Let a=2 i+3 j+4 k, b= i-2 j-2 k, c=- i+4 j+3 k.

Question

Let $\vec a=2\hat i+3\hat j+4\hat k$, $\vec b=\hat i-2\hat j-2\hat k$, $\vec c=-\hat i+4\hat j+3\hat k$. If $\vec d$ is perpendicular to both $\vec b$ and $\vec c$, and $\vec a\cdot\vec d=18$, then $|\vec a	imes\vec d|^2$ is:

Accepted Answer

$\vec d = \lambda(\vec b\times\vec c)$. $\vec a\cdot\vec d = \lambda[\vec a\ \vec b\ \vec c] = 18$:
$$\lambda\begin{vmatrix}2&3&4\1&-2&-2\-1&4&3\end{vmatrix}=18 \implies \lambda(4-3+8)=18\cdot(\text{scale}) \implies \lambda=2$$

$\vec d = 2(2\hat i-\hat j+2\hat k)$ (per source), so $|\vec d|^2=4(4+1+4)=36$, $|\vec a|^2=4+9+16=29$.

$$|\vec a\times\vec d|^2 = |\vec a|^2|\vec d|^2 - (\vec a\cdot\vec d)^2 = 29(36)-18^2 = 1044-324=720$$

Solutions · 1