Three Dimensional Geometry Test

Result

Three Dimensional Geometry Test - 8

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Question 1
2 / -0.83

If a₁ , b₁ , c₁ and a₂ , b₂ , c₂ are the direction ratios of two lines and θ is the acute angle between the two lines; then

A

B

C

D

Solution

If a₁ , b₁ , c₁ and a₂ , b₂ , c₂ are the direction ratios of two lines and θθ is the acute angle between the two lines; then , the cosine of the angle between these two lines is given by :
Question 2
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Find the equation of the line in cartesian form that passes through the point (–2, 4, –5) and parallel to the line given by

A

B

C

D

Solution

Find the equation of the line in cartesian form that passes through the point (–2, 4, –5) and parallel to the line given by

is given by:
And l = 3 , m = 5 and n = 6 .
Question 3
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Vector equation of a plane that contains three non collinear points having position vectors

A

B

C

D

Solution

Vector equation of a plane that contains three non collinear points having position vectors
Question 4
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The vector and cartesian equations of the planes that passes through the point (1, 0, –2) and the normal to the plane is

A

B

C

D

Solution

Let
be the position vector of the point here,
. Therefore, the required vector equation of the plane is:
Question 5
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Find the distance of the point (3, –2, 1) from the plane 2x –y + 2z + 3 = 0

A
17/3

B
19/3

C
13/3

D
15/3

Solution

As we know that the length of the perpendicular from point P(x₁ ,y₁ ,z₁ ) from the plane

Here, P(3, - 2,1) is the point and equation of Plane is 2x - y + 2z+3 = 0

Therefore, the perpendicular distance is :
Question 6
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Vector equation of a line that passes through the given point whose position vector is and parallel to a given vector is

A

B

C

D

Solution

Vector equation of a line that passes through the given point whose position vector is and parallel to a given vector is given by :
Question 7
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Find the values of p so that the lines are at right angles.

A
p =72/15

B
p =70/11

C
p =70/12

D
p =71/13

Solution

Give lines are :
and
The D.R 's of the lines are -3, 2p/7, 2 and -3p/7, 1, -5
Question 8
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Vector equation of a plane that passes through the intersection of planes expressed in terms of a non –zero constant λ is

A

B

C

D

Solution

In vector form:
Vector equation of a plane that passes through the intersection of planes
expressed in terms of a non –zero constant λ is given by :
Question 9
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Find the equations of the planes that passes through three points (1, 1, 0), (1, 2, 1), (–2, 2, –1)

A
2x + 3y –7z = 5

B
2x + 5y –3z = 5

C
3x + 3y –3z = 5

D
2x + 3y –3z = 5

Solution

In cartesian co-ordinate system :
Equation of a plane passing through three non collinear

Points (x₁ , y₁ , z₁ ) , (x₂ , y₂ , z₂ ) and (x₃ , y₃ , z₃ ) is given by :

Therefore, the equations of the planes that passes through three points (1,1,0), (1,2,1), (-2,2,-1) is given by :

⇒(x-1)(-2) - (y-1) (3) + 3z = 0
⇒2x+3y - 3z = 5
Question 10
2 / -0.83

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

A
4

B
2

C
3

D
5

Solution

As we know that the length of the perpendicular from point P(x₁ ,y₁ ,z₁ ) from the plane

Here, P(2,3,-5) is the point and equation of plane is x+2y - 2z = 9

Therefore, the perpendicular distance is :