Find the equation of a plane which is at a di | vedclass

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(C) Let the normal vector to the plane be $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$.Since the normal is equally inclined to the coordinate axes,the d

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$x+y+z=9$

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(C) Let the normal vector to the plane be $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$.Since the normal is equally inclined to the coordinate axes,the direction cosines are equal,i.e.,$\cos \alpha = \cos \beta = \cos \gamma$.We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,so $3 \cos^2 \alpha = 1$,which gives $\cos \alpha = \cos \beta = \cos \gamma = \frac{1}{\sqrt{3}}$.Thus,the unit normal vector is $\hat{n} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$.The equation of a plane at a distance $p$ from the origin is given by $\vec{r} \cdot \hat{n} = p$.Here,$p = 3\sqrt{3}$.Substituting the values,we get $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}) = 3\sqrt{3}$.$\frac{x}{\sqrt{3}} + \frac{y}{\sqrt{3}} + \frac{z}{\sqrt{3}} = 3\sqrt{3}$.Multiplying both sides by $\sqrt{3}$,we get $x + y + z = 3\sqrt{3} 	imes \sqrt{3} = 9$.Therefore,the equation of the plane is $x + y + z = 9$.

Find the equation of a 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.

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