Let x lie in the plane of a=2 i- j+ k and b= i+2 j- k.

Question

Let $\vec x$ lie in the plane of $\vec a=2\hat i-\hat j+\hat k$ and $\vec b=\hat i+2\hat j-\hat k$. If $\vec x\perp(3\hat i+2\hat j-\hat k)$ and its projection on $\vec a$ is $\dfrac{17\sqrt6}2$, then $|\vec x|^2$ is:

Answer: 486

Accepted Answer

$\vec x=\lambda\vec a+\mu\vec b = (2\lambda+\mu)\hat i+(2\mu-\lambda)\hat j+(\lambda-\mu)\hat k$.

Perpendicular to $(3,2,-1)$: $3(2\lambda+\mu)+2(2\mu-\lambda)-(\lambda-\mu)=0 \Rightarrow 3\lambda+8\mu=0$.

Projection on $\vec a$ ($|\vec a|=\sqrt6$): $\dfrac{\vec x\cdot\vec a}{\sqrt6}=\dfrac{17\sqrt6}2 \Rightarrow \vec x\cdot\vec a=51 \Rightarrow 6\lambda-\mu=51$.

Solving: $\lambda=8,\mu=-3$: $\vec x = 13\hat i-14\hat j+11\hat k$.

$$|\vec x|^2 = 169+196+121 = 486$$

Solutions · 1