Find the shortest distance from the point $( 3,9,8 )$ to the plane $3 x + 9 y + 4 z = 16$ .
A) $D = 61$
B) $D = 90$
C) $D = 10.29$
D) $D = \sqrt { 61 }$
E) $D = 106$

Question

Quizplus · Accepted Answer

The shortest distance from a point to a plane is given by the formula:$$d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$$where $ (x_0, y_0, z_0) $ is the given point, $ (a, b, c) $ is the normal vector to the plane, and $ d $ is the constant term in the plane equation.In this case, the given point is $ (3, 9, 8) $, the normal vector to the plane is $ (3, 9, 4) $, and the constant term is 16. Plugging these values into the formula, we get:$$d = \frac{|3(3) + 9(9) + 4(8) - 16|}{\sqrt{3^2 + 9^2 + 4^2}} = \frac{106}{\sqrt{98}} = 10.29$$Therefore, the shortest distance from the point $ (3, 9, 8) $ to the plane $ 3x + 9y + 4z = 16 $ is $ 10.29 $.

Find the Shortest Distance from the Point (3,9,8)( 3,9,8 )(3,9,8) To the Plane

Find the Shortest Distance from the Point $( 3,9,8 )$ To the Plane