Question

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

Answer 1

Let the direction cosines of the normal line be \(\cos \alpha, \cos \beta, \cos \gamma\)
"Normal is equally inclined to coordinate axis, so
\(\alpha=\beta=\gamma\)
We know, \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\)
\(\Rightarrow3\cos^{2}\alpha=1[\therefore\alpha=\beta=\gamma]\)

\(\Rightarrow \cos ^{2} \alpha=\dfrac{1}{3} \)

\(\Rightarrow \cos \alpha=\pm \dfrac{1}{\sqrt{3}}\)

Now, the unit normal vector is
\(\hat{n}=\pm \dfrac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)

Equation of a plane which is at a distance of\(\text{ p}\) from the origin and \(\hat{n}\) is the unit vector in the direction of normal to the plane is : \(\overrightarrow{r} \cdot \hat{n}=p\) Therefore, the equation of plane is
\(\overrightarrow{r} \cdot \hat{n}=p\)

\(\Rightarrow(x \hat{i}+y \hat{j}+z \hat{k}) \cdot\left[\pm \dfrac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\right]=3\sqrt{3}\)

\(\Rightarrow(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=\pm 3 \sqrt{3} \times \sqrt{3} \)

\(\Rightarrow x+y+z=\pm 9\)

Hence, required equation of plane is \(x+y+z=\pm 9\)