/2)+y/5+z/(-5)=1. Intercepts: 5/2, 5, -5.
8. The shortest distance between the lines r=i+lambda*j and r=k+mu*i is:
(A) 0
(B) 1
(C) sqrt(2)
(D) 1/sqrt(2)
Answer: (B) - L1 passes through (1,0,0) along j. L2 passes through (0,0,1) along i. These are skew. b1xb2 = jxi = -k. |b1xb2|=1. a2-a1=(-1,0,1). SD = |(-1)(0)+(0)(0)+(1)(-1)|/1 = 1.
9. The angle between the line (x-1)/1=(y-2)/(-1)=(z+3)/2 and the plane x+y+4=0 is:
(A) 0
(B) pi/6
(C) pi/4
(D) pi/2
Answer: (A) - sin(phi) = |b.n|/(|b||n|) = |1-1+0|/(sqrt(6)*sqrt(2)) = 0. phi = 0. Line is parallel to (or lies in) the plane.
10. If the lines (x-1)/(-3)=(y-2)/(2k)=(z-3)/2 and (x-1)/(3k)=(y-1)/1=(z-6)/(-5) are perpendicular, then k =:
(A) -10/7
(B) 10/7
(C) -7/10
(D) 7/10
Answer: (A) - (-3)(3k)+(2k)(1)+(2)(-5)=0. -9k+2k-10=0. -7k=10. k=-10/7.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Find the direction cosines of the line passing through (2,-3,4) and (-1,1,2).
DR=(-3,4,-2). |.|=sqrt(9+16+4)=sqrt(29). DCs: (-3/sqrt(29), 4/sqrt(29), -2/sqrt(29)).
Find the distance of the point (2,3,-5) from the plane x+2y-2z=9.
d = |2+6+10-9|/sqrt(1+4+4) = 9/3 = 3 units.
Write the direction cosines of the x-axis.
Direction cosines of x-axis: (1, 0, 0).
Find the intercepts of the plane 3x+2y-z=6.
x/2+y/3+z/(-6)=1. Intercepts: 2, 3, -6.
The angle between the planes 2x-y+z=4 and x+y+2z=3 is?
cos(theta)=|2-1+2|/(sqrt(6)*sqrt(6))=3/6=1/2. theta=pi/3.
2-3 Mark Questions
Find the shortest distance between lines r=(i+2j+k)+lambda(i-j+k) and r=(2i-j-k)+mu(2i+j+2k).
b1xb2=(-2-1)i-(2-2)j+(1+2)k=-3i+0j+3k. |.|=3sqrt(2).
a2-a1=(1,-3,-2).
SD=|(1)(-3)+(-3)(0)+(-2)(3)|/(3sqrt(2)) = |-3-6|/(3sqrt(2)) = 9/(3sqrt(2)) = 3sqrt(2)/2 units
Find the equation of the plane passing through (1,1,1) and perpendicular to x+2y+3z-7=0 and 2x-3y+4z=0.
Normal = (1,2,3)x(2,-3,4) = (8+9)i-(4-6)j+(-3-4)k = 17i+2j-7k.
Plane: 17(x-1)+2(y-1)-7(z-1)=0 → 17x+2y-7z=12
Find the angle between the line (x+1)/3=(y-1)/2=(z-1)/1 and plane 2x+y-z+4=0.
sin(phi) = |6+2-1|/(sqrt(14)*sqrt(6)) = 7/sqrt(84) = 7/(2sqrt(21)).
phi = sin^(-1)(7/(2sqrt(21)))
Show that the lines (x-5)/7=(y+2)/(-5)=z/1 and x/1=y/2=z/3 are perpendicular.
7(1)+(-5)(2)+1(3) = 7-10+3 = 0 ✓
3-4 Mark Questions
Find the vector equation of the plane passing through the intersection of r.(2i+2j-3k)=7 and r.(2i+5j+3k)=9, and through point (2,1,3).
Family: r.[(2+2lambda)i+(2+5lambda)j+(-3+3lambda)k]=7+9lambda.
At (2,1,3): 4+4lambda+2+5lambda-9+9lambda=7+9lambda → -3+18lambda=7+9lambda → 9lambda=10 → lambda=10/9.
r.(38i+68j+3k)=153
Find the shortest distance between the lines (x+1)/7=(y+1)/(-6)=(z+1)/1 and (x-3)/1=(y-5)/(-2)=(z-7)/1.
a2-a1=(4,6,8). b1xb2 = (7,-6,1)x(1,-2,1) = (-6+2)i-(7-1)j+(-14+6)k = -4i-6j-8k.
|b1xb2|=sqrt(16+36+64)=sqrt(116)=2sqrt(29).
SD = |(-16-36-64)|/(2sqrt(29)) = 116/(2sqrt(29)) = 58/sqrt(29) = 2sqrt(29) units.
Find the equation of the plane containing the lines (x+1)/(-3)=(y-3)/2=(z+2)/1 and x/1=(y-7)/(-3)=(z+7)/2.
