Let c be perpendicular to a= i+ j- k and b= i+2 j+ k.

Question

Let $\vec c$ be perpendicular to $\vec a=\hat i+\hat j-\hat k$ and $\vec b=\hat i+2\hat j+\hat k$. If $\vec c\cdot(\hat i+\hat j+3\hat k)=8$, then $\vec c\cdot(\vec a	imes\vec b)$ is \_\_\_\_\_.

Answer: 28

Accepted Answer

$\vec c=\lambda(\vec a\times\vec b)$. $\vec a\times\vec b = (1+2,\,-(1+1),\,2-1) = (3,-2,1)$.

$\vec c\cdot(1,1,3)=8$: $\lambda(3-2+3)=8 \Rightarrow \lambda=2$, so $\vec c=(6,-4,2)$.

$$\vec c\cdot(\vec a\times\vec b) = 2(3,-2,1)\cdot(3,-2,1) = 2(9+4+1)=28$$

Solutions · 1