The Exercise 11.3 of NCERT Solutions for Class 12 Maths Chapter 11- Three Dimensional Geometry is based on the following topics:

- Plane
- Equation of a plane in normal form
- Equation of a plane perpendicular to a given vector, passing through given point
- Equation of a plane passing through three non collinear points
- Intercept form of the equation of a plane
- Plane passing through the intersection of two given planes

- Coplanarity of Two Lines
- Angle between Two Planes
- Distance of a Point from a Plane
- Angle between a Line and a Plane

Solving the problems of this exercise will help the students in understanding the problem solving methods related to the topics mentioned above.

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### Access other exercises of Class 12 Maths Chapter 11

Exercise 11.1 Solutions 5 Questions

Exercise 11.2 Solutions 17 Questions

Miscellaneous Exercise On Chapter 11 Solutions 23 Questions

#### Access Answers of Maths NCERT Class 12 Chapter 11.3

**1. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.(a) z = 2**

**(b) x + y + z = 1**

**(c) 2x + 3y â€“ z = 5**

**(d) 5y + 8 = 0**

**Solution:**

**(a) **z = 2

Given:

The equation of the plane, z = 2 or 0x + 0y + z = 2 â€¦. (1)

Direction ratio of the normal (0, 0, 1)

By using the formula,**âˆš**[(0)^{2} + (0)^{2} + (1)^{2}] = **âˆš**1

= 1

Now,

Divide both the sides of equation (1) by 1, we get

0x/(1) + 0y/(1) + z/1 = 2

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

âˆ´ The direction cosines are 0, 0, 1

Distance (d) from the origin is 2 units

**(b) **x + y + z = 1

Given:

The equation of the plane, x + y + z = 1â€¦. (1)

Direction ratio of the normal (1, 1, 1)

By using the formula,**âˆš**[(1)^{2} + (1)^{2} + (1)^{2}] = **âˆš**3

Now,

Divide both the sides of equation (1) by **âˆš**3, we get

x/(**âˆš**3) + y/(**âˆš**3) + z/(**âˆš**3) = 1/**âˆš**3

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

âˆ´ The direction cosines are 1/**âˆš**3, 1/**âˆš**3, 1/**âˆš**3

Distance (d) from the origin is 1/**âˆš**3 units

**(c) **2x + 3y â€“ z = 5

Given:

The equation of the plane, 2x + 3y â€“ z = 5â€¦. (1)

Direction ratio of the normal (2, 3, -1)

By using the formula,**âˆš**[(2)^{2} + (3)^{2} + (-1)^{2}] = **âˆš**14

Now,

Divide both the sides of equation (1) by **âˆš**14, we get

2x/(**âˆš**14) + 3y/(**âˆš**14) â€“ z/(**âˆš**14) = 5/**âˆš**14

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

âˆ´ The direction cosines are 2/**âˆš**14, 3/**âˆš**14, -1/**âˆš**14

Distance (d) from the origin is 5/**âˆš**14 units

**(d) **5y + 8 = 0

Given:

The equation of the plane, 5y + 8 = 0

-5y = 8 or

0x â€“ 5y + 0z = 8â€¦. (1)

Direction ratio of the normal (0, -5, 0)

By using the formula,**âˆš**[(0)^{2} + (-5)^{2} + (0)^{2}] = **âˆš**25

= 5

Now,

Divide both the sides of equation (1) by 5, we get

0x/(5) â€“ 5y/(5) â€“ 0z/(5) = 8/5

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

âˆ´ The direction cosines are 0, -1, 0

Distance (d) from the origin is 8/5 units

**2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector**

**Solution:**

**3.** **Find the Cartesian equation of the following planes: (a) **

**Solution:**

Given:

The equation of the plane.

**4. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.(a) 2x + 3y + 4z â€“ 12 = 0**

**(b) 3y + 4z â€“ 6 = 0**

**(c) x + y + z = 1**

**(d) 5y + 8 = 0**

**Solution:**

**(a) **2x + 3y + 4z â€“ 12 = 0

Let the coordinate of the foot ofÂ âŠ¥Â P from the origin to the given plane be P(x, y, z).

2x + 3y + 4z = 12 â€¦. (1)

Direction ratio are (2, 3, 4)

**âˆš**[(2)^{2} + (3)^{2} + (4)^{2}] = **âˆš**(4 + 9 + 16)

= **âˆš**29

Now,

Divide both the sides of equation (1) by **âˆš**29, we get

2x/(**âˆš**29) + 3y/(**âˆš**29) + 4z/(**âˆš**29) = 12/**âˆš**29

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

âˆ´ The direction cosines are 2/**âˆš**29, 3/**âˆš**29, 4/**âˆš**29

Coordinate of the foot (ld, md, nd) =

= [(2/**âˆš**29) (12/**âˆš**29), (3/**âˆš**29) (12/**âˆš**29), (4/**âˆš**29) (12/**âˆš**29)]

= 24/29, 36/29, 48/29

**(b) **3y + 4z â€“ 6 = 0

Let the coordinate of the foot ofÂ âŠ¥Â P from the origin to the given plane be P(x, y, z).

