RD Sharma Class 12 Ex 29.9 Solutions Chapter 29 The Plane

Here we provide RD Sharma Class 12 Ex 29.9 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 29.9 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.9
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 29.9 Solutions Chapter 29 The Plane

Question 1. Find the distance of the point 2\hat{i}-\hat{j}-4\hat{k}   from the plane \vec{r}.(3\hat{i}-4\hat{j}+12\hat{k})-9=0.

Solution:

As we know the distance of a point \vec{a}   from a plane \vec{r}.\vec{n}=d   is given by:

D=|\frac{\vec{a}\vec{n}-d}{|\vec{n}|}|units

Here, a = 2\hat{i}-\hat{j}-4\hat{k}   and \vec{r}.(3\hat{i}-4\hat{j}+12\hat{k})-9=0   is the plane.

Hence, D=|\frac{(2\hat{i}-\hat{j}-4\hat{k})(3\hat{i}-4\hat{j}+12\hat{k})-9}{\sqrt{(3)^2+(-4)^2+(12)^2}}|

⇒ D=|\frac{(2)(3)+(-1)(-4)+(-4)(12)-9}{\sqrt{9+16+144}}|

= |-47/13| units 

= 47/13 units 

Hence, the distance of the point from the plane is 47/13 units.

Question 2. Show that the points \hat{i}-\hat{j}+3\hat{k}   and 3\hat{i}+3\hat{j}+3\hat{k}   are equidistant from the plane \vec{r}.(5\hat{i}+2\hat{j}-7\hat{k})+9=0.

Solution:

As we know the distance of a point \vec{a}   from a plane \vec{r}.\vec{n}=d   is given by:

D=|\frac{\vec{a}\vec{n}-d}{|\vec{n}|}|units

Let Dbe the distance of \hat{i}-\hat{j}+3\hat{k}   from the plane \vec{r}.(5\hat{i}+2\hat{j}-7\hat{k})+9=0.

⇒ D_1=|\frac{(\hat{i}-\hat{j}+3\hat{k})(5\hat{i}+2\hat{j}-7\hat{k})+9}{\sqrt{5)^2+(2)^2+(-7)^2}}|

|\frac{5-2-21+9}{\sqrt{78}}|

= 9/√78 units                  …….(1)

Now, let Dbe the distance between point 3\hat{i}+3\hat{j}+3\hat{k}   and the plane \vec{r}.(5\hat{i}+2\hat{j}-7\hat{k})+9=0  .

⇒ D_2=|\frac{(3\hat{i}+3\hat{j}+3\hat{k})(5\hat{i}+2\hat{j}-7\hat{k})+9)}{\sqrt{5)^2+(2)^2+(-7)^2}}|

|\frac{15+6-21+9}{\sqrt{78}}|

= 9/√78 units                    ……(2)

From eq(1) and (2), we have

The given points are equidistant from the given plane.

Question 3. Find the distance of the point (2, 3, −5) from the plane x + 2y − 2z − 9 = 0.

Solution:

As we know the distance is given by:

|\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|

⇒ D=|\frac{2+(2)(3)-2(-5)-9}{\sqrt{(1)^2+(2)^2+(-2)^2}}|

|\frac{2+6+10-9}{\sqrt{1+4+4}}|

= 9/√9 

D = 3 units.

Question 4. Find the equations of the planes parallel to the plane x + 2y − 2z + 8 = 0 which are at a distance of 2 units from the point (2, 1, 1).

Solution:

The equation of a plane parallel to the given plane is x + 2y − 2z + p = 0.

As we know that the distance between a point and plane is given by:

D=|\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|

Given, D = 2 units. Hence,

2=|\frac{2+2-2+p}{\sqrt{1+4+4}}|

⇒ 2=|\frac{2+p}{\sqrt9}|

On squaring both sides, we have

4=\frac{(2+p)^2}{9}

⇒ 36 = (2 + p)2

⇒ 2 + p = 6           or         2 + p = −6

⇒ p = 4                 or           p = −8

Hence, the equations of the required planes are:

x + 2y − 2z + 4 = 0 and x + 2y − 2z − 8 = 0.

Question 5. Show that the points (1, 1, 1) and(−3, 0, 1) are equidistant from the plane 3x + 4y − 12z +13 = 0.

Solution:

We know that the distance between a point and plane is given by:

D=|\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|

Let Dbe the distance of the point (1,1,1) from the plane.

⇒ D_1=|\frac{(3)(1)+(4)(1)-(12)(1)+13}{\sqrt{(3)^2+(4)^2+(-12)^2}}|

= 8/13 units 

Let D2 be the distance of the point.

⇒ D_2=|\frac{(3)(-3)+(4)(0)-(12)(1)+13}{\sqrt{(3)^2+(4)^2+(-12)^2}}|

= 8/13 units 

Hence, the points are equidistant from the plane.

