Question

Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0.

 

Answer 1

\[ \text{ We know that the distance of the point } \left( x_1 , y_1 , z_1 \right) \text{ from the plane ax + by + cz + d = 0 is given by } \]
\[\frac{\left| a x_1 + b y_1 + c z_1 + d \right|}{\sqrt{a^2 + b^2 + c^2}}\]
\[ \text{ Distance of the point (1, 1, 1) from the plane 3x + 4y - 12z + 13 = 0 } \]
\[\text{ The required distance } \]
\[ = \frac{\left| 3 \left( 1 \right) + 4 \left( 1 \right) - 12 \left( 1 \right) + 13 \right|}{\sqrt{3^2 + 4^2 + \left( - 12 \right)^2}}\]
\[ = \frac{\left| 3 + 4 - 12 + 13 \right|}{\sqrt{9 + 16 + 144}}\]
\[ = \frac{8}{13}  \text{ units } ... (1)\]
\[ \text{ Distance of the point (-3, 0, 1) from the plane 3x + 4y - 12z + 13 = 0 } \]
\[ \text{ The required distance } \]
\[ = \frac{\left| 3 \left( - 3 \right) + 4 \left( 0 \right) - 12 \left( 1 \right) + 13 \right|}{\sqrt{3^2 + 4^2 + \left( - 12 \right)^2}}\]
\[ = \frac{\left| - 9 + 0 - 12 + 13 \right|}{\sqrt{9 + 16 + 144}}\]
\[ = \frac{8}{13} \text{ units } .... (2)\]
\[ \text{ From (1) and (2), we can say that the given points are equidistant from the given plane } .\]