Why is three-dimensional geometry important?

In our daily lives, whatever objects we see are mostly three-dimensional. For example, chairs, tables, books, stationery objects like pens and pencils, etc. So, when we study mathematics as a discipline and learn measurement of common objects in mensuration, trigonometry, and geometry, we should also learn measurement related to three-dimensional geometry. Again, three-dimensional geometry is a crucial part of analytical geometry. This chapter is also important from the perspective of the board exam. Therefore, students must prepare the chapter very well for a better performance in exams.

Why should students practise Chapter 11 Class 12 Mathematics Important Questions?

Students must learn to solve questions quickly. The more questions they solve, the more skilled they become at calculation. By solving the important questions, students will get a fair idea of the question pattern, the types of questions that are important from an exam perspective, and thus they will feel more confident. So, students should practise more questions before the exam.

How will Important Questions provided by Extramarks benefit students?

Mathematics Class 12 Chapter 11 Important Questions provided by Extramarks are prepared after consulting the past years’ Important Questions and the CBSE Syllabus. The question set is divided into three sections: questions carrying 1 mark, 4 marks, and 6 marks. The solutions are also provided along with the questions. So, students can focus on solving questions instead of collecting questions from every possible source.

Home > CBSE > CBSE Important Questions Class 12 Maths Chapter 11 – Three Dimensional Geometry

CBSE Important Questions Class 12 Maths Chapter 11 – Three Dimensional Geometry

Class 12 Maths Chapter 11 Three Dimensional Geometry is part of the CBSE Class 12 Maths syllabus. It extends coordinate geometry into space using vectors. Students learn direction cosines and direction ratios of a line, the vector and cartesian equations of a line and a plane, the angle between two lines, between two planes and between a line and a plane, and the distance of a point from a plane.

These CBSE important questions for Class 12 Maths Chapter 11 cover the equation of a line through two points, the angle between two lines, the equation of a plane in normal form and through three points, the distance of a point from a plane, the shortest distance between two skew lines, and the condition for a line to lie in a plane. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Write the direction cosines of the x-axis, y-axis and z-axis. What relation do direction cosines satisfy?

Answer: • x-axis: 1, 0, 0
• y-axis: 0, 1, 0
• z-axis: 0, 0, 1
The direction cosines l, m, n of any line always satisfy
l² + m² + n² = 1
If a, b, c are the direction ratios, then l = a/√(a²+b²+c²), and similarly for m and n.

Q2. Find the vector and cartesian equations of the line passing through the points (1, 2, 3) and (4, 5, 6).

Answer: The direction ratios are obtained by subtracting the coordinates:
(4 − 1, 5 − 2, 6 − 3) = (3, 3, 3), which simplifies to (1, 1, 1).
Vector equation: r⃗ = a⃗ + λb⃗
r⃗ = (î + 2ĵ + 3k̂) + λ(î + ĵ + k̂)
Cartesian equation: (x − x₁)/a = (y − y₁)/b = (z − z₁)/c
(x − 1)/1 = (y − 2)/1 = (z − 3)/1

Q3. Find the angle between the lines whose direction ratios are (1, 1, 2) and (−3, −4, 1).

Answer: The angle between two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) is
cosθ = |a₁a₂ + b₁b₂ + c₁c₂| / [√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²)]
Numerator: |(1)(−3) + (1)(−4) + (2)(1)| = |−3 − 4 + 2| = 5
Denominator: √(1 + 1 + 4) × √(9 + 16 + 1) = √6 × √26 = √156
cosθ = 5/√156
θ = cos⁻¹(5/√156)

Q4. Write the different forms of the equation of a plane.

Answer: • Normal form: r⃗ · n̂ = d, or in cartesian form lx + my + nz = d, where l, m, n are the direction cosines of the normal and d is the distance from the origin.
• General form: Ax + By + Cz + D = 0, where (A, B, C) are the direction ratios of the normal.
• Through a point with a given normal: A(x − x₁) + B(y − y₁) + C(z − z₁) = 0
• Intercept form: x/a + y/b + z/c = 1, where a, b, c are the intercepts on the axes.