0x + 3y + 4z = 6 â€¦. (1)

Direction ratio are (0, 3, 4)

**âˆš**[(0)^{2} + (3)^{2} + (4)^{2}] = **âˆš**(0 + 9 + 16)

= **âˆš**25

= 5

Now,

Divide both the sides of equation (1) by 5, we get

0x/(5) + 3y/(5) + 4z/(5) = 6/5

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

âˆ´ The direction cosines are 0/5, 3/5, 4/5

Coordinate of the foot (ld, md, nd) =

= [(0/5) (6/5), (3/5) (6/5), (4/5) (6/5)]

= 0, 18/25, 24/25

**(c) **x + y + z = 1

Let the coordinate of the foot ofÂ âŠ¥Â P from the origin to the given plane be P(x, y, z).

x + y + z = 1 â€¦. (1)

Direction ratio are (1, 1, 1)

**âˆš**[(1)^{2} + (1)^{2} + (1)^{2}] = **âˆš**(1 + 1 + 1)

= **âˆš**3

Now,

Divide both the sides of equation (1) by **âˆš**3, we get

1x/(**âˆš**3) + 1y/(**âˆš**3) + 1z/(**âˆš**3) = 1/**âˆš**3

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

âˆ´ The direction cosines are 1/**âˆš**3, 1/**âˆš**3, 1/**âˆš**3

Coordinate of the foot (ld, md, nd) =

= [(1/**âˆš**3) (1/**âˆš**3), (1/**âˆš**3) (1/**âˆš**3), (1/**âˆš**3) (1/**âˆš**3)]

= 1/3, 1/3, 1/3

**(d) **5y + 8 = 0

Let the coordinate of the foot ofÂ âŠ¥Â P from the origin to the given plane be P(x, y, z).

0x â€“ 5y + 0z = 8 â€¦. (1)

Direction ratio are (0, -5, 0)

**âˆš**[(0)^{2} + (-5)^{2} + (0)^{2}] = **âˆš**(0 + 25 + 0)

= **âˆš**25

= 5

Now,

Divide both the sides of equation (1) by 5, we get

0x/(5) â€“ 5y/(5) + 0z/(5) = 8/5

So this is of the form lx + my + nz = d

Where, l, m, n are the direction cosines and d is the distance

âˆ´ The direction cosines are 0, -1, 0

Coordinate of the foot (ld, md, nd) =

= [(0/5) (8/5), (-5/5) (8/5), (0/5) (8/5)]

= 0, -8/5, 0

**5. Find the vector and Cartesian equations of the planes**

**(a) that passes through the point (1, 0, â€“2) and the normal to the plane is**

**(b) that passes through the point (1,4, 6) and the normal vector to the plane is**

**Â **

**Solution:**

x â€“ 1 â€“ 2y + 8 + z â€“ 6 = 0

x â€“ 2y + z + 1 = 0

x â€“ 2y + z = -1

âˆ´ The required Cartesian equation of the plane is x â€“ 2y + z = -1

x â€“ 1 â€“ 2y + 8 + z â€“ 6 = 0

x â€“ 2y + z + 1 = 0

x â€“ 2y + z = -1

âˆ´ The required Cartesian equation of the plane is x â€“ 2y + z = -1

**6. Find the equations of the planes that passes through three points.(a) (1, 1, â€“1), (6, 4, â€“5), (â€“4, â€“2, 3)**

**(b) (1, 1, 0), (1, 2, 1), (â€“2, 2, â€“1)**

**Solution: **

Given:

The points are (1, 1, -1), (6, 4, -5), (-4, -2, 3).

Let,

= 1(12 â€“ 10) â€“ 1(18 â€“ 20) -1 (-12 + 16)

= 2 + 2 â€“ 4

= 0

Since, the value of determinant is 0.

âˆ´ The points are collinear as there will be infinite planes passing through the given 3 points.

**(b) **(1, 1, 0), (1, 2, 1), (â€“2, 2, â€“1)

**7. Find the intercepts cut off by the plane 2x + y â€“ z = 5.**

**Solution:**

Given:

The plane 2x + y â€“ z = 5

Let us express the equation of the plane in intercept form

x/a + y/b + z/c = 1

Where a, b, c are the intercepts cut-off by the plane at x, y and z axes respectively.

