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Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.9

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  • Last Updated : 12 Oct, 2021
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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 D1 be 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 D2 be 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 D1 be 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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