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

Question

Let $\vec a=\hat i+2\hat j+3\hat k$, $\vec b=\hat i+\hat j-\hat k$. If $\vec c$ satisfies $\vec a\cdot\vec c=11$, $\vec b\cdot(\vec a	imes\vec c)=27$, and $\vec b\perp(\vec a	imes\vec c)$ decomposition gives $|\vec b	imes(\vec a	imes\vec c)|=3\sqrt{14}$ (angle relation), then $|\vec a	imes\vec c|^2$ is \_\_\_\_\_\_\_.

Answer: 285

Accepted Answer

Since $\vec a\cdot\vec b=1+2-3=0$, $\vec b\perp\vec a$. Let $\theta$ be the angle between $\vec b$ and $\vec a\times\vec c$. Then:
$$|\vec b||\vec a\times\vec c|\sin\theta = 3\sqrt{14}\quad\text{(perpendicular component)},\qquad |\vec b||\vec a\times\vec c|\cos\theta = \vec b\cdot(\vec a\times\vec c)=27$$

Working through the source's relation $\sin\theta=\dfrac{\sqrt{14}}{\sqrt{95}}$ and $|\vec b\times(\vec a\times\vec c)|=3\sqrt{95}$, one obtains
$$|\vec a\times\vec c| = \sqrt3\times\sqrt{95} = \sqrt{285} \implies |\vec a\times\vec c|^2 = 285$$

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