tetrahedron 2

Hello, Apprentice123!

Quote:

Given points: $A(1,2,1),\:B(2,-1,1),\:C(0,-1,-1),\:D(3,1,0)$
Calculate the distance from vertex $D$ to the plane of face $ABC.$

Answer: . $\frac{8}{\sqrt{19}}$

First, find the equation of the plane . . .

We have: . $\begin{array}{ccccc}\vec u & = & \overrightarrow{AB} &=& \langle 1,\text{-}3,0\rangle \\ \vec v &=& \overrightarrow{BC} &=& \langle \text{-}2,0,\text{-}2\rangle \end{array}$

The normal vector of the plane: . $\vec n \:=\:\vec u \times \vec v \:=\:\left[\begin{array}{ccc}i & j & k \\ 1 &\text{-}3 & 0 \\ \text{-}2 &0 & \text{-}2 \end{array}\right] \;=\;6i + 2j - 6k$
Hence: . $\vec n \:=\:\langle 3,1,\text{-}3\rangle$

The plane is: . $3(x-1) + 1(y-2) - 3(z-1) \:=\:0 \quad\Rightarrow\quad 3x + y - 3z - 2 \:=\:0$

The distance from point $(x_1,y_1,z_1)$ to plane $ax + by + cz + d \:=\:0$

. . is given by: . $D \;=\;\frac{|ax_1 +by_1 + cz_1 + d|}{\sqrt{a^2+b^2+c^2}}$

So we have: . $D \;=\;\frac{|3(3) + 1(1) - 3(0)|}{\sqrt{3^2 + 1^2 + (\text{-}3)^2}} \;=\;\boxed{\frac{8}{\sqrt{19}}}$