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Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.10

  • Last Updated : 19 Jan, 2021

Question 1. Find the distance between the parallel planes 2x – y + 3z – 4 = 0 and 6x – 3y + 9z + 13 = 0

Solution:

Let P(x1, y1, z1) be any point on plane 2x – y + 3z – 4 = 0.

⟹ 2x1 – y1 + 3z1 = 4 (equation-1)

Distance between (x1, y1, z1) and the plane

6x – 3y + 9z + 13 = 0:

As we know that, the distance of point (x1, y1, z1) from the plane π: ax + by + cz + d = 0 is given by:

p = \left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|

Now, substitute the values, we get

p = \left|\frac{(6)(x_1)+(-3)(y_1)+(9)(z_1)+13}{\sqrt{6^2+(-3)^2+9^2}}\right|

\left|\frac{3(2x_1-y_1+3z_1)+13}{\sqrt{6^2+(-3)^2+9^2}}\right|

 \left|\frac{3(4)+13}{3\sqrt{14}}\right|   [by using equation 1]

\frac{25}{3\sqrt{14}}

Therefore, the distance between the parallel planes 2x – y + 3z – 4 = 0 and 6x – 3y + 9z + 13 = 0 is \frac{25}{3\sqrt{14}}  units.

Question 2. Find the equation of the plane which passes through the point (3, 4, -1) and is parallel to the plane 2x – 3y + 5z + 7 = 0. Also find the distance between the two planes.

Solution:

Since the plane is parallel to 2x – 3y + 5z + 7 = 0, it must be of the form:

2x – 3y + 5z + θ = 0

It is given that,

The plane passes through (3, 4, –1)

⟹ 2(3) – 3(4) +5(–1) + θ = 0

θ = -11

Thus, 

The equation of the plane is as follows:

2x – 3y + 5z – 11 = 0

Distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1):

As we know that, the distance of point (x1, y1, z1) from the plane π: ax + by + cz + d = 0 is given by:

p = \left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|

Now, after substituting the values, we will get

\left|\frac{(2)(3)+(-3)(4)+(5)(-1)+7}{\sqrt{2^2+(-3)^2+5^2}}\right|

\frac{4}{\sqrt{38}}

Therefore, the distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, -1) is \frac{4}{\sqrt{38}}

Question 3. Find the equation of the plane mid-parallel to the planes 2x – 2y + z + 3 = 0 and 2x – 2y + z + 9 = 0

Solution:

Given:

Equation of planes: 

π1= 2x – 2y + z + 3 = 0

π2= 2x – 2y + z + 9 = 0

Let the equation of the plane mid–parallel to these planes be:

π3: 2x – 2y + z + θ = 0

Now,

Let P(x1, y1, z1) be any point on this plane,

⟹ 2(x1) – 2(y1) + (z1) + θ = 0 —(equation-1)

As we know that, the distance of point (x1, y1, z1) from the plane π: ax + by + cz + d = 0 is given by:

p = \left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|

Distance of P from π1:

p = \left|\frac{(2x_1-2y_2+z_1)+3}{\sqrt{2^2+(-2)^2+1^2}}\right|

=  \left|\frac{(-θ)+3}{3}\right|   (By using equation 1)

Similarly,

DIstance of q from π2:

q = \left|\frac{(2x_1-2y_2+z_1)+9}{\sqrt{2^2+(-2)^2+1^2}}\right|

\left|\frac{(-θ)+9}{3}\right|    (By using equation 1)

As π3 is mid-parallel is π1 and π2:

p = q 

So,

\left|\frac{(-θ)+3}{3}\right|=\left|\frac{-θ +9}{3}\right|

Now square on both sides, we get

\left(\frac{(-θ)+3}{3}\right)^2=\left(\frac{-θ+9}{3}\right)^2

(3 – θ)2 = (9 – θ)2

9 – 2×3×θ + θ2 = 81 – 2×9×θ + θ2

θ = 6

Now, substitute the value of θ = 6 in equation 2x – 2y + z + θ = 0, we get

Hence, the equation of the mid-parallel plane is 2x – 2y + z + 6 = 0

Question 4. Find the distance between the planes \vec{r}.(\hat{i}+2\hat{j}+3\hat{k})+7=0  and \vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0

Solution:

Let \vec{a}  be the position vector of any pont P on the plane

\vec{r}(\hat{i}+2\hat{j}+3\hat{k})+7=0

So, 

\vec{a}(\hat{i}+2\hat{j}+3\hat{k})+7=0  —(equation 1)

As we know that, the distance of from  \vec{a}  the plane  \vec{r}.\vec{n}-d=0  is given by:

p = \left|\frac{\vec{a}.\vec{n}-d}{|n|}\right|

Length of perpendicular from P(\vec{a}) to plane \vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0  is given by substituting the values of \vec{a}\  and\  \vec{n} , we get

p = \left|\frac{2\vec{a}.(2\hat{i}+4\hat{j}+6\hat{k})+7}{|2\hat{i}+4\hat{j}+6\hat{k}|}\right|

\left|\frac{2\vec{a}.(\hat{i}+2\hat{j}+3\hat{k})+7}{\sqrt{2^2+4^2+6^2}}\right|

\left|\frac{2(-7)+7}{\sqrt{56}}\right|

p = \frac{7}{\sqrt{56}}

Therefore, the distance between the planes

\vec{r}.(\hat{i}+2\hat{j}+3\hat{k})+7=0  and  \vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0  is \frac{7}{\sqrt{56}}


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