Q5. Find the equation of the plane passing through the point (2, −1, 3) with normal vector 8î − 7ĵ − 5k̂.

Answer: The equation of a plane through (x₁, y₁, z₁) with normal direction ratios (A, B, C) is
A(x − x₁) + B(y − y₁) + C(z − z₁) = 0
Substituting the given values:
8(x − 2) − 7(y + 1) − 5(z − 3) = 0
8x − 16 − 7y − 7 − 5z + 15 = 0
8x − 7y − 5z − 8 = 0

Q6. Find the distance of the point (0, 0, 0) from the plane 8x − 7y − 5z − 8 = 0.

Answer: The distance of a point (x₁, y₁, z₁) from the plane Ax + By + Cz + D = 0 is
d = |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²)
Substituting the origin:
d = |0 − 0 − 0 − 8| / √(64 + 49 + 25)
= 8 / √138
= 8/√138 units, which is about 0.68 units.

Q7. Find the equation of the plane passing through the three points (2, 5, −3), (−2, −3, 5) and (5, 3, −3).

Answer: First find two vectors in the plane:
a⃗₁ = (−2 − 2, −3 − 5, 5 + 3) = (−4, −8, 8)
a⃗₂ = (5 − 2, 3 − 5, −3 + 3) = (3, −2, 0)
Normal vector = a⃗₁ × a⃗₂:
î: (−8)(0) − (8)(−2) = 16
ĵ: −[(−4)(0) − (8)(3)] = 24
k̂: (−4)(−2) − (−8)(3) = 32
Normal = (16, 24, 32), which simplifies to (2, 3, 4).
Using the point (2, 5, −3):
2(x − 2) + 3(y − 5) + 4(z + 3) = 0
2x + 3y + 4z − 7 = 0

Q8. Write the condition for two planes to be (i) parallel and (ii) perpendicular.

Answer: Let the planes be A₁x + B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂z + D₂ = 0.
(i) Parallel: Their normals are parallel, so the direction ratios are proportional:
A₁/A₂ = B₁/B₂ = C₁/C₂
(ii) Perpendicular: Their normals are perpendicular, so the dot product is zero:
A₁A₂ + B₁B₂ + C₁C₂ = 0
The angle between two planes is defined as the angle between their normals.

Q9. Find the shortest distance between the lines r⃗ = (î + 2ĵ + k̂) + λ(î − ĵ + k̂) and r⃗ = (2î − ĵ − k̂) + μ(2î + ĵ + 2k̂).

Answer: For two skew lines r⃗ = a⃗₁ + λb⃗₁ and r⃗ = a⃗₂ + μb⃗₂, the shortest distance is
d = |(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂)| / |b⃗₁ × b⃗₂|
a⃗₂ − a⃗₁ = (2 − 1, −1 − 2, −1 − 1) = (1, −3, −2)
b⃗₁ × b⃗₂:
î: (−1)(2) − (1)(1) = −3
ĵ: −[(1)(2) − (1)(2)] = 0
k̂: (1)(1) − (−1)(2) = 3
So b⃗₁ × b⃗₂ = (−3, 0, 3), with magnitude √18 = 3√2.
Dot product: (1)(−3) + (−3)(0) + (−2)(3) = −9
d = |−9| / (3√2) = 9/(3√2)
= 3/√2 units

Q10. Find the angle between the line (x − 2)/3 = (y + 1)/−1 = (z − 3)/2 and the plane x + y + 4 = 0.