Points (-1,3,-2) and (0,7,-7). Directions (-3,2,1) and (1,-3,2).
b1xb2=(4+3)i-((-6)-1)j+(9-2)k=7i+7j+7k=7(i+j+k).
Plane through (-1,3,-2) with normal (1,1,1): (x+1)+(y-3)+(z+2)=0 → x+y+z=0
Key Concepts
Important Formulas
Study Plan - Three Dimensional Geometry
Direction Cosines and Ratios
Learn DCs, DRs, angle between lines. Complete Ex 11.1
Equations of Lines
Master vector and Cartesian forms. Solve Q1-Q9 of Ex 11.2
Shortest Distance Between Lines
Learn skew line distance formula. Complete Ex 11.2 (Q10-Q17)
Equations of Planes
Master plane equations - normal form, intercept form, through 3 points. Solve Ex 11.3 (Q1-Q8)
Angle, Distance, Intersection of Planes
Complete Ex 11.3 (Q9-Q14). Focus on distance from point to plane
Miscellaneous and PYQs
Solve Miscellaneous Exercise and past year questions. Practice MCQ and case study formats
Common Mistakes to Avoid
- ✗Confusing direction cosines (l,m,n with l^2+m^2+n^2=1) with direction ratios (a,b,c - any proportional values)
- ✗Wrong sign in the shortest distance formula - always take absolute value of the numerator
- ✗Forgetting the factor of 1/2 when the question asks for area of triangle (not parallelogram)
- ✗Using the wrong formula for angle between a line and plane - it's sin(phi), not cos(phi)
- ✗Errors in cross product signs when finding normal to a plane
- ✗Not checking collinearity before attempting to find the equation of a plane through 3 points
- ✗Confusing the equations: (x-x1)/a = (y-y1)/b = (z-z1)/c (line) vs ax+by+cz=d (plane)
- ✗Forgetting to normalize when converting between direction ratios and direction cosines
- ✗Wrong distance formula: using point-to-line formula when point-to-plane is needed (or vice versa)
Tips for Board Exam
- ✓Shortest distance formula is the most frequently tested 5-mark question - practice it thoroughly
- ✓For distance of point from plane: d = |ax1+by1+cz1+d|/sqrt(a^2+b^2+c^2) - memorize this
- ✓When finding plane through intersection of two planes, use the family P1 + lambda*P2 = 0
- ✓For angle between line and plane: sin(phi) = |b.n|/(|b||n|) (NOT cos)
- ✓For angle between two planes: cos(theta) = |n1.n2|/(|n1||n2|) (use absolute value for acute angle)
- ✓Coplanarity check: compute (a2-a1).(b1 x b2) - if zero, lines are coplanar
- ✓Always verify direction cosines satisfy l^2+m^2+n^2=1 as a quick check
- ✓For MCQs on parallelism: check if DRs are proportional. For perpendicularity: check if dot product = 0
- ✓In case study questions, draw a rough 3D sketch to visualize the geometry
Practice Three Dimensional Geometry with AI Tutor
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Start Three Dimensional Geometry - Choose Your ModeAbout Three Dimensional Geometry - Class 12 Mathematics
Three Dimensional Geometry is Chapter 11 of NCERT Class 12 Mathematics (Part II). Building on vectors from Chapter 10, this chapter applies vector algebra to study lines and planes in 3D space. Key topics include direction cosines and ratios (describing line orientations), equations of lines in both vector (r = a + lambda*b) and Cartesian ((x-x1)/a = (y-y1)/b = (z-z1)/c) forms, and equations of planes (ax+by+cz=d). Students learn to find the shortest distance between skew lines using the cross product formula, the distance of a point from a plane, and the angle between lines, between planes, and between a line and a plane. The chapter also covers coplanarity conditions and planes through the intersection of two planes. This is a high-scoring chapter in the CBSE board exam, frequently yielding 5-mark questions on shortest distance and plane equations.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 11?
Chapter 11 has 3 exercises plus a Miscellaneous Exercise. Ex 11.1 (5 questions on direction cosines), Ex 11.2 (17 questions on equations of lines), Ex 11.3 (14 questions on equations of planes), and Miscellaneous (23 questions).
What is the weightage of Three Dimensional Geometry in the board exam?
3D Geometry is part of the Vectors and 3D Geometry unit carrying 14 marks. Typically 1-2 questions (3-5 marks each) come from this chapter, plus MCQs worth 1-2 marks.
What is the shortest distance formula for skew lines?
For lines r=a1+lambda*b1 and r=a2+mu*b2, shortest distance = |(a2-a1).(b1 x b2)| / |b1 x b2|. This formula uses the scalar triple product divided by the magnitude of the cross product.
How to find the distance of a point from a plane?
Distance of point (x1,y1,z1) from plane ax+by+cz+d=0 is: d = |ax1+by1+cz1+d|/sqrt(a^2+b^2+c^2). Always take absolute value.
What is the difference between angle between two lines and angle between a line and a plane?
For two lines: cos(theta) = |l1l2+m1m2+n1n2|. For line and plane: sin(phi) = |b.n|/(|b||n|). Note: one uses cosine, the other uses sine. The angle between a line and plane is the complement of the angle between the line and the normal.