Question

A plane makes intercepts −6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.

Answer 1

We know that the equation of the plane which makes intercepts a, b and c on the coordinate axes is 

\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\]

So, the equation of the plane which makes intercepts −6, 3, 4 on the x-axis, y-axis and z-axis, respectively is \[\frac{x}{- 6} + \frac{y}{3} + \frac{z}{4} = 1\]
\[ \Rightarrow - 2x + 4y + 3z = 12\]
\[ \Rightarrow 2x - 4y - 3z + 12 = 0\]

∴ Length of the perpendicular from (0, 0, 0) to the plane

\[2x - 4y - 3z + 12 = 0\]

\[= \left| \frac{2 \times 0 - 4 \times 0 - 3 \times 0 + 12}{\sqrt{2^2 + \left( - 4 \right)^2 + \left( - 3 \right)^2}} \right|\]
\[ = \left| \frac{12}{\sqrt{4 + 16 + 9}} \right|\]
\[ = \frac{12}{\sqrt{29}} \text{ units } \]

Thus, the length of the perpendicular from the origin to the plane is \[\frac{12}{\sqrt{29}}\]  units .