Answer: For a line with direction ratios (a, b, c) and a plane with normal (A, B, C), the angle φ between them satisfies
sinφ = |aA + bB + cC| / [√(a² + b² + c²) √(A² + B² + C²)]
Here the line has (3, −1, 2) and the plane normal is (1, 1, 0).
Numerator: |(3)(1) + (−1)(1) + (2)(0)| = 2
Denominator: √14 × √2 = √28 = 2√7
sinφ = 2/(2√7) = 1/√7
φ = sin⁻¹(1/√7)

Q11. Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7.

Answer: Direction ratios = (2 − 3, −3 + 4, 1 + 5) = (−1, 1, 6)
The parametric form of the line is:
x = 3 − λ, y = −4 + λ, z = −5 + 6λ
Substitute into the plane equation:
2(3 − λ) + (−4 + λ) + (−5 + 6λ) = 7
6 − 2λ − 4 + λ − 5 + 6λ = 7
5λ − 3 = 7
λ = 2
Putting λ = 2 back:
x = 1, y = −2, z = 7
The point is (1, −2, 7).

Q12. Write the condition for a line to lie in a plane.

Answer: A line r⃗ = a⃗ + λb⃗ lies in the plane r⃗ · n⃗ = d if two conditions hold together:
1. The line is parallel to the plane, that is the direction vector of the line is perpendicular to the normal:
b⃗ · n⃗ = 0
2. A point on the line lies in the plane:
a⃗ · n⃗ = d
If only the first condition holds, the line is parallel to the plane but outside it. Both conditions together mean the line lies completely within the plane.

Related Study Material on Extramarks

Q.1 Find the angle between the two planes 3x-6y+2z-7 = 0 and 2x+2y-2z = 5.

Option:

$\begin{array}{l} The equation of planes are 3 x -6 y +2 z -7 = 0 and \\ 6 x +3 y -2 z = 5. \\ Here a_{1} = 3, b_{1} = 6, c_{1} = 2 \\ a_{1} = 6,</m \end{array}$

Ans.

$\begin{array}{l} The equation of planes are 3 x -6 y +2 z -7 = 0 and \\ 6 x +3 y -2 z = 5. \\ Here a_{1} = 3, b_{1} = 6, c_{1} = 2 \\ a_{1} = 6, b_{1} = 3, c_{1} = 2 \\ Let Î¸ be the angle between the planes . \\ cosÎ¸ = |\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{(a_{1})}^{2} + {(b_{1})}^{2} + {(c_{1})}^{2}} \sqrt{} {(a_{2})}^{2} + {(b_{2})}^{2} + {(c_{3})}^{2}}| \\ = |\frac{3 Ã— 6 6 Ã— 3 + 2 Ã— 2}{\sqrt{{(3)}^{2} + {(6)}^{2} + {(2)}^{2}} \sqrt{{(6)}^{2} + {(3)}^{2} + {(2)}^{2}}}| \\ = |\frac{18 18 4}{\sqrt{{(3)}^{2} + {(6)}^{2} + {(2)}^{2}} \sqrt{{(6)}^{2} + {(3)}^{2} + {(2)}^{2}}}| \\ = |\frac{4}{\sqrt{9 + 36 + 4} \sqrt{9 + 36 + 4}}| \\ = |\frac{4}{\sqrt{49} \sqrt{49}}| \\ = |\frac{4}{7 Ã— 7}| \\ = |\frac{4}{49}| \\ ˆµ cosÎ¸ = \frac{4}{49} \\ Î¸ = \cos^{1} (\frac{4}{49}) \end{array}$

Q.2

$\begin{array}{l} Find the equation of the line in cartesian and vector form \\ that passes through the point (1, 2, 3) and perpendicular \\ to the plane \overset{}{r .} (\hat{i} + \hat{2 j} \hat{5 k}) + 9 = 0 . \end{array}$

Ans.

