Answer to Question #209780 in Analytic Geometry for Jaguar

Question

Let ~u =< −2, 1, −1, ~v =< −3, 2, −1 > and w~ =< 1, 3, 5 >. Compute:

(6.1) ~u × w~ ,

(6.2) ~u × (w~ × ~v) and (~u × w~ ) × ~v.

Answer 1

(6.1)

$$\vec u\times \vec w=\begin{vmatrix} \vec i & \vec j & \vec k \\ -2 & 1 & -1 \\ 1 & 3 & 5 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix}-\vec j\begin{vmatrix} -2 & -1 \\ 1 & 5 \end{vmatrix}+\vec k\begin{vmatrix} -2 & 1 \\ 1 & 3 \end{vmatrix}$$

$$=8\vec i+9\vec j-7\vec k$$

$$\vec u\times \vec w=\langle8, 9, -7\rangle$$

(6.2)

$$\vec w\times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ 1 & 3 & 5 \\ -3 & 2 & -1 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 3 & 5 \\ 2 & -1 \end{vmatrix}-\vec j\begin{vmatrix} 1 & 5 \\ -3 & -1 \end{vmatrix}+\vec k\begin{vmatrix} 1 & 3 \\ -3 & 2 \end{vmatrix}$$

$$=-13\vec i-14\vec j+11\vec k$$

$$\vec u\times (\vec w\times \vec v)=\begin{vmatrix} \vec i & \vec j & \vec k \\ -2 & 1& -1 \\ -13 & -14 & 11 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 1 & -1 \\ -14 & 11 \end{vmatrix}-\vec j\begin{vmatrix} -2 & -1 \\ -13 & 11 \end{vmatrix}+\vec k\begin{vmatrix} -2 & 1 \\ -13 & -14 \end{vmatrix}$$

$$=-3\vec i+35\vec j+41\vec k$$

$$\vec u\times (\vec w\times \vec v)=\langle -3, 35, 41 \rangle$$

$$(\vec u\times \vec w)\times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ 8 & 9 & -7 \\ -3 & 2 & -1 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 9 & -7 \\ 2 & -1 \end{vmatrix}-\vec j\begin{vmatrix} 8 & -7 \\ -3 & -1 \end{vmatrix}+\vec k\begin{vmatrix} 8 & 9 \\ -3 & 2 \end{vmatrix}$$

$$=5\vec i+29\vec j+43\vec k$$

$$(\vec u\times \vec w)\times \vec v=\langle 5, 29, 43 \rangle$$