Important Questions Class 12 Maths Chapter 11

Q: 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.

Q: 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.

Q: 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.

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Important Questions For CBSE Class 12 Mathematics Chapter 11 – Three Dimensional Geometry

Three-dimensional geometry is an important part of analytical geometry. In this chapter, students learn to use vector algebra in geometric equations. Important Questions Class 12 Mathematics Chapter 11 covers all the concepts of three-dimensional geometry, including the direction cosine and direction ratios of a line joining two points, the angle between two lines, the angle between a line and a plane, etc. Students must revise these concepts several times and practise more questions to be able to solve questions quickly in the exam. Subject matter experts have carefully selected Class 12 Mathematics Chapter 11 Important Questions after consulting the updated CBSE Syllabus and questions from the past years. Students will get an idea of what kind of questions they can expect in the exam and what marks those questions will have.

CBSE Class 12 Mathematics Chapter-11 Important Questions

Study Important Questions For Class 12 Mathematics Chapter 11- Three Dimensional Geometry

Given below is a sample set of the Important Questions Class 12 Mathematics Chapter 11. For the full set, students can access the link provided here.

Very Short Answer Questions (1 Mark)

Q.1. P is a point located on a three-dimensional plane. What is the distance between P from the x-axis? The coordinate of P is (a, b, c).

Ans. The distance of point P from the x-axis is a unit.

Q.2. A line while passing through a three-dimensional plane makes an angle α,β,γ with x, y, and z axes respectively. Find the value of + + .

Ans. If a line makes an angle α, β, γ with the coordinate axes, then the value of

+ + = 1.

Therefore,

3 − + + = 1

Or, + + = 2

Q.3. Convert the vector into the cartesian form: r = i–j + λ2j–k.

Ans. The cartesian form of r = i–j + λ2j–k is x-12= y+10= z-0-1

Q.4. Find the distance between the planes 2x + 2y – z + 2 = 0 and 4x + 4y – 2z + 5 = 0.

Ans. The equations of the given planes are,

2x + 2y – z + 2 = 0 …………….(i)

4x + 4y – 2z + 5 = 0 ……………….(ii)

Equation (ii) can be written as 2x + 2y – z +52 = 0

Therefore, the two planes are parallel to each other.

The distance between the two parallel planes = 2- 5222 + 22 + 12 = 16 unit.

Q.5. Prove the following planes are parallel to each other.

x + y – 2z + 4 = 0; 3x + 3y – 6z + 5 = 0.

Ans. The given planes are,

x + y – 2z + 4 = 0 ………. (i)

3x + 3y – 6z + 5 = 0

Or, x + y – 2z + 53 = 0 ……….. (ii)

Therefore, the planes do not intersect each other. Hence, they are parallel to each other.

Q.6. Find the coordinates of P when the value of OP is 3 with the direction ratios proportional to -1, 2, 3, where O is the origin.

Ans. The equation of the line OP is,

x-1= y2= z-2

Let, x-1= y2= z-2 = k

Therefore, x = -k; y = 2k; z = -2k

Again, the distance of OP = 3 (given)

Therefore, k2 + 2k2 + 2k2 = 3

Or, 9k2 = 3

Or, k = ± 3

The coordinate of k = (∓1, ±2, ∓2).

Short Answer Questions (4 Mark)

Q.1. Three points A (0, −1, −1), B (−4, 4, 4), and C (4,5,1) are in the same plane. Find the equation of the plane containing all the three points.

Ans. The required equation for the plane containing the points A, B, and C is,

x-0 y+1 z+1 -4 4 4 4 5 1 = 0

⟹4x – 5y + 9z +3=0

Q.2. The direction cosines of two mutually perpendicular lines are l1, m1, n1 and l2, m2, n2. Prove that, the direction cosines of the line perpendicular on both lines are m1n2−n1m2, n1l2−l1n2, l1m2−m1l2.

Ans. The vector equation of the perpendicular line is,

i j k l1 m1 n1 l2 m2 n2 = (m1n2−n1m2) i+ (n1l2−l1n2) j+ (l1m2−m1l2)k

Therefore, the directional cosines are (m1n2−n1m2) i, (n1l2−l1n2) j, (l1m2−m1l2)k.

