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  • Explain solution RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 4 maths

Explain solution RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 4 maths

Answers (1)

Answer:

 Option (c).

Hint:

 Distance between two parallel planes is 0.

Given:

 2x+2y-z=0 \text { and } 4x+4y-2z+5=0

Solution:

We know that, the distance between two parallel planes:

Ax+By+Cz+D_1=0 \qquad \qquad \dots(1)

and

Ax+By+Cz+D_2=0 \qquad \qquad \dots(2)

is given by

D=\frac{\left | D_2-D_{1} \right |}{\sqrt{A^2+B^2+C^2}}

Here, the two parallel planes are given as:

2x+2y-z+2=0 \qquad \qquad \dots (3)

and

4x+4y-2z+5=0

i.e.

2x+2y-z+\frac{5}{2}=0 \qquad \qquad \dots (4)

Compiling equation (3) with equation (1), and equation (4) with equation (2), we get;

A = 2,B=2,C=-1,D_1=2,D_2=\frac{5}{2}.

So, the distance between the given two parallel planes are,

\begin{aligned} &D=\frac{\left | \frac{5}{2}-2 \right |}{\sqrt{2^2+2^2+(-1)^2}}\\ &D=\frac{\frac{1}{2}}{\sqrt{4+4+1}}\\ &D=\frac{1}{2\sqrt{9}}\\ &D=\frac{1}{2\times 3}\\ &D=\frac{1}{6} \end{aligned}

Hence, the distance between the parallel planes

2x+2y-z=0 \text { and } 4x+4y-2z+5=0 \text { is } \frac{1}{6}.

Posted by

Gurleen Kaur

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