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

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Given, the plane is at a distance of 3\sqrt{3} from the origin, and the normal is equally inclined to coordinate axes.

We need to find the equation of this plane.

We know, the vector equation of a plane located at a distance d from the origin is represented by:

\vec{r}.\hat{n}=d\\ \Rightarrow \left ( x\hat{i}+y\hat{j}+z\hat{k} \right ).\left ( l\hat{i}+m\hat{j}+n\hat{k} \right )=d

lx + my + nz = d ….(i)   , where l, m and n are the direction cosines of  the normal of the plane.

Since the normal is equally inclined to the coordinate axes,

l = m = n \\ \cos \alpha =\cos\beta =\cos \gamma …(ii)

Also, we know,

\cos^{2} \alpha =\cos^{2}\beta =\cos^{2} \gamma=1\\ \Rightarrow \cos^{2} \alpha =\cos^{2}\alpha =\cos^{2} \alpha =1 \: \: \left ( from(ii) \right )\\ \Rightarrow 3\cos^{2}\alpha=1\\ \Rightarrow \cos^{2}\alpha=\frac{1}{3}\\ \Rightarrow \cos \alpha =\frac{1}{\sqrt{3}}

This means, l=m=n =\frac{1}{\sqrt{3}}

if we substitute the values of l, m and n in equation (i),

\left (\frac{1}{\sqrt{3}} \right )x+\left (\frac{1}{\sqrt{3}} \right )y+\left (\frac{1}{\sqrt{3}} \right )z=d\: \: \left [where\: d=3\sqrt{3} \right ]


\left (\frac{1}{\sqrt{3}} \right )x+\left (\frac{1}{\sqrt{3}} \right )y+\left (\frac{1}{\sqrt{3}} \right )z=3\sqrt{3} \\ \Rightarrow \frac{x+y+z}{\sqrt{3}}=3\sqrt{3}\\ \Rightarrow x+y+z=3\sqrt{3}\times \sqrt{3}\\ \Rightarrow x+y+z=3 \times 3=9

Therefore, the required equation of the plane is x + y + z = 9.


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