Question

Find the distance of the point (2, 3, 5) from the xy - plane.

 

Answer 1

\[ \text{ We know that the distance of the point } \left( x_1 , y_1 , z_1 \right)\text{ from the planeax + by + cz + d = 0 is given by } \]
\[\frac{\left| a x_1 + b y_1 + c z_1 + d \right|}{\sqrt{a^2 + b^2 + c^2}}\]
\[ \text{ Equation of the xy- plane is z = 0, which means 0x + 0y + z = 0 }\]
\[\text{ So, the required distance } \]
\[ = \frac{\left| 0 \left( 2 \right) + 0 \left( 3 \right) + \left( 5 \right) \right|}{\sqrt{0^2 + 0^2 + 1^2}}\]
\[ = \frac{5}{1}\]
\[ = 5  \text{ units } \]