Answer to Question #104504 in Analytic Geometry for havefun7741

Question

Let A(3,2,-1), B(5,0,-2), C(3,3,0) and D(4,-6,8) be four points in the plane,Donate the plane containing A, B, and C.Use vector methods to solve the following.
(a)Find a unit vector perpendicular to the plane.
(b)WITHOUT finding the equation of the plane , calculate the shortest distance between D and the plane

Answer 1

Let $\alpha$ be the plane containing points A, B, and C. The vector $\vec{n}$ is a unit vector perpendicular to the plane. $\overrightarrow{AB} = (2,-2,-1), \overrightarrow{AC} = (0,1,1), \overrightarrow{AD} = (1,-8,9).\\$

a) The unit vector $\vec{n}$ can be found as a cross-product of vectors $\overrightarrow{AC}$ and $\overrightarrow{AB}$ normalized by its length.

$\vec{n} =\dfrac{\overrightarrow{AB} \times \overrightarrow{AC} }{|\overrightarrow{AB} \times \overrightarrow{AC}|}\\$ .

$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ 2 & -2 & -1 \\ 0 & 1 & 1 \end{vmatrix} = (-1, 2, 2).\\$

$|\overrightarrow{AC} \times \overrightarrow{AB}| = \sqrt{(-1)^2 + (2)^2 + (2)^2} = 3.\\$

Thus, $\vec{n} = \dfrac{1}{3}(-1, 2, 2).\\$

b) The shortest distance $DD'$ between D and the plane can be found as a projection of vector $\overrightarrow{AD}$ onto the $\vec{n}$ direction. Namely,

$DD'=\overrightarrow{AD} \cdot \vec{n}.\\$

$DD' = \dfrac{1}{3}( 1\cdot(-1) + (-8)\cdot2 + 9\cdot 2) = \dfrac{1}{3}$

Answer. a) $\vec{n} = \dfrac{1}{3}(-1, 2, 2)$ , b) $DD'= \dfrac{1}{3}$ .