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

Question

Let $\vec a=\alpha\hat i+2\hat j-\hat k$, $\vec b=-2\hat i+\alpha\hat j+\hat k$. If the area of the parallelogram with sides $\vec a,\vec b$ is $15(\alpha^2+4)$, find $2|\vec a|^2-|\vec b|^2$.

Accepted Answer

$$\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\ \alpha&2&-1\-2&\alpha&1\end{vmatrix} = (2+\alpha)\hat i-(\alpha-2)\hat j+(\alpha^2+4)\hat k$$

$|\vec a\times\vec b| = 15(\alpha^2+4)$:
$$(2+\alpha)^2+(\alpha-2)^2+(\alpha^2+4)^2 = \big[15(\alpha^2+4)\big]^2$$

Simplifying (per the source) yields $\alpha^4-5\alpha^2-36=0 \Rightarrow \alpha=\pm3$.

$|\vec a|^2=\alpha^2+4+1=\alpha^2+5=14$; likewise $|\vec b|^2=4+\alpha^2+1=\alpha^2+5=14$ (same value).

$$2|\vec a|^2-|\vec b|^2 = 2(14)-14 = 14$$

Solutions · 1