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

Question

Let $\vec a=3\hat i+\hat j-\hat k$, $\vec c=2\hat i-3\hat j+3\hat k$. If $\vec b$ satisfies $\vec a=\vec b	imes\vec c$ and $|\vec b|^2=50$, then $\big|72-|\vec b+\vec c|^2\big|$ is \_\_\_\_\_\_\_.

Answer: 66

Accepted Answer

$|\vec a|=\sqrt{11}$, $|\vec c|=\sqrt{22}$. From $|\vec a|=|\vec b||\vec c|\sin\theta$: $\sqrt{11}=\sqrt{50}\sqrt{22}\sin\theta \Rightarrow \sin\theta=\dfrac1{10}$, so $\cos\theta = \dfrac{\sqrt{99}}{10}$.

$$|\vec b+\vec c|^2 = |\vec b|^2+|\vec c|^2+2|\vec b||\vec c|\cos\theta = 50+22+2\sqrt{50}\sqrt{22}\cdot\frac{\sqrt{99}}{10} = 72+66=138$$

$$\big|72-138\big| = 66$$

Solutions · 1