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All questions of Chapter 11 - Three Dimensional Geometry for Commerce Exam

Find the equation of the set of points which are equidistant from the points (1, 2 , 3) and (3, 2, -1)​
a) x + 2z = 0
b) y + 2z = 0
c) x – 2z = 0
d) x – 2y = 0 
Correct answer is option 'C'. Can you explain this answer?

Aryan Khanna answered
Pt. A(1, 2 , 3)
Pt. B(3, 2, -1)
Let P(x,y,z)
So, AP = BP
((x-1)2 + (y-2)2 + (z-3)2)1/2 = ((x-3)2 + (y-2)2 + (z+1)2)1/2
(x-1)2 + (y-2)2 + (z-3)2) = (x-3)2 + (y-2)2 + (z+1)2
x2 +1 -2x + y2 + 4 - 4y + z2 + 9 – 6z = x2 +9 -6x + y2 + 4 - 4y + z2 + 1 + 2z
4x – 8z = 0
x – 2z = 0
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The direction cosines of the line joining the points (2, -1, 8) and (-4, -3, 5) are:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Geetika Shah answered
Pt. A(2, -1, 8)
Pt. B(-4, -3, 5)
Direction Ratio DR of AB : ( -4-2 , -3+1 , 5-8 )
: (-6,-2,-3)
Direction cosine of AB : ( -6/(62+22+32)1/2 , -2/(62+22+32)1/2 , -3/(62+22+32)1/2)
: ( -6/7, -2/7, -3/7)
 

If a line has the direction ratios -4, 18, -12 then what are its direction cosines?​
  • a)
    -2, 9, -6
  • b)
    -4, 18, -12
  • c)
    2/11, 9/11, 6/11
  • d)
    -2/11, 9/11, -6/11
Correct answer is option 'D'. Can you explain this answer?

Vikas Kapoor answered
DR of the line :  (-4, 18 -12)
DC of the line : (-4/k, 18/k, -12/k)
where k = ((42) + (182) + (12)2)1/2
= (16 + 324 + 144)1/2
= (484)1/2
= 22
So, DC : (-4/22, 18/22, -12/22)
: (-2/11 , 9/11 , -6/11)

For which value of a lines  and  are perpendicular?
  • a)
    11/70
  • b)
    5
  • c)
    1
  • d)
    70/11
Correct answer is option 'B'. Can you explain this answer?

Tejas Verma answered
(x-1)/(-3) = (y-2)/(2p/7) = (z-3)/2 
(x-1)(-3p/7) = (y - 5)/1 = (z - 6)/(-5)
The direction ratio of the line are -3, 2p/7, -2 and (-3p)/7, 1, -5
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other if a1a2 + b1b2 + c1c2 = 0
Therefore, (-3)(-3p/7) + (2p/7)(1) + 2(-5) = 0
(9p/7) + (2p/7) = 10
11p = 70
p = 70/1

Equation of a plane which is at a distance d from the origin and the direction cosines of the normal to the plane are l, m, n is.
  • a)
    lx – my + nz = d
  • b)
    – lx + my + nz = d
  • c)
    lx + my + nz = d
  • d)
    lx + my + nz = – d
Correct answer is option 'C'. Can you explain this answer?

Bhargavi Sen answered
Equation of a Plane with Given Normal Vector
To find the equation of a plane in 3D space, we need a point on the plane and a normal vector to the plane. In this case, we are given the direction cosines of the normal vector and the distance of the plane from the origin.

Given Information:
- Direction cosines of the normal to the plane: l, m, n
- Distance of the plane from the origin: d

Equation of the Plane:
The general form of the equation of a plane in 3D space is:
lx + my + nz = d

Explanation:
- The direction cosines of the normal vector to the plane are l, m, n. This means that the coefficients of x, y, and z in the equation of the plane will be l, m, and n respectively.
- The distance of the plane from the origin is represented by the constant term d in the equation.

Correct Answer:
The correct equation of the plane in this case is:
lx + my + nz = d

Conclusion:
When given the direction cosines of the normal vector and the distance of the plane from the origin, the equation of the plane can be determined by using the general form lx + my + nz = d.

