Question

Determine the points in the plane 3x-4y+5z=50 nearest to the origin

Answer 1

Answer:

That will be foot of perpendicular from origin

$\frac{x - 0}{3} = \frac{y - 0}{ - 4} = \frac{z - 0}{5} = - ( \frac{ - 50}{ {3}^{2} + {4}^{2} + {5}^{2} } ) \\ \\ \frac{x}{3} = \frac { - y}{4} = \frac{z}{5} = 1 \\ (x \: \: y \: \: z) = (3 \: \: \: \: - 4 \: \: \: \: \: \: 5)$