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  • Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise very short answer type question 23

Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise very short answer type question 23

Answers (1)

Answer:

 x + y + z = 15

Hint:

 we will use equation of plane as lx + my + nz = d

Given:

 Distance from the origin 5\sqrt{3}

Solution:

Since the normal of plane equation inclined to co-ordinate axes

\therefore cos \: \alpha =cos\: \beta =cos\: \gamma

We know that

\begin{aligned} &cos^2 \: \alpha +cos^2\: \beta +cos^2\: \gamma =1\\ &\Rightarrow 3cos^2\alpha =1\\ &\Rightarrow cos\: \alpha =\frac{1}{\sqrt{3}}\\ &cos \: \alpha =cos\: \beta =cos\: \gamma =\frac{1}{\sqrt{3}}\\ &l=\frac{1}{\sqrt{3}},\; m=\frac{1}{\sqrt{3}},\;n= \frac{1}{\sqrt{3}} \end{aligned}

∴Equation of plane  lx + my + nz = d, d = distance from origin \begin{aligned} &(5\sqrt{3}) \end{aligned}

\begin{aligned} &\Rightarrow \frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=5\sqrt{3}\\ &\Rightarrow x + y+z =15 \end{aligned}

Posted by

Gurleen Kaur

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