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  • Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.9 Question 13 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.9 Question 13 maths textbook solution.

Answers (1)

Answer : 2\sqrt{2}

Hint :

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

Given : We have to find the distance of the point (1,-2,4) from the plane passing through the point (1,2,2)

Solution : Let us consider the equation of plane passing through the point (1,2,2) be

a(x-1)+b(y-2)+c(z-2)=0                      (1)

The direction ratios of the normal to the plane are a, b and c

The equation of the plane are

\begin{aligned} &x-y+2 z=3 \\ &2 x-2 y+z+12=0 \end{aligned}

Since the plane passing through the point (1,2,2) is perpendicular the given plane

Therefore,

\begin{aligned} &a-b+2 c=0\\ &2 a-2 b+c=0 \end{aligned}

Now, eliminating a, b and c from (1), (2) and (3)

We get

\begin{aligned} &\left|\begin{array}{ccc} x-1 & y-2 & z-2 \\ 1 & -1 & 2 \\ 2 & -2 & 1 \end{array}\right|=0 \\ &(x-1)(-1+4)-(y-2)(1-4)+(z-2)(-2+2)=0 \end{aligned}

\begin{aligned} &3(x-1)+3(y-2)+0=0 \\ &3 x-3+3 y-6=0 \\ &3 x+3 y-9=0 \\ &x+y-3=0 \end{aligned}

Now using distance formula

The distance of the point (1,-2,4) from the plane x+y-3=0 is

\begin{aligned} &=\left|\frac{1-2-3}{\sqrt{1^{2}+1^{2}+0^{2}}}\right| \\ &=\left|\frac{-4}{\sqrt{2}}\right| \\ &=2 \sqrt{2} \text { Units } \end{aligned}

 

Posted by

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