2x + yÂ â€“Â z = 5 â€¦. (1)

Now divide both the sides of equation (1) by 5, we get

2x/5 + y/5 â€“ z/5 = 5/5

2x/5 + y/5 â€“ z/5 = 1

x/(5/2) + y/5 + z/(-5) = 1

Here, a = 5/2, b = 5 and c = -5

âˆ´ The intercepts cut-off by the plane are 5/2, 5 and -5.

**8. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.**

**Solution:**

We know that the equation of the plane ZOX is y = 0

So, the equation of plane parallel to ZOX is of the form, y = a

Since the y-intercept of the plane is 3, a = 3

âˆ´ The required equation of the plane is y = 3

**9. Find the equation of the plane through the intersection of the planes 3x â€“ y + 2z â€“ 4 = 0 and x + y + z â€“ 2 = 0 and the point (2, 2, 1).**

**Solution:**

Given:

Equation of the plane passes through the intersection of the plane is given by

(3x â€“ y + 2z â€“ 4) +Â Î» (x + y + z â€“ 2) = 0 and the plane passes through the points (2, 2, 1).

So, (3 Ã— 2 â€“ 2 + 2 Ã— 1 â€“ 4) +Â Î» (2 + 2 + 1Â â€“Â 2) = 0

2 + 3Î»Â = 0

3Î»Â = -2

Î» = -2/3 â€¦. (1)

Upon simplification, the required equation of the plane is given as

(3x â€“ y + 2z â€“ 4) â€“ 2/3 (x + y + z â€“ 2) = 0

(9x â€“ 3y + 6z â€“ 12 â€“ 2x â€“ 2y â€“ 2z + 4)/3 = 0

7x â€“ 5y + 4z â€“ 8 = 0

âˆ´ The required equation of the plane is 7x â€“ 5y + 4z â€“ 8 = 0

**10.** **Find the vector equation of the plane passing through the intersection of the planes and through the point (2, 1, 3).**

**Solution:**

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

**11. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x â€“ y + z = 0.**

**Solution:**

Let the equation of the plane that passes through the two-given planes

x + y + z = 1 and 2x + 3y + 4z = 5 is

(x + y + z â€“ 1) +Â Î» (2x + 3y + 4zÂ â€“Â 5) = 0

(2Î»Â + 1) x + (3Î»Â + 1) y + (4Î»Â + 1) z -1 â€“ 5Î»Â = 0â€¦â€¦ (1)

So the direction ratio of the plane is (2Î»Â + 1, 3Î»Â + 1, 4Î»Â + 1)

And direction ratio of another plane is (1, -1, 1)

Since, both the planes areÂ âŠ¥Â

So by substituting in a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

(2Î»Â + 1Â Ã—Â 1) + (3Î»Â + 1Â Ã—Â (-1)) + (4Î»Â + 1 Ã— 1) = 0

2Î»Â + 1 â€“ 3Î»Â â€“ 1 + 4Î»Â + 1 = 0

3Î» + 1 = 0

Î» = -1/3

Substitute the value ofÂ Î»Â in equation (1) we get,

x â€“ z + 2 = 0

âˆ´ The required equation of the plane is x â€“ z + 2 = 0

**12. Find the angle between the planes whose vector equations are**

**Solution:**

**13. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.(a) 7x + 5y + 6z + 30 = 0 and 3x â€“ y â€“ 10z + 4 = 0**

**(b) 2x + y + 3z â€“ 2 = 0 and x â€“ 2y + 5 = 0**

**(c) 2x â€“ 2y + 4z + 5 = 0 and 3x â€“ 3y + 6z â€“ 1 = 0**

**(d) 2x â€“ 2y + 4z + 5 = 0 and 3x â€“ 3y + 6z â€“ 1 = 0**

**(e) 4x + 8y + z â€“ 8 = 0 and y + z â€“ 4 = 0**

**Solution:**

**(a) **7x + 5y + 6z + 30 = 0 and 3x â€“ y â€“ 10z + 4 = 0

Given:

The equation of the given planes are

7x + 5y + 6z + 30 = 0 and 3x â€“ y â€“ 10z + 4 = 0

Two planes areÂ âŠ¥Â if the direction ratio of the normal to the plane is

a_{1}a_{2}Â + b_{1}b_{2}Â + c_{1}c_{2}Â = 0

21 â€“ 5 â€“ 60

-44 â‰ 0

Both the planes are notÂ âŠ¥Â to each other.

