Important Questions Class 12 Maths Chapter 11

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.

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 = ij + λ2jk.

Ans. The cartesian form of r = ij + λ2jk 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).

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 . 3ij + 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 . 3ij + 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, x2x1PQ, y2y1PQ, z2z1PQ

PQ is the distance between the two points.

Therefore, PQ = x2x12+y2y12+ z2z12

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,

x2x1PQ, y2y1PQ, z2z1PQ

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

=377, -277, 877

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

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)

a2a1 . 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 a2a1 . b2 × b1 b2 × b1

By putting the values, we get

a2a1 . 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 2ij+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 − (2ij+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+ 3k2ij+ 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:

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Q.2

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Q.3

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

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Q.4

Option:

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Q.5

Option:

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

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