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

Question

Let $\vec a=\hat i+\alpha\hat j+3\hat k$, $\vec b=3\hat i-\alpha\hat j+\hat k$. If the area of the parallelogram with sides $\vec a,\vec b$ is $8\sqrt3$, then $\vec a\cdot\vec b$ is \_\_\_\_\_.

Answer: 2

Accepted Answer

$$\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\1&\alpha&3\3&-\alpha&1\end{vmatrix} = (\alpha+3\alpha)\hat i-(1-9)\hat j+(-\alpha-3\alpha)\hat k = 4\alpha\hat i+8\hat j-4\alpha\hat k$$

$$|\vec a\times\vec b|^2 = 16\alpha^2+64+16\alpha^2 = (8\sqrt3)^2=192 \implies 32\alpha^2=128 \implies \alpha^2=4$$

$$\vec a\cdot\vec b = 3-\alpha^2+3 = 6-4=2$$

Solutions · 1