Find the distance of the plane having intercept 3, 4 and -1 from the origin.

Concept:

For a plane having intercept a, b and c then equation of the plane is given by:

\(\rm {x\over a}+{y\over b}+{z\over c} = 1\)

The distance of a point (p, q, r) from the plane ax + by + cz + d = 0

D = \(\rm \left|ap+bq+cr + d\over \sqrt{a^2+b^2+c^2}\right|\)

Calculation:

Given intercepts are 3, 4 and -1

The equation of the plane is 

\(\rm {x\over 3}+{y\over 4}+{z\over -1} = 1\)

4x + 3y - 12z - 12 = 0

The distance of the plane from the origin (0, 0, 0)

D = \(\rm \left|4\times0+3\times 0-12\times0 -12\over \sqrt{4^2+3^2+(-12)^2}\right|\)

D = \(\rm \left|-12\over \sqrt{9+16+144}\right|\) = \(\boldsymbol{12\over 13}\)