If a line in the ZX-plane makes an angle 60o with Z-axis, the direction cosines of this line are:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Saumya Gupta answered
The angle between the line and z axis is 60 and the line in ZX plane so the angle between line and X axis is (90-60)=30 and line is perpendicular to Y axis so the angle between line and Y axis is 90. So direction cosine of line is cos30,cos90,cos60 =√3/2,0,1/2.

The direction cosines of the line equally inclined with the axes, are:​
  • a)
    1, 1, 1,
  • b)
    0, 0, 0
  • c)
    1/3, 1/3, 1/3
  • d)
    1/√3, 1/√3, 1/√3
Correct answer is option 'D'. Can you explain this answer?

Geetika Mehta answered
Direction cosines are the ratios of the direction ratios of a line. In other words, they represent the cosines of the angles that the line makes with the positive x, y, and z axes.

Let's consider a line equally inclined with the axes. This means that the line makes equal angles with each axis.

To find the direction cosines of this line, we need to determine the cosines of the angles it makes with the x, y, and z axes.

Let's assume that the line makes an angle of θ with each axis. Since the line is equally inclined with the axes, all the angles are equal.

Let's calculate the direction cosine along the x-axis:

cos(θ) = adjacent side / hypotenuse
cos(θ) = 1 / hypotenuse
hypotenuse = 1 / cos(θ)

Similarly, the direction cosine along the y-axis and z-axis would also be 1 / cos(θ).

Therefore, the direction cosines of the line equally inclined with the axes are (1 / cos(θ), 1 / cos(θ), 1 / cos(θ)).

Since the line is equally inclined with the axes, the angle θ is the same for all three direction cosines.

Hence, the correct answer is option D: 1/3, 1/3, 1/3.

Find the distance of the point (0, 0, 0) from the plane 3x – 4y + 12 z = 3
  • a)
    9/13
  • b)
    7/13
  • c)
    5/13
  • d)
    3/13
Correct answer is option 'D'. Can you explain this answer?

Krish Ghoshal answered
As we know that the length of the perpendicular from point 
P(x1,y1,z1) from the plane a1x+b1y+c1z+d1 = 0 is given by: 

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)
Correct answer is option 'A'. Can you explain this answer?

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 :

The equation of a plane through a point whose position vector is  perpendicular to the vector  . is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

In vector form The equation of a plane through a point whose position vector is  perpendicular to the vector  Is given by : 

If l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines; and θ is the acute angle between the two lines; then
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Ipsita Sen answered
If l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines; and θ is the acute angle between the two lines; then the cosine of the angle between these two lines is given by : 

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)
Correct answer is option 'D'. Can you explain this answer?

Nandita Basak answered
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 :

In the vector form, equation of a plane which is at a distance d from the origin, and  is the unit vector normal to the plane through the origin is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Anuj Saini answered
In the vector form, equation of a plane which is at a distance d from the origin, and  is the unit vector normal to the plane through the origin is given by : 

Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2
  • a)
    0, 1, 0; 2
  • b)
    1, 0, 0; 3
  • c)
    1, 0, 1; 3
  • d)
    0, 0, 1; 2
Correct answer is option 'D'. Can you explain this answer?

We have z = 2 . , it can be written as : 0x+0y+1z = 2. Compare it with lx+my+nz = d , we get ; l = 0 , m = 0 , n = 1 and d = 2 . therefore , D.C.’s of normal to the plane are 0 , 0 , 1 and distance from the origin = 2.

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
Correct answer is option 'C'. Can you explain this answer?

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 :

 

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
Correct answer is option 'D'. Can you explain this answer?

Anand Khanna answered
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

If l, m, n are the direction cosines and a, b, c are the direction ratios of a line then
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Avantika Saha answered
If l, m, n are the direction cosines and a, b, c are the direction ratios of a line then , the directions cosines of the line are given by :

Determine the direction cosines of the normal to the plane and the distance from the origin. Plane x + y + z = 1
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Here , D.R’s of normal to the plane are 1, 1 , 1 ,its D.C ‘s are :

On dividing x + y + z = 1 by √3 , we get :
 It is of the form : lx+my+nz = d , therefore , d = 1/√3 .

Distance between 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Aryan Sen answered
In vector form Distance between two parallel lines  given by :

If a line makes angles 90, 135, 45 with the x, y and z – axes respectively, find its direction cosines.
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Pranav Pillai answered
If a line makes angles 90, 135, 45 with the x, y and z – axes respectively, then the direction cosines of this line is given by :

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