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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.4 question 8 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.4 question 8 maths

Answers (1)

Answer:

 x + y + z = 9

Hint:

You must know the rules of solving vector functions 

Given:

 Find the equation of the plane which is at a distance of  3\sqrt{3}  units from the origin and the normal to which is equally inclined to the coordinate axes.

Solution:

Let α, β and γ be the angles made by n with x,y and z axis

It is given that,

\begin{aligned} &\alpha =\beta =\gamma \\ &cos\: \alpha =cos\: \beta =cos\: \gamma \\ &\Rightarrow l=m=n \end{aligned}

But

\begin{aligned} &l^2+m^2+n^2=1\\ &l^2+l^2+l^2=1\\ &3l^2=1\\ &l^2=\frac{1}{3}\\ &l=\frac{1}{\sqrt{3}}\\ &l=m=n=\frac{1}{\sqrt{3}} \end{aligned}

It is given that the length of perpendicular of the plane
from the origin = p = 3\sqrt{3}

The normal form of plane is

\begin{aligned} &lx+my+nz=p\\ &\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=3\sqrt{3}\\ &\frac{x+y+z}{\sqrt{3}}=3\sqrt{3}\\ &x+y+z=3\sqrt{3}(\sqrt{3})\\ &x+y+z=9 \end{aligned}

Posted by

Gurleen Kaur

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