**NCERT Solutions for Maths Class 12 Exercise 11.3**

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**NCERT Solutions for Maths Class 12 Exercise 11.3** helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

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**NCERT Solutions for Maths Class 12 Exercise 11.1**

**NCERT Solutions for Maths Class 12 Exercise 11.2**

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**NCERT Solutions for Maths Class 12 Exercise 11.3**

**Maths Class
12 Ex 11.3 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.

Here, a = 0, b = 0, c = 1

**Maths Class 12 Ex 11.3 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** i** + 5

**– 6**

*j***.**

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

Find the
Cartesian equation of the following planes.

**Solution:**

**(a)** Let ** r** is
the position vector of any arbitrary point P(x, y, z) on the plane.

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

In the following
cases, find the coordinates of the foot of the 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(x_{1}, y_{1}, z_{1})
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 (x_{1}, y_{1}, z_{1})

**Maths Class
12 Ex 11.3 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 ** i**
+

**–**

*j***.**

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

**– 2**

*i***+**

*j***.**

*k***Solution:**

**(a)** Normal to the plane is ** i** +

**–**

*j***and passes through (1, 0, –2)**

*k***Maths Class
12 Ex 11.3 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

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

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

**Solution:**

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

Dividing by 5:

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

**Maths Class
12 Ex 11.3 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.

**Maths Class
12 Ex 11.3 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 + Î»(x + y + z – 2) = 0
… (i)

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

∴
3 × 2 – 2 + 2 × 1
– 4 + Î»(2 + 2 + 1 – 2) = 0

Thus, Î» = –2/3

Putting the value of Î» in equation (i), we get the required equation is 7x – 5y
+ 4z – 8 = 0.

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

Find the vector
equation of the plane passing through the intersection of the planes ** r**.(2

**+ 2**

*i***– 3**

*j***) = 7 and**

*k***.(2**

*r***+ 5**

*i***+ 3**

*j***) = 9 and through the point (2, 1, 3).**

*k***Solution:**

The equation of
the plane passing through the line of intersection of the planes

**Maths Class
12 Ex 11.3 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)

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

Find the angle
between the planes whose vector equations are ** r**.(2

**+ 2**

*i***– 3**

*j***) = 5 and**

*k***.(3**

*r***– 3**

*i***+ 5**

*j***) = 3.**

*k***Solution:**

The angle Î¸
between the given planes is

**Maths Class
12 Ex 11.3 Question 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 – 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)

**Maths Class
12 Ex 11.3 Question 14.**

In the following
cases, find the distance of each of the given points from the corresponding given
plane.

(a) Point (0, 0, 0); Plane 3x – 4y + 12z = 3

(b) Point (3, –2, 1); Plane 2x – y + 2z + 3 = 0

(c) Point (2, 3, –5); Plane x + 2y – 2z = 9

(d) Point (–6, 0, 0); Plane 2x – 3y + 6z – 2 = 0

**Solution:**

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