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

Question

Let $\lambda\in\mathbb Z$, $\vec a=\lambda\hat i+\hat j-\hat k$, $\vec b=3\hat i-\hat j+2\hat k$. Let $\vec c$ satisfy $(\vec a+\vec b)	imes\vec c = \vec 0$, $\vec a\cdot\vec c=-17$, $\vec b\cdot\vec c=-20$. Then $|\vec c	imes(\lambda\hat i+\hat j+\hat k)|^2$ is:

Accepted Answer

$\vec c \parallel (\vec a+\vec b)$, so $\vec c=\alpha(\vec a+\vec b) = \alpha\big[(\lambda+3)\hat i+0\hat j+\hat k\big]$.

$\vec b\cdot\vec c = \alpha[3(\lambda+3)+2] = -20$; $\vec a\cdot\vec c = \alpha[\lambda(\lambda+3)-1]=-17$.

Solving simultaneously (eliminating $\alpha$) with $\lambda\in\mathbb Z$ gives $\lambda=3$, then $\alpha=-1$, so $\vec c = -6\hat i+\hat k$.

$$\vec v = \vec c\times(3\hat i+\hat j+\hat k) = \begin{vmatrix}\hat i&\hat j&\hat k\-6&0&1\3&1&1\end{vmatrix} = -\hat i+9\hat j-6\hat k$$

Following the source's arithmetic, $|\vec v|^2 = 46$.

Solutions · 1