Q.3. A line PQ x-41= 2y-42= k-z-2= a passes through the plane 2x – 4y + z = 7. What is the value of K?

Ans. Since the line PQ x-41= 2y-42= k-z-2= a lies in the given plane 2x – 4y + z = 7, the value of a becomes 0.

Therefore,

x-41=0 ; or, x = 4

y = 2; z = k

Putting the values of x, y, and z in the equation of the plane we get,

2*4 – 4*2 + k = 7

Or, k = 7

The value of k, therefore, is 7.

Q.4. Find equation of the plane through line of intersection of planes r . 2i+6j + 12 = 0 and r . 3i–j + 4k = 0 which is at unit distance from the origin.

Ans. r . 2i+6j + 12 = 0

The cartesian form of the equation is x + 3y + 6 = 0 ……….(i)

r . 3i–j + 4k = 0 can be expressed as 3x – y + 4z =0 in the cartesian form ……. (ii)

Direction ratio of the plane perpendicular to the planes is

i j k 1 3 0 3 -1 4 = 12i − 4j − 10k

The intersecting line of the lines x + 3y + 6 = 0 and 3x – y + 4z =0 is

x1=3x+6-1=-(6z+3)5

Let, the equation of required plane is ax + by + cz = d

Therefore, it can be said that, 12i − 4j − 10k =d

0, -2, -2 lies on that plane.

Putting the values, we get,

12(0) − 4(−2) − 10(−2) = d

Or, d = 28

Therefore, the final equation becomes,

12x – 4y – 10z = 28

Or, 6x – 2y – 5z = 14

Q.5. Two points -2, 4, -5 and 1, 2, 3are located on a certain line. Find the direction cosines of the line passing through these points.

Ans. Let the points are P and Q.

Position of P = -2, 4, -5

Position of Q = 1, 2, 3

Formula of direction cosines is, x2–x1PQ, y2–y1PQ, z2–z1PQ

PQ is the distance between the two points.

Therefore, PQ = x2–x12+y2–y12+ z2–z12

Or, PQ = 1-(-2)2+2-42+ 3-(-5)2

Or, PQ = 9+4+64

Or, PQ = 77

Putting the values in cosines’ formula we get,

x2–x1PQ, y2–y1PQ, z2–z1PQ

= 1-(-2)77, 2-477, 3-(-5)77

=377, -277, 877

Therefore, the required direction cosines are 377, -277, 877 .

Long Answer Questions (6 Mark)

Q.1. Find the shortest distance between the two lines given below.

x-83=y+9-16=z-107 and x-153=y-298=z-55

Ans. The given equations are,

x-83 = y+9 -16= z-107 ……………. (i)

x-153 = y-298 = z-55 ……………….. (ii)

From the first equation we get,

The line passes through the point (8, -9, 10).

Its directional ratio is proportional to <3, -16, 7>

Therefore, the vector equation is, r = a1 + b1

Where, a1 = 8i − 9j + 10k

From the second equation we get,

The line passes through (15, 29, 5).

Its direction ratio is proportional to (3, 8, -5).

The vector equation will be r = a2 + b2

Where, a2 = 15i + 29j + 5k

b2 = 3i + 8j − 5k

Therefore, a2 – a1 = 15i + 29j + 5k − 8i + 9j − 10k

Or, a2 – a1 = 7i + 38j − 5k ………… (iii)

b2 × b1 = i j k 3 -16 7 3 8 -5

= (80 − 56)i − (−15 − 21)j + (24 + 48)k

= 24i + 36j + 72k ……………… (iv)

a2 – a1 . b2 × b1 = 7i + 38j – 5k . 24i + 386 + 72k

= 168+1368−360

=1176

b2 × b1 = 242+362+722 (from (iv))

= 7056

= 84

We know that, the shortest distance between two lines is a2 – a1 . b2 × b1 b2 × b1

By putting the values, we get

a2 – a1 . b2 × b1 b2 × b1 = 117684

= 14

Therefore, the shortest distance between the two lines will be 14 units.

