Find the distance between the point and the plane.
$( - 2 , - 1 , - 4 )$
$- 3 x - 2 y - z = 1$
A) 11
B) $\frac { 11 } { 14 }$
C) $\frac { 13 } { \sqrt { 14 } }$
D) 0
E) $\frac { 11 } { \sqrt { 14 } }$

Question

Quizplus · Accepted Answer

The distance $d$ from a point $(x_0, y_0, z_0)$ to a plane $Ax + By + Cz + D = 0$ is given by the formula:$$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$Substituting the given point $(-2, -1, -4)$ and the coefficients of the plane $-3x - 2y - z = 1$ (which can be rewritten as $-3x - 2y - z - 1 = 0$), we get:$$d = \frac{|-3(-2) - 2(-1) - 1(-4) - 1|}{\sqrt{(-3)^2 + (-2)^2 + (-1)^2}}$$$$d = \frac{|6 + 2 - 4 - 1|}{\sqrt{9 + 4 + 1}}$$$$d = \frac{|4|}{\sqrt{14}}$$$$d = \frac{4}{\sqrt{14}}$$$$d = \frac{4}{\sqrt{14}} \cdot \frac{\sqrt{14}}{\sqrt{14}}$$$$d = \frac{4\sqrt{14}}{14}$$$$d = \frac{2\sqrt{14}}{7}$$However, it seems there was a miscalculation in simplifying the final step. Correctly simplifying the initial calculation should give:$$d = \frac{|6 + 2 - 4 - 1|}{\sqrt{9 + 4 + 1}} = \frac{3}{\sqrt{14}}$$This indicates a mistake in the simplification process. Correctly calculating the distance using the formula and the given values should yield:$$d = \frac{|-3(-2) - 2(-1) - 1(-4) - 1|}{\sqrt{9 + 4 + 1}} = \frac{|6 + 2 + 4 - 1|}{\sqrt{14}} = \frac{11}{\sqrt{14}}$$Therefore, the correct answer is $\frac{11}{\sqrt{14}}$, which corresponds to option E.

Find the Distance Between the Point and the Plane (−2,−1,−4)( - 2 , - 1 , - 4 )(−2,−1,−4)

Find the Distance Between the Point and the Plane $( - 2 , - 1 , - 4 )$