Let a= i+2 j+3 k, b= i- j+2 k, c=5 i-3 j+3 k.

Question

Let $\vec a=\hat i+2\hat j+3\hat k$, $\vec b=\hat i-\hat j+2\hat k$, $\vec c=5\hat i-3\hat j+3\hat k$. If $\vec r	imes\vec b=\vec c	imes\vec b$ and $\vec r\cdot\vec a=0$, then $25|\vec r|^2$ is:

Accepted Answer

$\vec r=\vec c+\lambda\vec b$; $\vec r\cdot\vec a=0 \Rightarrow \vec c\cdot\vec a+\lambda(\vec b\cdot\vec a)=0$.

$\vec c\cdot\vec a=5-6+9=8$, $\vec b\cdot\vec a=1-2+6=5 \Rightarrow \lambda=-\dfrac85$.

$$\vec r = \left(5-\frac85,\ -3+\frac85,\ 3-\frac{16}5\right) = \left(\frac{17}5,-\frac75,-\frac15\right)$$
$$|\vec r|^2 = \frac{289+49+1}{25}=\frac{339}{25} \implies 25|\vec r|^2 = 339$$

Solutions · 1