Answer to Question #200633 in Analytic Geometry for Vina

Question

Compute and find a relation between the expressions

(8.1) ~u · (~v × w~ ), w~ · (~u × ~v), and ~v · (w~ × ~u)

Find an expression for :

(8.2) ~u · (~v × w~ ) and ~u × (~v × w~ )

(8.3) Find the side lengths and angles of the triangle with vertices the tips the vectors in Question 6 above.

(8.4) Use ~u and ~v from Question 6 to find the area of the parallelogram formed by ~u and ~v.

(8.5) Use ~u, ~v and w~ from Question 6 to find the volume of the parallelepiped with edges deter- mined by the three vectors ~u, ~v and w~ 

Answer 1

(8.1)

The scalar triple product is unchanged under a circular shift of its three operands

$$\vec u \cdot(\vec v \times \vec w)=\vec w \cdot(\vec u \times \vec v)=\vec v \cdot(\vec w \times \vec u)$$

$$\vec u=\lang-1, 1, 2\rang, \vec v=\lang2, -1, 0\rang, \vec w=\lang1, 1, 3\rang$$

$$\vec u \cdot(\vec v \times \vec w)=\begin{vmatrix} -1 & 1 & 2 \\ 2 & -1 & 0 \\ 1 & 1 & 3 \\ \end{vmatrix}$$

$$=-1\begin{vmatrix} -1 & 0 \\ 1 & 3 \end{vmatrix}-1\begin{vmatrix} 2 & 0 \\ 1 & 3 \end{vmatrix}+2\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}$$

$$=3-6+6=3$$

$$\vec u \cdot(\vec v \times \vec w)=\vec w \cdot(\vec u \times \vec v)=\vec v \cdot(\vec w \times \vec u)=3$$

(8.2)

$$\vec u \cdot(\vec v \times \vec w)=3$$

$$\vec v \times \vec w=\begin{vmatrix} \vec i & \vec j & \vec k \\ 2 & -1 & 0 \\ 1 & 1 & 3 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} -1 & 0 \\ 1 & 3 \end{vmatrix}-\vec j\begin{vmatrix} 2 & 0 \\ 1 & 3 \end{vmatrix}+\vec k\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}$$

$$=-3\vec i-6\vec j+3\vec k$$

$$\vec u \times(\vec v \times \vec w)=\begin{vmatrix} \vec i & \vec j & \vec k \\ -1 & 1 & 2 \\ -3 & -6 & 3 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 1 & 2 \\ -6 & 3 \end{vmatrix}-\vec j\begin{vmatrix} -1 & 2 \\ -3 & 3 \end{vmatrix}+\vec k\begin{vmatrix} -1 & 1 \\ -3 & -6 \end{vmatrix}$$

$$=15\vec i-3\vec j+9\vec k$$

(8.3)

$\vec u=\lang-1, 1, 2\rang, \vec v=\lang2, -1, 0\rang, \vec w=\lang1, 1, 3\rang$

$$\overrightarrow{AB}=\lang3, -2, -2\rang, \overrightarrow{BC}=\lang-1, 2, 3\rang, \overrightarrow{AC}=\lang2, 0, 1\rang$$

$$AB=\sqrt{(3)^2+(-2)^2+(-2)^2}=\sqrt{17}$$

$$BC=\sqrt{(-1)^2+(2)^2+(3)^2}=\sqrt{14}$$

$$AC=\sqrt{(2)^2+(1)^2+(0)^2}=\sqrt{5}$$

$$\overrightarrow{AB}\cdot \overrightarrow{AC}=3(2)-2(0)-2(1)=4$$

$$\overrightarrow{BA}\cdot \overrightarrow{BC}=-3(-1)+2(2)+2(3)=13$$

$$\overrightarrow{CA}\cdot \overrightarrow{CB}=-2(1)+0(-2)-1(-3)=1$$

$$\cos\alpha=\dfrac{-3}{\sqrt{6}\sqrt{5}}=>\alpha=180\degree-\cos^{-1}\dfrac{\sqrt{30}}{10}$$

$$\approx64.3\degree$$

$$\cos\beta=\dfrac{13}{\sqrt{17}\sqrt{14}}=>\beta=\cos^{-1}\dfrac{13\sqrt{238}}{238}$$

$$\approx32.6\degree$$

$$\cos\gamma=\dfrac{1}{\sqrt{5}\sqrt{14}}=>\gamma=\cos^{-1}\dfrac{\sqrt{70}}{70}$$

$$\approx83.1\degree$$

(8.4)

$$\vec u \times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ -1 & 1 & 2 \\ 2 & -1 & 0 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 1 & 2 \\ -1 & 0 \end{vmatrix}-\vec j\begin{vmatrix} -1 & 2 \\ 2 & 0 \end{vmatrix}+\vec k\begin{vmatrix} -1 & 1 \\ 2 & -1 \end{vmatrix}$$

$$=2\vec i+\vec j-\vec k$$

$$Area=\|\vec u \times \vec v\|=\sqrt{(2)^2+(1)^2+(-1)^2}=\sqrt{6}\ (units^2)$$

(8.5)

$$V=|\vec u \cdot(\vec v \times \vec w)|=\bigg|\begin{vmatrix} -1 & 1 & 2 \\ 2 & -1 & 0 \\ 1 & 1 & 3 \\ \end{vmatrix}\bigg|=3 (units^3\ )$$