Question

The distance of the plane 3x + 2y + 6z – 49=0 from the point (6, 9, 1) is :​

Answer 1

Answer:

we have,

distance = ax+ by + cz + d/√(a^2 + b^2 + c^2)

or, D = 3*6+2*9+6*1-49/√(3^2+2^2+6^2)

or, D = -7/7

or,D = -1 => D = 1 Units. (Hence, distance is never negative)