Let a=2 i+7 j- k, b=3 i+5 k, c= i- j+2 k.

Question

Let $\vec a=2\hat i+7\hat j-\hat k$, $\vec b=3\hat i+5\hat k$, $\vec c=\hat i-\hat j+2\hat k$. Let $\vec d$ be perpendicular to both $\vec a,\vec b$, with $\vec c\cdot\vec d=12$. Then $(-\hat i+\hat j-\hat k)\cdot(\vec c	imes\vec d)$ is:

Accepted Answer

$\vec d = \lambda(\vec a\times\vec b) = \lambda(35\hat i-13\hat j-21\hat k)$.

$\vec c\cdot\vec d = \lambda(35+13-42)=6\lambda = 12 \Rightarrow \lambda=2$, so $\vec d = 70\hat i-26\hat j-42\hat k$.

$$\vec c\times\vec d = \begin{vmatrix}\hat i&\hat j&\hat k\1&-1&2\70&-26&-42\end{vmatrix}$$

Computing and dotting with $(-1,1,-1)$ gives the final result 44 (as in the source's worked computation).

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