Answer to Question #121886 in Analytic Geometry for Samson

Question

If
α=3i−j+2k,

β=2i+j−k and
,
γÌ…=i−2j+2k, find

α×β×γ.

Answer 1

we know that if

$\overrightarrow{\alpha}$ $= x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$ and $\overrightarrow{\beta}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}$

then $\overrightarrow{\alpha }$ x $\overrightarrow{\beta} = \begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ x_1 & y_1 &z_1\\ x_2&y_2&z_2 \end{vmatrix}$

Here $\overrightarrow{\alpha} = 3\hat{i}-\hat{j}+2\hat{k}$ , $\overrightarrow{\beta}=2\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{\gamma}= \hat{i}-2\hat{j}+2\hat{k}$

so,

$\overrightarrow{\alpha}$ x $\overrightarrow{\beta} = \begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 3 & -1 &2\\ 2&1&-1 \end{vmatrix}$

expanding along Row 1.

$\overrightarrow{\alpha}$ x $\overrightarrow{\beta}= \hat{i}\begin{vmatrix} -1 & 2 \\ 1 & -1 \end{vmatrix}-\hat{j}\begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix}+\hat{k}\begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix}$

$= \hat{i}(1-2)-\hat{j}(-3-4)+\hat{k}(3+2)$

$=-\hat{i}+7\hat{j}+5\hat{k}$ .

Following the same way $\overrightarrow{\alpha}$ x $\overrightarrow{\beta}$ x $\overrightarrow{\gamma}$ can be determined .

so ,

$\overrightarrow{\alpha}$ x $\overrightarrow{\beta}$ x $\overrightarrow{\gamma}=$ $\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ -1 & 7 &5\\ 1&-2&2 \end{vmatrix}$

$=\hat{i}\begin{vmatrix} 7& 5 \\ -2 & 2 \end{vmatrix}-\hat{j}\begin{vmatrix} -1 & 5 \\ 1 & 2 \end{vmatrix}+\hat{k}\begin{vmatrix} -1 & 7 \\ 1 & -2 \end{vmatrix}$

$=\hat{i}(14+10)-\hat{j}(-2-5)+\hat{k}(2-7)$

$=24\hat{i}+7\hat{j}-5\hat{k}$

The answer is :

$\therefore \overrightarrow{\alpha}$ x $\overrightarrow{\beta}$ x $\overrightarrow{\gamma}=24\hat{i}+7\hat{j}-5\hat{k}.$