$\begin{array}{l} The equation of plane in vector form is \\ \overset{}{r .} (\hat{i} + \hat{2 j} \hat{5 k}) + €‹ 9 = 0, where \overset{}{r .} = \hat{xi} + \hat{yj} + \hat{zk} \\ (\hat{xi} + \hat{yj} + \hat{zk}), (\hat{i} + \hat{2 j} \hat{5 k}) + 9 = 0 \\ In cartesian form, the equation becomes \\ x +2 y -5 z + 9 = 0 \\ The direction ratio of this plane is (1, 2, €‹ 5), which is directed \\ along a line perpendicular to the plane . \\ ˆ´ The equation of a line that passes through the point \\ (1, 2, 3) and having direction ratio (1, 2, €‹ 5) is \\ \frac{x 1}{1} = \frac{y 2}{2} = €‹ \frac{z 3}{5} \\ In vector from, \\ the position vector of a point P (1, 2, €‹ 3) is \overset{}{a} = \hat{i} + \hat{2 j} + \hat{3 k} \\ The vector parallel to line is \overset{}{b} = \hat{i} + \hat{2 j} \hat{5 k} . \\ We have the equation of a line which passes through the point \\ whose position vector is \overset{}{a} and parallel to \overset{}{b} is \\ \overset{}{r} = \overset{}{a} + \overset{}{b} \\ Thus, equation of line in vector form is \\ \overset{}{r} = (\hat{i} + \hat{2 j} + \hat{3 k}) + (\hat{i} + \hat{2 j} \hat{5 k}) \end{array}$

Q.3

$\begin{array}{l} Find the distance between the two skew lines . \\ \overset{}{r} = (\hat{i} + \hat{j} + \hat{k}) + (3 \hat{i} + 5 \hat{j} + 2 \hat{k}) and \\ \overset{}{r} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) + (3 \hat{i} + 4 \hat{j} + 2 \hat{k}) \end{array}$

Option: We have the vector equation of linesr=i^+j^+k^+3i^+5j^+2k^ and r=3i^+2j^+4

Ans.

$\begin{array}{l} W e h a v e the vector equation of lines \\ \overset{}{r} = (\hat{i} + \hat{j} + \hat{k}) + (3 \hat{i} + 5 \hat{j} + 2 \hat{k}) and \\ \overset{}{r} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) + Î¼ (3 \hat{i} + 4 \hat{j} + 2 \hat{k}) \\ Here, \overset{}{a_{1}} = \hat{i} + \hat{j} + \hat{k}, \\ {\overset{}{a}}_{2} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \\ {\overset{}{a}}_{2} \overset{}{a_{1}} = 2 \hat{i} + \hat{j} + 3 \hat{k} \\ a n d {\overset{}{b}}_{1} = 3 \hat{i} + 5 \hat{j} + 2 \hat{k}, \\ {\overset{}{b}}_{2} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \\ ˆ´ {\overset{}{b}}_{1} Ã— {\overset{}{b}}_{2} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ 3 & 4 & 2 \end{matrix}| \\ = \hat{i} (10 8) \hat{j} (6 6) + \hat{k} (12 15) \\ = \hat{i} (2) \hat{j} (0) + \hat{k} (3) \\ = 2 \hat{i} 3 \hat{k} \\ |{\overset{}{b}}_{1} Ã— {\overset{}{b}}_{2}| = \sqrt{4 + 0 + 9} \\ = \sqrt{13} \\ ˆµ ({\overset{}{a}}_{2} \overset{}{a_{1}}) . ({\overset{}{b}}_{1} Ã— {\overset{}{b}}_{2}) = (2 \hat{i} + \hat{j} + 3 \hat{k}) . (2 \hat{i} 3 \hat{k}) \\ =4+0-9 \\ =-5 \end{array}$

$\begin{array}{l} The distance between two skew lines is given by, \\ d = \frac{|({\overset{}{a}}_{2} \overset{}{a_{1}}) . ({\overset{}{b}}_{1} Ã— {\overset{}{b}}_{2})|}{|({\overset{}{b}}_{1} Ã— {\overset{}{b}}_{2})|} \\ = \frac{|5|}{13} \\ = \frac{5}{13} \end{array}$

