# NCERT Solutions For Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 12 |

Subject | Maths |

Chapter | Chapter 11 |

Exercise | Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.3 |

Number of Questions Solved | 14 |

## NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

**NCERT TEXTBOOK EXERCISES**

**Ex 11.3 Class 12 Maths Question 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 – z = 5

(d) 5y+8 = 0

**Solution:**

(a) Direction ratios of the normal to the plane are 0,0,1

=> a = 0, b = 0, c = 1

**Ex 11.3 Class 12 Maths Question 2.**

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3\hat { i } +5\hat { j } -6\hat { k }$

**Solution:**

**Ex 11.3 Class 12 Maths Question 3.**

Find the Cartesian equation of the following planes.

(a) $\overrightarrow { r } \cdot (\hat { i } +\hat { j } -\hat { k) } =2$

(b) $\overrightarrow { r } \cdot (\hat { 2i } +3\hat { j } -4\hat { k) } =1$

(c) $\overrightarrow { r } \cdot [(s-2t)\hat { i } +(3-t)\hat { j } +(2s+t)\hat { k) } =15$

**Solution:**

(a) $\overrightarrow { r }$ is the position vector of any arbitrary point P (x, y, z) on the plane.

**Ex 11.3 Class 12 Maths Question 4.**

In the following cases find the coordinates of the foot of perpendicular drawn from the origin

(a) 2x + 3y + 4z – 12 = 0

(b) 3y + 4z – 6 = 0

(c) x + y + z = 1

(d) 5y + 8 = 0

**Solution:**

(a) Let N (x1, y1, z1) be the foot of the perpendicular from the origin to the plane 2x+3y+4z-12 = 0

∴ Direction ratios of the normal are 2, 3, 4.

Also the direction ratios of ON are (x1,y1,z1)

**Ex 11.3 Class 12 Maths Question 5.**

Find the vector and cartesian equation of the planes

(a) that passes through the point (1,0, -2) and the normal to the plane is $\hat { i } +\hat { j } -\hat { k }$

(b) that passes through the point (1,4,6) and the normal vector to the plane is $\hat { i } -2\hat { j } +\hat { k }$

**Solution:**

(a) Normal to the plane is i + j – k and passes through (1,0,-2)

**Ex 11.3 Class 12 Maths Question 6.**

Find the equations of the planes that passes through three points

(a) (1,1,-1) (6,4,-5), (-4, -2,3)

(b) (1,1,0), (1,2,1), (-2,2,-1)

**Solution:**

(a) The plane passes through the points (1,1,-1) (6,4,-5), (-4,-2,3)

Let the equation of the plane passing through(1,1,-1)be

**Ex 11.3 Class 12 Maths Question 7.**

Find the intercepts cut off by the plane 2x+y-z = 5.

**Solution:**

Equation of the plane is 2x + y- z = 5 x y z

Dividing by 5: $\Rightarrow \frac { x }{ \frac { 5 }{ 2 } } +\frac { y }{ 5 } -\frac { z }{ -5 } =1$

∴ The intercepts on the axes OX, OY, OZ are 5/2, 5, -5 respectively

**Ex 11.3 Class 12 Maths Question 8.**

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

**Solution:**

Any plane parallel to ZOX plane is y=b where b is the intercept on y-axis.

∴ b = 3.

Hence equation of the required plane is y = 3.

**Ex 11.3 Class 12 Maths Question 9.**

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

**Solution:**

Given planes are:

3x – y + 2z – 4 = 0 and x + y + z – 2 = 0

Any plane through their intersection is

3x – y + 2z – 4 + λ(x + y + z – 2) = 0

point (2,2,1) lies on it,

∴3 x 2 – 2 + 2 x 1 – 4 +λ(2+2+1-2)=0

=>λ = −2/3

Now required equation is 7x – 5y + 4z – 8 = 0

**Ex 11.3 Class 12 Maths Question 10.**

Find the vector equation of the plane passing through the intersection of the planes $\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =7,\overrightarrow { r } \cdot \left( 2\hat { i } +5\hat { j } +3\hat { k } \right) =9$ and through the point (2,1,3).

**Solution:**

Equation of the plane passing through the line of intersection of the planes

**Ex 11.3 Class 12 Maths Question 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 – y + z = 0.

**Solution:**

Given planes are

x + y + z – 1 = 0 …(i)

2x + 3y + 4z – 5 = 0 …(ii)

x – y + z = 0 ….(iii)

**Ex 11.3 Class 12 Maths Question 12.**

Find the angle between the planes whose vector equations are $\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =5,\overrightarrow { r } \cdot \left( 3\hat { i } -3\hat { j } +5\hat { k } \right) =3$

**Solution:**

The angle θ between the given planes is

**Ex 11.3 Class 12 Maths Question 13.**

In the following determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angle between them.

(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0.

**Solution:**

(a) Direction ratios of the normal of the planes 7x + 5y + 6z + 30 = 0 are 7,5,6

Direction ratios of the normal of the plane 3x – y – 10z + 4 = 0 are 3,-1,-10

The plane 7x + 5y + 6z + 30 = 0 …(i)

3x – y – 10z + y = 0 …(ii)

**Ex 11.3 Class 12 Maths Question 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 – 4y + 12z = 3

(b) (3,-2,1) 2x – y + 2z + 3 = 0.

(c) (2,3,-5) x + 2y – 2z = 9

(d) (-6,0,0) 2x – 3y + 6z – 2 = 0

**Solution:**

(a) Given plane: 3x – 4y + 12z – 3 = 0

## NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.3 PDF

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