Let a= i+ j+ k, c= j- k, and b satisfy a b= c, a b=1.

Question

Let $\vec a=\hat i+\hat j+\hat k$, $\vec c=\hat j-\hat k$, and $\vec b$ satisfy $\vec a	imes\vec b=\vec c$, $\vec a\cdot\vec b=1$. If $l$ is the length of the projection of $\vec b$ on $\vec a	imes\vec c$, find $3l^2$.

Answer: 2

Accepted Answer

$$\vec a\times\vec c = \begin{vmatrix}\hat i&\hat j&\hat k\1&1&1\0&1&-1\end{vmatrix} = (-2,1,1)$$

$\vec b\cdot(\vec a\times\vec c) = [\vec b\ \vec a\ \vec c] = -[\vec a\ \vec b\ \vec c] = -(\vec a\times\vec b)\cdot\vec c = -\vec c\cdot\vec c = -2$.

$$l = \frac{|\vec b\cdot(\vec a\times\vec c)|}{|\vec a\times\vec c|} = \frac2{\sqrt6} \implies l^2=\frac46=\frac23 \implies 3l^2=2$$

Solutions · 1