Now, two planes are || to each other if the direction ratio of the normal to the plane is

âˆ´ The angle is cos^{-1} (2/5)

**(b) **2x + y + 3z â€“ 2 = 0 and x â€“ 2y + 5 = 0

Given:

The equation of the given planes are

2x + y + 3z â€“ 2 = 0 and x â€“ 2y + 5 = 0

Two planes areÂ âŠ¥Â if the direction ratio of the normal to the plane is

a_{1}a_{2}Â + b_{1}b_{2}Â + c_{1}c_{2}Â = 0

2 Ã— 1 + 1 Ã— (-2) + 3 Ã— 0

= 0

âˆ´ The given planes areÂ âŠ¥Â to each other.

**(c) **2x â€“ 2y + 4z + 5 = 0 and 3x â€“ 3y + 6z â€“ 1 = 0

Given:

The equation of the given planes are

2x â€“ 2y + 4z + 5 =0 and x â€“ 2y + 5 = 0

We know that, two planes areÂ âŠ¥Â if the direction ratio of the normal to the plane is

a_{1}a_{2}Â + b_{1}b_{2}Â + c_{1}c_{2}Â = 0

6 + 6 + 24

36 â‰ 0

âˆ´Â Both the planes are notÂ âŠ¥Â to each other.

Now let us check, both planes are || to each other if the direction ratio of the normal to the plane is

âˆ´ The given planes are || to each other.

**(d) **2x â€“ 2y + 4z + 5 = 0 and 3x â€“ 3y + 6z â€“ 1 = 0

Given:

The equation of the given planes are

2x â€“ y + 3z â€“ 1 = 0 and 2x â€“ y + 3z + 3 = 0

We know that, two planes areÂ âŠ¥Â if the direction ratio of the normal to the plane is

a_{1}a_{2}Â + b_{1}b_{2}Â + c_{1}c_{2}Â = 0

2 Ã— 2 + (-1) Ã— (-1) + 3 Ã— 3

14 â‰ 0

âˆ´Â Both the planes are notÂ âŠ¥Â to each other.

Now, let us check two planes are || to each other if the direction ratio of the normal to the plane is

âˆ´ The given planes are || to each other.

**(e) **4x + 8y + z â€“ 8 = 0 and y + z â€“ 4 = 0

Given:

The equation of the given planes are

4x + 8y + z â€“ 8 = 0 and y + z â€“ 4 = 0

We know that, two planes areÂ âŠ¥Â if the direction ratio of the normal to the plane is

a_{1}a_{2}Â + b_{1}b_{2}Â + c_{1}c_{2}Â = 0

0 + 8 + 1

9 â‰ 0

âˆ´Â Both the planes are notÂ âŠ¥Â to each other.

Now let us check, two planes are || to each other if the direction ratio of the normal to the plane is

âˆ´Â Both the planes are not || to each other.

Now let us find the angle between them which is given as

âˆ´ The angle is 45^{o}.

**14. In the following cases, find the distance of each of the given points from the corresponding given plane.Point Plane(a) (0, 0, 0) 3x â€“ 4y + 12 z = 3**

**(b) (3, -2, 1) 2x â€“ y + 2z + 3 = 0**

**(c) (2, 3, -5) x + 2y â€“ 2z = 9**

**(d) (-6, 0, 0) 2x â€“ 3y + 6z â€“ 2 = 0**

**Solution:**

(a) Point Plane

(0, 0, 0) 3x â€“ 4y + 12 z = 3

We know that, distance of point P(x_{1}, y_{1}, z_{1}) from the plane Ax + By + Cz â€“ D = 0 is given as:

Given point is (0, 0, 0) and the plane is 3x â€“ 4y + 12z = 3

= |3/**âˆš**169|

= 3/13

âˆ´ The distance is 3/13.

**(b)** Point Plane

(3, -2, 1) 2x â€“ y + 2z + 3 = 0

We know that, distance of point P(x_{1}, y_{1}, z_{1}) from the plane Ax + By + Cz â€“ D = 0 is given as:

Given point is (3, -2, 1) and the plane is 2x â€“ y + 2z + 3 = 0

= |13/**âˆš**9|

= 13/3

âˆ´ The distance is 13/3.

**(c)** Point Plane

(2, 3, -5) x + 2y â€“ 2z = 9

We know that, distance of point P(x_{1}, y_{1}, z_{1}) from the plane Ax + By + Cz â€“ D = 0 is given as:

Given point is (2, 3, -5) and the plane is x + 2y â€“ 2z = 9

= |9/**âˆš**9|

= 9/3

= 3

âˆ´ The distance is 3.

**(d)** Point Plane

(-6, 0, 0) 2x â€“ 3y + 6z â€“ 2 = 0

_{1}, y_{1}, z_{1}) from the plane Ax + By + Cz â€“ D = 0 is given as:

Given point is (-6, 0, 0) and the plane is 2x â€“ 3y + 6z â€“ 2 = 0

= |14/**âˆš**49|

= 14/7

= 2

âˆ´ The distance is 2.