Question

Find the distance of the point (2, 3, −5) from the plane x + 2y − 2z − 9 = 0.

 

Answer 1

\[ \text{ We know that the distance of the point } \left( x_1 , y_1 , z_1 \right) \text{ from the plane ax + by + cz + d = 0 is given by } \]
\[\frac{\left| a x_1 + b y_1 + c z_1 + d \right|}{\sqrt{a^2 + b^2 + c^2}}\]
\[\text{ So, the required distance } \]
\[ = \frac{\left| \left( 2 \right) + 2 \left( 3 \right) - 2 \left( - 5 \right) - 9 \right|}{\sqrt{1^2 + 2^2 + \left( - 2 \right)^2}}\]
\[ = \frac{\left| 2 + 6 + 10 - 9 \right|}{\sqrt{1 + 4 + 4}}\]
\[ = \frac{9}{3}\]
\[ = 3 \text{ units } \]