Answer to Question #105096 in Analytic Geometry for Lam

Question

Let A(3,2,1), B(5,0,2), C(3,3,0) and D(1,-6,8) be four points in R3. use vector methods to solve the following.

a) Find the smaller angle between the side AB and AC.
b) Find the shortest distance between D and ABC plane, without finding the equation of the plane

Answer 1

$A(3,2,1), B(5,0,2), C(3,3,0), D(1,-6,8)$

1) Form vectors

$\overrightarrow{AB}=(5-3,0-2,2-1)=(2,-2,1)\\ \overrightarrow{AC}=(3-3,3-2,0-1)=(0,1,-1)$

We will find a scalar product

$\overrightarrow{AB}\cdot\overrightarrow{AC}=|\overrightarrow{AB}|\cdot |\overrightarrow{AC}|\cdot cos\alpha=\\ =x_{\overrightarrow{AB}}\cdot x_{\overrightarrow{AC}}+y_{\overrightarrow{AB}}\cdot y_{\overrightarrow{AC}}+z_{\overrightarrow{AB}}\cdot z_{\overrightarrow{AC}}$

$|\overrightarrow{AB}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=3\\ |\overrightarrow{AC}|=\sqrt{0^2+1^2+(-1)^2}=\sqrt{0+1+1}=\sqrt{2}\\ \overrightarrow{AB}\cdot\overrightarrow{AC}=2\cdot0+(-2)\cdot1+1\cdot(-1)=-3\\ \overrightarrow{AB}\cdot\overrightarrow{AC}=|\overrightarrow{AB}|\cdot |\overrightarrow{AC}|\cdot cos\alpha\\ -3=3\cdot\sqrt{2}\cdot cos\alpha\\ cos\alpha=-\frac{1}{\sqrt2}\\ \alpha=135^0$

The smaller angle between the side AB and AC

$180^0-\alpha=180^0-135^0=45^0$

2) Form vectors

$\overrightarrow{AB}=(2,-2,1)\\ \overrightarrow{AC}=(0,1,-1)\\ \overrightarrow{AD}=(1-3,-6-2,8-1)=(-2,-8,7)$

Find the volume of the pyramid $DABC$

$V=\frac {1}{3}S\cdot H\\ V=\frac{1}{6}|\overrightarrow{AB}\cdot\overrightarrow{AC}\cdot\overrightarrow{AD}|$

where $\overrightarrow{AB}\cdot\overrightarrow{AC}\cdot\overrightarrow{AD}$ is the scalar triple product,

$S=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$

where $\overrightarrow{AC}\times\overrightarrow{AC}$ is the cross product

$\overrightarrow{AB}\cdot\overrightarrow{AC}\cdot\overrightarrow{AD}=\begin{vmatrix} 2 & -2&1 \\ 0 &1&-1\\ -2&-8&7 \end{vmatrix}=\\ =2\cdot1\cdot7+(-2)\cdot(-1)\cdot(-2)+0\cdot(-8)\cdot1-\\ -1\cdot1\cdot(-2)-(-2)\cdot0\cdot7-(-8)\cdot(-1)\cdot2=\\ =14-4+0+2-0-16=-4$

$V=\frac{1}{6}|-4|=\frac{2}{3}$

$\overrightarrow{AB}\times\overrightarrow{AC}=(\begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix},-\begin{vmatrix} 2 & 1 \\ 0 & -1 \end{vmatrix},\begin{vmatrix} 2 & -2 \\ 0 & 1 \end{vmatrix})=\\ =(2-1,-(-2-0),2-0)=(1,2,2)\\ |\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{1^2+2^2+2^2}=\sqrt9=3$

$S=\frac{1}{2}\cdot3=\frac{3}{2}\\$

$V=\frac {1}{3}S\cdot H\\ \frac{2}{3}=\frac{1}{3}\cdot\frac{3}{2}\cdot H\\ H=\frac{4}{3}$

The shortest distance between D and ABC plane is $\frac{4}{3}$