Question

Compute a · (b ✕ c), where a = i − 3j + k, b = 4i + j + k and c = 9i − j + 3k.

Answer 1

$\mathbf a=\mathbf i-3\,\mathbf j+\mathbf k$
$\mathbf b=4\,\mathbf i+\mathbf j+\mathbf k$
$\mathbf c=9\,\mathbf i-\mathbf j+3\,\mathbf k$

$\mathbf b\times\mathbf c=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\4&1&1\\9&-1&3\end{vmatrix}=4\,\mathbf i-3\,\mathbf j-13\,\mathbf k$

$\mathbf a\cdot(\mathbf b\times\mathbf c)=(\mathbf i-3\,\mathbf j+\mathbf k)\cdot(2\,\mathbf i-3\,\mathbf j-13\,\mathbf k)=4+(-3)^2-13=0$

Answer 2

We need to find the answer of the expression $a\cdot (b\times c)$

The required expression is $a\cdot (b\times c)=0$

The vectors are

$a=\hat{i}-3\hat{j}+1\hat{k}$

$b=4\hat{i}+\hat{j}+\hat{k}$

$c=9\hat{i}-\hat{j}+3\hat{k}$

Solving the term in the bracket

$b\times c=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 1 \\ 9 & -1 & 3 \end{vmatrix}\\ =\hat{i}(3+1)-\hat{j}(4\times3-9\times1)+\hat{k}(4\times-1-9\times 1)\\ =4\hat{i}-3\hat{j}-13\hat{k}$

$a\cdot (b\times c)=(\hat{i}-3\hat{j}+1\hat{k})(4\hat{i}-3\hat{j}-13\hat{k})\\ =1\cdot 4+(-3)\cdot (-3)+1\cdot (-13)\\ =4+9-13\\ =13-13=0$

So, $a\cdot (b\times c)=0$

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