Answer to Question #104504 in Analytic Geometry for havefun7741

Question #104504
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
1
Expert's answer
2020-03-03T16:37:31-0500



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

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

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


AB×AC=ijk221011=(1,2,2).\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ 2 & -2 & -1 \\ 0 & 1 & 1 \end{vmatrix} = (-1, 2, 2).\\

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

Thus, n=13(1,2,2).\vec{n} = \dfrac{1}{3}(-1, 2, 2).\\


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

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

DD=13(1(1)+(8)2+92)=13DD' = \dfrac{1}{3}( 1\cdot(-1) + (-8)\cdot2 + 9\cdot 2) = \dfrac{1}{3}


Answer. a) n=13(1,2,2)\vec{n} = \dfrac{1}{3}(-1, 2, 2) , b) DD=13DD'= \dfrac{1}{3} .


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