Let u be coplanar with a=2 i+3 j- k, b= j+ k.

Question

Let $\vec u$ be coplanar with $\vec a=2\hat i+3\hat j-\hat k$, $\vec b=\hat j+\hat k$. If $\vec u\perp\vec a$ and $\vec u\cdot\vec b=24$, then $|\vec u|^2$ is:

Accepted Answer

Since $\vec u\perp\vec a$ and coplanar with $\vec a,\vec b$: $\vec u = \lambda\big[(\vec a\times\vec b)\times\vec a\big] = \lambda\big[|\vec a|^2\vec b-(\vec a\cdot\vec b)\vec a\big]$.

$|\vec a|^2=14$, $\vec a\cdot\vec b=3-1=2$: $\vec u=\lambda[14(0,1,1)-2(2,3,-1)] = \lambda(-4,8,16)$.

$\vec u\cdot\vec b=24$: $\lambda(8+16)=24 \Rightarrow \lambda=1$, so $\vec u=(-4,8,16)$.

$$|\vec u|^2 = 16+64+256=336$$

Solutions · 1