Let a = √7 i+ j- k and b= j+2 k.

Question

Let $\vec a = \sqrt7\hat i+\hat j-\hat k$ and $\vec b=\hat j+2\hat k$. Let $\vec r$ satisfy $\vec r	imes\vec a + \vec a	imes\vec b = \vec 0$ and $\vec r\cdot\vec a = 0$. Then $|3\vec r|^2$ is equal to:

Answer: 44

Accepted Answer

$\vec r\times\vec a = \vec b\times\vec a \Rightarrow (\vec r-\vec b)\times\vec a=\vec 0 \Rightarrow \vec r = \vec b+\lambda\vec a$.

$\vec r\cdot\vec a=0 \Rightarrow \vec b\cdot\vec a+\lambda(\vec a\cdot\vec a)=0$. With $\vec b\cdot\vec a = 0+1-2=-1$ and $\vec a\cdot\vec a=7+1+1=9$: $-1+9\lambda=0\Rightarrow\lambda=\frac19$.

$|\vec r|^2 = |\vec b|^2+\frac{1}{81}|\vec a|^2+\frac29(\vec a\cdot\vec b) = 5+\frac{9}{81}+\frac29(-1) = 5+\frac19-\frac29 = \frac{44}{9}$

$$|3\vec r|^2 = 9|\vec r|^2 = 44$$

Solutions · 1