Question

A plane makes intercepts −6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it. [CBSE 2014]

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

$$\begin{gathered} \frac{x}{-6}+\frac{y}{3}+\frac{z}{4}=1 \\ \Rightarrow -2x+4y+3z=12 \\ \Rightarrow 2x-4y-3z+12=0 \end{gathered}$$

∴ Length of the perpendicular from (0, 0, 0) to the plane $2x-4y-3z+12=0$

$$\begin{gathered} =\left|\frac{2\times 0-4\times 0-3\times 0+12}{\sqrt{2^2+{\left(-4\right)}^2+{\left(-3\right)}^2}}\right| \\ =\left|\frac{12}{\sqrt{4+16+9}}\right| \\ =\frac{12}{\sqrt{29}}\mathrm{units} \end{gathered}$$

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