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

Question

Let $\vec a = \hat i-2\hat j+3\hat k$, $\vec b = 2\hat i+\hat j-\hat k$, $\vec c = \lambda\hat i+\hat j+\hat k$, and $\vec v = \vec a	imes\vec b$. If $\vec v\cdot\vec c = 11$, and the length of the projection of $\vec b$ on $\vec c$ is $p$, then $9p^2$ is equal to

Accepted Answer

$$\vec v = \vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\1&-2&3\2&1&-1\end{vmatrix} = -\hat i+7\hat j+5\hat k$$

$\vec v\cdot\vec c = -\lambda+7+5=11 \Rightarrow \lambda=1$, so $\vec c = \hat i+\hat j+\hat k$.

$$p = \frac{|\vec b\cdot\vec c|}{|\vec c|} = \frac{|2+1-1|}{\sqrt3} = \frac{2}{\sqrt3} \implies p^2 = \frac43 \implies 9p^2 = 12$$

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