Three Dimensional Geometry Online Objective Test | Three Dimensional Geometry Online Objective Test

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q1. Determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

A. sin−1(25)

B. tan−1(25)

C. cot−1(25)

D. cos−1(25)

Answer : Option D
Explaination / Solution:

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q2. In the following case, determine whether the given planes are parallel orperpendicular, and in case they are neither, find the angles between them. 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

A. The planes are parallel

B. The planes are at 55∘

C. The planes are perpendicular

D. The planes are at 45∘

Answer : Option C
Explaination / Solution:

We have ,
2x + y + 3z – 2 = 0 and x – 2y + 5 = 0. Let

θ

be the angle between the planes , then
As

a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 (1) + 1 (- 2) + 3 (0) = 0

Therefore , the given planes are perpendicular to each other.

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q3. In the following case, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

A. The planes are at 55∘

B. The planes are parallel

C. The planes are perpendicular

D. The planes are at 45∘

Answer : Option B
Explaination / Solution:

We have,
2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z = 0 . Here ,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = \frac{2}{3}

Therefore , the given planes are parallel.

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q4. In the following case, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

A. The planes are at 45∘

B. The planes are parallel

C. The planes are at 55∘

D. The planes are perpendicular

Answer : Option B
Explaination / Solution:

We have ,
2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0 . Here ,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = 1

Therefore , the given planes are parallel.

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q5. In the following case, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 4x + 8y + z – 8 = 0 and y + z – 4 = 0

A. 43∘

B. 47∘

C. 49∘

D. 45∘

Answer : Option D
Explaination / Solution:

We have, 4x + 8y + z – 8 = 0 and y + z – 4 = 0. Let be the angle between the planes, then

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q6. Find the distance of the point (3, – 2, 1) from the plane 2x – y + 2z + 3 = 0

A. 153

B. 133

C. 193

D. 173

Answer : Option B
Explaination / Solution:

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q7. Find the distance of the point (2, 3, – 5) from the plane x + 2y – 2z = 9

A. 2

B. 4

C. 3

D. 5

Answer : Option C
Explaination / Solution:

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q8. Find the distance of the point (– 6, 0, 0) from the plane 2x – 3y + 6z – 2 = 0

A. 4

B. 2

C. 3

D. 5

Answer : Option B
Explaination / Solution:

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q9. Equation of a plane perpendicular to a given line with direction ratios A, B, C and passing through a given point (x1, y1, z1) is

A. A (x + x1) + B (y – y1) + C (z – z1) = 0

B. A (x - x1) + B (y – y1) + C (z – z1) = 0

C. A (x – x1) + B (y – y1) + C (z – z1) = 1

D. A (x – x1) + B (y + y1) + C (z – z1) = 1

Answer : Option B
Explaination / Solution:

In Cartesian co – ordinate system Equation of a plane perpendicular to a given line with direction ratios A, B, C and passing through a given point (x1, y1, z1) is given by : A (x – x1) + B (y – y1) + C (z – z1) = 0 .

# Three Dimensional Geometry # CBSE 12th Mathematics Prepare / Learn

Q10. Find the coordinates of the foot of the perpendicular drawn from the origin to 2x + 3y + 4z – 12 = 0

A. (2429,3629,4929)

B. (2729,3629,4829)

C. (2429,3929,4829)

D. (2429,3629,4829)

Answer : Option D
Explaination / Solution:

D.R.’s of the line are < 2 , 3 , 4 > .
Therefore , equation of the line is :

\frac{x - 0}{2} = \frac{y - 0}{3} = \frac{z - 0}{4} = λ

Thus , the coordinates of any point P on the above line are P ( 2

λ

, 3

λ

λ

) .
But , this point P also lies on the given plane:
2(2

λ

) + 3(3

λ

) +4(4

λ

) – 12 = 0.

\Rightarrow 29 λ = 12 \Rightarrow λ = \frac{12}{29}

Therefore , the coordinates of the foot of perpendicular are given by :

(2 \times \frac{12}{29}, 3 \times \frac{12}{29}, 4 \times \frac{12}{29})

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