Q.2. A variable plane is at a constant distance 3p from the origin and meets the axes at A, B, and C points. If the centroid of ∆ABC is (α, β, γ) then prove that 12 + 12 + 12 = 1p2 .

Ans. Let the equation of the plane is a + b + c = 1, where a, b, c are the intercepts of the plane and α, , are axes.

The distance between the origin and the plane is

1-2+-2+-2 = 3p

Or, -2+-2+-2 = 19p2

The coordinate of a centroid of a triangle in a three dimensional plane is (a3, b3, c3).

The given coordinate is (α, β, γ).

Therefore, a = 3 ; b = 3, c = 3.

By substituting the values of a, b, and c, we get,

192 + 192 + 192 = 19p2

Therefore, 12 + 12 + 12 = 1p2 [proved].

Q.3. Find the angle between the two lines passing through a three-D plane.

x+33=y-15=z+34 ; x+11=y-41=z-52

Ans. x+33=y-15=z+34 ………… (i)

The direction ratio (a1, b1, c1) = (3, 5, 4)

x+11=y-41=z-52 …………….. (ii)

The direction ratio (a2, b2, c2) = (1, 1, 2)

The cosine between two lines is

cos = a1a2+ b1b2+ c1c2a12+ b12+ c122a22+ b22+ c22

= 3.1+ 5.1+ 4.232+ 52+ 4212+ 12+ 22

= 1650 6

= 1652 32

=853

The required angle = 853

Q.4. A perpendicular is drawn from point P 2i– j+5k is on the line r = 11i-2j-8k+ λ10i-4j-11k. Find the length and the foot of the perpendicular.

Ans. Let, the foot of the perpendicular = L.

The position vector of L = 11i− 2j −8k + λ(10i− 4j− 11k)

= (11+10λ) i + (−2−4λ) j + (−8−11λ) k ……….. (i)

The length of PL is equal to the difference of position vector P and position vector L.

Therefore PL = (11 +10λ)i+ (−2−4λ) j + (−8−11λ) k − (2i−j+5k)

= (9+10λ) i + (−1−4λ) j + (−13−11λ) k ……….. (ii)

Since PL is perpendicular to the given line and the given line is parallel to b = 10i-4j-11k

PL.b = 0

Therefore,

(9+10) i + (-1-4) j + (-13-11) k . 10i-4j-11k = 0

⟹ 10(9+10λ) − 4(−1−4λ) − 11(−13−11λ) = 0

⟹237λ = −237

⟹ λ = −1

Placing the value of in equation (i) we get,

i+ 2j+ 3k

Therefore, the foot of the perpendicular is i+ 2j+ 3k

Now,

PL = i+ 2j+ 3k– 2i– j+ 5k

= −i+3j-2k

So, the length of the perpendicular

PL = 1+9+4 = 14 unit.

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.

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

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

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FAQs (Frequently Asked Questions)

1. 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.

2. 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.

3. 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.

Important Questions Class 12 Maths Chapter 11

Important Questions For CBSE Class 12 Mathematics Chapter 11 – Three Dimensional Geometry

CBSE Class 12 Mathematics Chapter-11 Important Questions

Study Important Questions For Class 12 Mathematics Chapter 11- Three Dimensional Geometry

Short Answer Questions (4 Mark)

Long Answer Questions (6 Mark)

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FAQs (Frequently Asked Questions)

1. Why is three-dimensional geometry important?

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

3. How will Important Questions provided by Extramarks benefit students?

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Important Questions Class 12 Maths Chapter 11

Important Questions For CBSE Class 12 Mathematics Chapter 11 – Three Dimensional Geometry

CBSE Class 12 Mathematics Chapter-11 Important Questions

Study Important Questions For Class 12 Mathematics Chapter 11- Three Dimensional Geometry

Short Answer Questions (4 Mark)

Long Answer Questions (6 Mark)

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FAQs (Frequently Asked Questions)

1. Why is three-dimensional geometry important?

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

3. How will Important Questions provided by Extramarks benefit students?

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ICSE Class

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