Q.4

$\begin{array}{l} Find the Cartesian equation of the plane which \\ passing through the intersection of the planes \\ \overset{}{r} . (\hat{i} + \hat{j} + \hat{k}) = 1 and \overset{}{r} . (3 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 8 and the point \\ (1, 1, 1) . \end{array}$

Option:

$\begin{array}{l} We have \overset{}{r} = x \hat{i} + y \hat{j} + z \hat{k} \\ The equation of planes are \\ (x \hat{i} + y \hat{j} + z \overset{^</}{k}) \end{array}$

Ans.

$\begin{array}{l} We have \overset{}{r} = x \hat{i} + y \hat{j} + z \hat{k} \\ The equation of planes are \\ (x \hat{i} + y \hat{j} + z \hat{k}) . (\hat{i} + \hat{j} + \hat{k}) = 1 and (x \hat{i} + y \hat{j} + z \hat{k}) . (3 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 8 \\ x + y + z 1 = 0 and 3 x + 2 y + 2 z 8 = 0 \\ Let Equation of the plane passing through the intersection \\ of the planes 3 x + 2 y + 2 z 8 = 0 and x + y + z 1 = 0 is \\ 3 x + 2 y + 2 z 8 + (x + y + z 1) = 0 \\ (3 +) x + (2 +) y + (2 +) z = 8 + (1) \\ Since this plane passes through point (1, 1, 1) \\ ˆ´ (3 +) + (2 +) + (2 +) = 8 + \\ 7 + 3 = 8 + \\ 2 = 1 \\ = \frac{1}{2} \\ On putting the value of in equation (1) we get, \\ (3 + \frac{1}{2}) x + (2 + \frac{1}{2}) y + (2 + \frac{1}{2}) z = 8 + \frac{1}{2} \\ \frac{7}{2} x + \frac{5}{2} y + \frac{5}{2} z = \frac{17}{2} \\ 7 x + 5 y + 5 z 17 = 0 \end{array}$

Q.5

$\begin{array}{l} If, and are the angles made by a line OP with \\ coordinate axes, then prove that \sin^{2} + \sin^{2} + \sin^{2} = 2 . \end{array}$

Option:

Since , and are the angles made by a line OP with coordinate axes, let Px,y,z be any point such that OP=r.OP=

Ans.

$\begin{array}{l} Since Î±, Î² and Î³ are the angles made by a line OP with \\ coordinate axes, let P (x, y, z) be any point such that OP = r . \\ OP = r \\ OR = x \\ QR = y \\ OQ = \sqrt{x^{2} + y^{2}} \\ PQ = z \\ ˆ´ OP = r = \sqrt{x^{2} + y^{2} + z^{2}} \\ ˆ´ cosÎ± = \frac{x}{r} \\ x = rcosÎ± \\ y = rcosÎ² \\ z = rcosÎ³ \end{array}$

$\begin{array}{l} We have, \\ \sqrt{x^{2} + y^{2} + z^{2}} = r \\ x^{2} + y^{2} + z^{2} = r^{2} \\ {(rcosÎ±)}^{2} + {(rcosÎ²)}^{2} + {(rcosÎ³)}^{2} = r^{2} \\ r^{2} (\cos^{2} Î± + \cos^{2} Î² + \cos^{2} Î³) = r^{2} \\ \cos^{2} Î± + \cos^{2} Î² + \cos^{2} Î³ = 1 \\ 1 \sin^{2} Î± + 1 \sin^{2} Î² + 1 \sin^{2} Î³ = 1 \\ 3 (\sin^{2} Î± + \sin^{2} Î² + \sin^{2} Î³) = 1 \\ \sin^{2} Î± + \sin^{2} Î² + \sin^{2} Î³ = 2 \end{array}$

CBSE Important Questions Class 12 Maths Chapter 11 – Three Dimensional Geometry

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