Let a = i+ j+ k, , R.

Question

Let $\vec a = \hat i+\alpha\hat j+\beta\hat k$, $\alpha,\beta\in\mathbb R$. Let $\vec b$ be such that the angle between $\vec a$ and $\vec b$ is $\dfrac\pi4$, $|\vec b|^2=6$, and $\vec a\cdot\vec b = 3\sqrt2$. Then $(\alpha^2+\beta^2)|\vec a	imes\vec b|^2$ is equal to

Accepted Answer

$\vec a\cdot\vec b = |\vec a||\vec b|\cos\frac\pi4 = |\vec a|\sqrt6\cdot\frac1{\sqrt2} = |\vec a|\sqrt3 = 3\sqrt2 \Rightarrow |\vec a|=\sqrt6$

So $|\vec a|^2 = 1+\alpha^2+\beta^2 = 6 \Rightarrow \alpha^2+\beta^2=5$.

$|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2 = 6\cdot6-18=18$.

$$(\alpha^2+\beta^2)|\vec a\times\vec b|^2 = 5\times18 = 90$$

Solutions · 1