A vector a= i+2 j+ k lies in the plane of b= i+ j, c= i- j+4 k.

Question

A vector $\vec a=\alpha\hat i+2\hat j+\beta\hat k$ lies in the plane of $\vec b=\hat i+\hat j$, $\vec c=\hat i-\hat j+4\hat k$. If $\vec a$ bisects the angle between $\vec b$ and $\vec c$, then:

Accepted Answer

The bisector direction is $\hat b\pm\hat c$ where $\hat b=\dfrac{\hat i+\hat j}{\sqrt2}$, $\hat c=\dfrac{\hat i-\hat j+4\hat k}{3\sqrt2}$.

Trying $\vec a=\mu(\hat b-\hat c)$: $\vec a = \dfrac\mu{3\sqrt2}\big[(3\hat i+3\hat j)-(\hat i-\hat j+4\hat k)\big] = \dfrac\mu{3\sqrt2}(2\hat i+4\hat j-4\hat k)$

Matching the $\hat j$-coefficient to 2: $\dfrac{4\mu}{3\sqrt2}=2 \Rightarrow \mu=\dfrac{3\sqrt2}2$, giving $\vec a=\hat i+2\hat j-2\hat k$.

$$\vec a\cdot\hat k = -2 \implies \vec a\cdot\hat k+2=0$$

Solutions · 1