RD Sharma Class 12 Ex 29.4 Solutions Chapter 29 The Plane

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.4
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 29.4 Solutions Chapter 29 The Plane

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Find the vector equation of a plane which is at a distance of 3 units from the origin and has\hat{k} as the unit vector normal to it.

Solution:

We know, the vector equation of a plane normal to unit vector\hat{n} and at a distance of d from origin is given as

\vec{r}.\hat{n} = d

Here, d = 3 units ,we get

\vec{r}.\hat{k} = 3

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\hat{i} - 2\hat{j} - 2\hat{k}.

Solution:

We know, the vector equation of a plane normal to unit vector\hat{n} and at a distance of d from origin is given as

\vec{r}.\hat{n} = d

Here, d = 5 units

and\vec{n} = \hat{i} - 2\hat{j} - 2\hat{k}.

\hat{n} = \frac{\vec{n}}{\vec{|n|}}
= \frac{\hat{i} - 2\hat{j} - 2\hat{k}}{\sqrt{1^2 + (-2)^2 + (-2)^2}}
= \frac{\hat{i} - 2\hat{j} - 2\hat{k}}{\sqrt{9}}
= \frac{\hat{i} - 2\hat{j} - 2\hat{k}}{3}

Hence the required equation is,

\vec{r}.\frac{1}{3}(\hat{i} - 2\hat{j} - 2\hat{k}) = 5

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:

(x\hat{i} + y\hat{j} + z\hat{k}).(2\hat{i} - 3\hat{j} - 6\hat{k}) = 14

Dividing the equation by\sqrt{2^2 + (- 3)^2 + (- 6)^2}

\vec{r}.\frac{(2\hat{i}- 3\hat{j} - 6\hat{k})}{\sqrt{4 + 9 + 36}} = \frac{14}{\sqrt{4 + 9 + 36}}

⇒ \vec{r}.[\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}] = 2 ……(1)

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

\vec{r}.\hat{n} = d ……(2)

Comparing (1) and (2), we get

Distance from origin = 2 units

Direction cosine of normal to plane =\frac{2}{7}, - \frac{3}{7}, \frac{6}{7}

Question 4. Reduce the equation\vec{r}.(\hat{i} - 2\hat{j} + 2\hat{k}) + 6 = 0 to the normal form and hence find the length of perpendicular from origin to the plane.

Solution:

Given: \vec{r}.(\hat{i} - 2\hat{j} + 2\hat{k}) = -6

Multiplying both sides by –1, we get

\vec{r}.(\hat{-i} + 2\hat{j} - 2\hat{k}) = 6

⇒ \vec{r}.\vec{n} = 6 …..(1)

|\vec{n}| = \sqrt{(-1)^2 + 2^2 + (-2)^2} = 3

Dividing (1) by 3 on both sides,

\vec{r}.[-\frac{\hat{i}}{3} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}] = 2

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

\vec{r}.\hat{n} = d ……(2)

Comparing (1) and (2), we get

d = 2

\hat{n} = [-\frac{\hat{i}}{3} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}]

Length of normal = 2 units.

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

Solution:

(x\hat{i} + y\hat{j} + z\hat{k}).(2\hat{i} - 3\hat{j} + 6\hat{k}) = -14
⇒ \hat{r}.(2\hat{i} - 3\hat{j} + 6\hat{k}) = -14

Multiplying both sides by –1, we get

\vec{r}.(-2\hat{i} + 3\hat{j} - 6\hat{k}) = 14

⇒ \vec{r}.\vec{n} = 14 …..(1)

|\vec{n}| = \sqrt{(-2)^2 + 3^2 + (-6)^2} = 7

Dividing (1) by 7 on both sides,

\vec{r}.[-\frac{2}{3}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}] = 2

Hence normal form of the equation is

-\frac{2}{7}x + \frac{3}{7}y - \frac{6}{7}z = 2

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 =\vec{n} = 12\hat{i} - 3\hat{j} + 4\hat{k}

|\vec{n}| = \sqrt{(12)^2 + (-3)^2 + (4)^2} = \sqrt{169} = 13

So, Normal unit vector= \hat{n} = \frac{\vec{n}}{ |\vec{n}|}

= \frac{12}{13}\hat{i} - \frac{3}{13}\hat{j} + \frac{4}{13}\hat{k} = 5

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

\vec{r}.\hat{n} = d
⇒ \vec{r}.[\frac{12}{13}\hat{i} - \frac{3}{13}\hat{j} + \frac{4}{13}\hat{k}] = 5

or,\frac{12}{13}x - \frac{3}{13}y + \frac{4}{13}z = 5

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

Solution:

