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  • Need solution for RD Sharma maths Class 12 Chapter 28 The Plane Exercise 28.9 Question 6 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 28 The Plane Exercise 28.9 Question 6 textbook solution.

Answers (1)

Answer : The answer of the given question are x-2 y+2 z+2=0, x-2 y+2 z-4=0

Hint :

       P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

Given :

       x-2 y+2 z-3=0 \text {, Point }(1,1,1)

Solution :

Since the planes are parallel to  x-2 y+2 z-3=0 they must be of the form x-2 y+2 z+\theta=0

We know, the distance of point \left(x_{1}, y_{1}, z_{1}\right) from the plane p: a x+b y+c z+d=0 is given by

P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

According to the question, the distance of the plane from  (1,1,1) is 1 Unit

\begin{aligned} &\left|\frac{1 \times 1+(-2) \times 1+2 \times 1+\theta}{\sqrt{1^{2}+(-2)^{2}+2^{2}}}\right|=1 \\ &\left|\frac{1+\theta}{3}\right|=1 \end{aligned}

\begin{aligned} &\frac{1+\theta}{3}=1 \text { Or } \frac{1+\theta}{3}=-1 \\ &\theta=2 \text { Or }-4 \end{aligned}

The required planes are

x-2 y+2 z+2=0 \& x-2 y+2 z-4=0

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