# RD Sharma Class 12 Ex 29.4 Solutions Chapter 29 The Plane

Here we provide RD Sharma Class 12 Ex 29.4 Solutions Chapter 29 The Plane for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 29.4 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 29.4 Solutions Chapter 29 The Plane

### Question 1. Find the vector equation of a plane which is at a distance of 3 units from the origin and hasas the unit vector normal to it.

Solution:

We know, the vector equation of a plane normal to unit vectorand at a distance of d from origin is given as

Here, d = 3 units ,we get

### Question 2. Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector

Solution:

We know, the vector equation of a plane normal to unit vectorand at a distance of d from origin is given as

Here, d = 5 units

and

Hence the required equation is,

### Question 3. Reduce the equation 2x – 3y – 6z = 14 to the normal form and hence find the length of perpendicular from the origin to the plane. Also, find the direction of cosines of normal to the plane.

Solution:

Dividing the equation by

……(1)

Since Vector equation of a plane with distance d and normal to the unit vector n is given by

……(2)

Comparing (1) and (2), we get

Distance from origin = 2 units

Direction cosine of normal to plane =

### Question 4. Reduce the equationto the normal form and hence find the length of perpendicular from origin to the plane.

Solution:

Multiplying both sides by –1, we get

…..(1)

Dividing (1) by 3 on both sides,

Since vector equation of a plane with distance d and normal to the unit vector n is given by

……(2)

Comparing (1) and (2), we get

d = 2

Length of normal = 2 units.

### Question 5. Write the normal form of the equation 2x – 3y + 6z + 14 = 0.

Solution:

Multiplying both sides by –1, we get

…..(1)

Dividing (1) by 7 on both sides,

Hence normal form of the equation is

### Question 6. The direction ratios of the perpendicular from the origin to a plane are 12, –3, 4 and the length of the perpendicular is 5. Find the equation of the plane.

Solution:

Normal vector =

So, Normal unit vector

Since vector equation of a plane with distance d and normal to the unit vector n is given by

or,

### Question 7. Find the unit normal vector to the plane x + 2y + 3z – 6 = 0.

Solution:

…….(1)

Dividing (1) by, we get

Since vector equation of a plane with distance d and normal to the unit vector n is given by

…..(2)

Thus, normal unit vector =

### Question 8. Find the equation of a plane which is at a distance of units from the origin and the normal to which is equally inclined with the coordinate axes.

Solution:

Since vector equation of a plane with distance d and normal to the unit vector n is given by

d =

Letbe a normal vector,

Sinceis equally inclined to the coordinate axes, let l, m, n be the cosines of. Also l = m = n.

We know, l2 + m2 + n2 = 1

or,

Now,

Vector equation of the required plane is

or,

or, x + y + z = 9.

### Question 9. Find the equation of the plane passing through the point (1,2,1) and perpendicular to the line joining the points (1,4,2) and (2,3,5). Find also the perpendicular distance of the origin from the plane.

Solution:

Vector equation of a plane is given by

…..(1)

We have,

and,

Puttingandin (1), we get

….(2)

Dividing (1) by

Hence, vector equation of plane is

and, cartesian form is x – y + 3z – 2 = 0.

### Question 10. Find the vector equation of a plane which is at a distance offrom the origin and its normal vector from the origin is. Also, find its cartesian form.

Solution:

Since vector equation of a plane with distance d and normal to the unit vector n is given by

Since,

Unit vector normal to the plane =

or,

Vector equation becomes,

Cartesian equation is 2x – 3y + 4z = 6.

### Question 11. Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin.

Solution:

2x – 3y + 4z – 6 = 0

or, 2x – 3y + 4z = 6

Vector equation becomes,

or,…..(1)

Dividing (1) by, normal form of equation becomes,

Hence, the perpendicular distance of the origin from the plane isunits.

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