(x\hat{i} + y\hat{j} + z\hat{k}).(\hat{i} + 2\hat{j} + 3\hat{k}) = 6
⇒ \hat{r}.(\hat{i} + 2\hat{j} + 3\hat{k}) = 6

⇒ \vec{r}.\vec{n} = 6 …….(1)

|\vec{n}| = \sqrt{(1)^2 + 2^2 + (3)^2} = \sqrt{14}

Dividing (1) by\sqrt{14} , we get

\vec{r}.[\frac{1}{\sqrt14}\hat{i} + \frac{2}{\sqrt14}\hat{j} + \frac{3}{\sqrt14}\hat{k}] = \frac{6}{\sqrt14}

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

\vec{r}.\hat{n} = d …..(2)

Thus, normal unit vector =\hat{n} = \frac{1}{\sqrt14}\hat{i} + \frac{2}{\sqrt14}\hat{j} + \frac{3}{\sqrt14}\hat{k}

Question 8. Find the equation of a plane which is at a distance of 3\sqrt{3} 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

\vec{r}.\hat{n} = d

d =3\sqrt{3} units

Let\vec{a} be a normal vector,\vec{a} = p\hat{i} + q\hat{j} + r\hat{k}

Since\vec{a} is equally inclined to the coordinate axes, let l, m, n be the cosines of\vec{n} . Also l = m = n.

We know, l2 + m2 + n2 = 1

or,l = m = n = \frac{1}{\sqrt{3}}

Now,\hat{a} = \frac{\hat{i}}{\sqrt{3}} + \frac{\hat{j}}{\sqrt{3}} + \frac{\hat{k}}{\sqrt{3}}

Vector equation of the required plane is

\vec{r}.\frac{1}{\sqrt3}[\hat{i} + \hat{j} + \hat{k}] = 3\sqrt{3}

or,\vec{r}(\hat{i} + \hat{j} + \hat{k}) = 9

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

(\vec{r} - \vec{a}).\vec{n} = 0 …..(1)

We have,\vec{a} = \hat{i} - 2\hat{j} + \hat{k}

and,\vec{n} = \vec{BC} = \hat{i} - \hat{j} + 3\hat{k}

Putting\vec{a} and\vec{n} in (1), we get

\vec{r}(\hat{i} - \hat{j} + 3\hat{k}) = 2 ….(2)

|\vec{n}| = \sqrt{(1)^2 + 1^2 + (3)^2} = \sqrt{11}

Dividing (1) by\sqrt{11}

\vec{r}.[\frac{1}{\sqrt11}\hat{i} - \frac{1}{\sqrt11}\hat{j} + \frac{3}{\sqrt11}\hat{k}] = \frac{2}{\sqrt11}

Hence, vector equation of plane is\vec{r}.(\hat{i} - \hat{j} + 3\hat{k}) = 2.

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

Question 10. Find the vector equation of a plane which is at a distance of\frac{6}{\sqrt{29}} from the origin and its normal vector from the origin is2\hat{i} - 3\hat{j} + 4\hat{k} 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

\vec{r}.\hat{n} = d

Since,|\vec{n}| = \sqrt{(2)^2 + (-3)^2 + (4)^2} = \sqrt{29}

Unit vector normal to the plane =\frac{\vec{n}}{|\vec{n}|}

or,\hat{n} = \frac{2}{\sqrt29}\hat{i} - \frac{3}{\sqrt29}\hat{j} + \frac{4}{\sqrt29}\hat{k}

d = \frac{6}{\sqrt29}

Vector equation becomes,

\vec{r}.[\frac{2}{\sqrt29}\hat{i} - \frac{3}{\sqrt29}\hat{j} + \frac{4}{\sqrt29}\hat{k}] = \frac{6}{\sqrt29}

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,

(x\hat{i} + y\hat{j} + z\hat{k}).(2\hat{i} - 3\hat{j} + 4\hat{k}) = 6

or,\vec{r}.(2\hat{i} - 3\hat{j} + 4\hat{k}) = 6 …..(1)

|\vec{n}| = \sqrt{(2)^2 + (-3)^2 + (4)^2} = \sqrt{29}

Dividing (1) by|\vec{n}| , normal form of equation becomes,

\vec{r}.[\frac{2}{\sqrt29}\hat{i} - \frac{3}{\sqrt29}\hat{j} + \frac{4}{\sqrt29}\hat{k}] = \frac{6}{\sqrt29}

Hence, the perpendicular distance of the origin from the plane is\frac{6}{\sqrt29} units.

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