Question

Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.

 

Answer 1

\[\text{ The given equation of the plane is } \]
\[2x + 2y + 2z = 8\]
\[ \Rightarrow \left( \text{  x } \hat{i} + \text{ y }\hat{j}+ z \hat{k}  \right) . \left( \text{ 2 }\hat{i}  + \text{  2 } \hat{j}  + 2 \hat{k}  \right) = 8\]
\[ \Rightarrow \vec{r} . \left( \text{ 2 } \hat{i}  + \text{  2 }\hat{j}  + 2 \hat{k}  \right) = 8, \text{ which is the vector equation of the plane } .\]
\[( \text{ Because the vector equation of the plane is } \vec{r} . \vec{n} = \vec{a} . \vec{n} ,\]
\[\text{ where the normal to the plane } , \vec{n} = \text{ 2 } \hat{i} + \text{  2 }\hat{j}  + 2 \hat{k}  . )\]
\[\left| \vec{n} \right| = \sqrt{4 + 4 + 4} = 2 \sqrt{3}\]
\[\text{ So, the unit vector perpendicular to }  \vec{n} =\frac{\vec{n}}{\left| \vec{n} \right|}=\frac{ \text{ 2 }\hat{i}  + \text{ 2 }\hat{j} + 2 \hat{k} }{2 \sqrt{3}}=\frac{1}{\sqrt{3}} \hat{i} +\frac{1}{\sqrt{3}} \hat{j} +\frac{1}{\sqrt{3}} \hat{k} \]
\[\text{ So, thedirection cosines of the normal to the plane are } l = \frac{1}{\sqrt{3}}, m = \frac{1}{\sqrt{3}}, n = \frac{1}{\sqrt{3}}\]
\[\text{ Let }  \alpha, \beta \text{ and } \gamma \text{ be the angles made by the given plane with the coordinate axes } .\]
\[\text{ Then } ,\]
\[l = \cos \alpha = \frac{1}{\sqrt{3}}; m = \cos \beta = \frac{1}{\sqrt{3}}; n = \cos \gamma = \frac{1}{\sqrt{3}}\]
\[ \Rightarrow \cos \alpha = \cos \beta = \cos \gamma\]
\[ \Rightarrow \alpha = \beta = \gamma\]
\[\text{ So, the given plane is equally inclined to the coordinate axes } .\]