Question 6. Find the equation of the planes parallel to the plane x − 2y + 2z − 3 = 0 which are at a unit distance from the point (2, 1, 1).

Solution:

Equation of a plane parallel to the given plane is x − 2y + 2z + p = 0.

As we know that the distance between a point and plane is given by:

D=|\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|

Given, D = 1 units. Hence,

1=|\frac{1+2-2+p}{\sqrt{1+4+4}}|

⇒ 1=|\frac{1+p}{\sqrt9}|

On squaring both sides, we have

1=\frac{(1+p)^2}{3}

⇒ 9 = (1 + p)2

⇒ 1 + p = 3           or         1 + p = −3

⇒ p = 2                 or           p = −4

Hence, the equations of the required planes are:

x − 2y + 2z + 2 = 0 and x − 2y + 2z − 4 = 0.

Question 7. Find the distance of the point (2, 3, 5) from the xy-plane.

Solution:

As we know the distance of the point from the plane is given by:

D = |\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|

|\frac{(2)(0)+(3)(0)+(5)(1)+0}{\sqrt{(0)^2+(0)^2+(1)^2}}|

|\frac{0+0+5}{1}|

D = 5 units

Question 8. Find the distance of the point (3, 3, 3) from the plane \vec{r}.(5\hat{i}+2\hat{j}+3\hat{k})+9=0.

Solution:

As we know the distance of a point \vec{a}  from a plane \vec{r}.\vec{n}=d  is given by:

D = |\frac{\vec{a}\vec{n}-d}{|\vec{n}|}|units

|\frac{(3\hat{i}+3\hat{j}+3\hat{k})(5\hat{i}+2\hat{j}+3\hat{k})+9}{\sqrt{25+4+49}}|

|\frac{15+6-21+9}{\sqrt{78}}|

D = 9/√78 units.

Question 9. If the product of distances of the point (1, 1, 1) from the origin and the plane x − y + z + p = 0 be 5, find p.

Solution:

The distance of the point (1, 1, 1) from the origin is \sqrt{3}.

Distance of (1, 1, 1) from the plane is |\frac{1+p}{\sqrt{3}}|

Given: \sqrt{3}.|\frac{1+p}{\sqrt{3}}|=5

⇒ |1 + p| = 5

⇒ p = 4 or −6.

Question 10. Find an equation of the set of all points that are equidistant from the planes 3x − 4y + 12 = 6 and 4x + 3z = 7.

Solution:

D_1=|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}|
=|\frac{3x_1-4y_1+12z_1-6}{\sqrt{3^2+)(-4)^2+12^2}}|

|\frac{3x_1-4y_1+12z_1-6}{13}|

Now, D_2=|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}|

|\frac{4x_1+3z_1-7}{5}|

Given, D1 = D2

⇒  |\frac{3x_1-4y_1+12z_1-6}{13}|=|\frac{4x_1+3z_1-7}{5}|

Hence, the equations become

37x1 + 20y1 − 21z1 − 61 = 0 and  67x1 + 20y1 + 99z1 − 121 = 0.

Question 11. Find the distance between the point (7, 2, 4) and the plane determined by the points A(−2, −3, 5) and C(5, 3, −3).

Solution:

The equations of the plane are given as:

−4a − 8b + 8c = 0 and 3a − 2b + 0c = 0

Solving the above set using cross multiplication method, we get

\frac{a}{(-8)(0)-(-2)(8)}=\frac{b}{3(-8)-(-4)(0)}=\frac{c}{(-4)(-2)-(-3)(-8)}=p

⇒ \frac{a}{0+16}=\frac{b}{24+0}=\frac{c}{8+24}=p

⇒ \frac{a}{16}=\frac{b}{24}=\frac{c}{32}=p

⇒ \frac{a}{2}=\frac{b}{3}=\frac{c}{4}=p

⇒ a = 2p, b = 3p, c = 4p

Thus, the equation of the plane becomes 2x + 3y + 4z − 7 = 0.

and, distance = √29 units.

Question 12. A plane makes intercepts −6, 3, 4 respectively on the coordinate axis. Find the length of the perpendicular from the origin on it.

Solution:

Since the plane makes intercepts −6, 3, 4, the equation becomes:

\frac{x}{-6}+\frac{y}{3}+\frac{z}{4}=1

Let p be the distance of perpendicular from the origin to the plane.

⇒ \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}

⇒ \frac{1}{p^2}=\frac{1}{(-6)^2}+\frac{1}{3^2}+\frac{1}{4^2}

⇒ \frac{1}{p^2}=\frac{4+16+9}{144}

⇒ 1/p2 = 29/144 

⇒ p2 = 144/29

⇒ p = 12/√29 units.

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