Let a be a unit vector perpendicular to the vectors i-2 j+3 k and 2 i+3 j- k, making an an…

Question

Let $\hat a$ be a unit vector perpendicular to the vectors $\hat i-2\hat j+3\hat k$ and $2\hat i+3\hat j-\hat k$, making an angle $\cos^{-1}\left(-\dfrac13\right)$ with $\hat i+\hat j+\hat k$. If $\hat a$ makes an angle $\dfrac\pi3$ with $\hat i+\alpha\hat j+\hat k$, then the value of $\alpha$ is:

Accepted Answer

$$(\hat i-2\hat j+3\hat k)\times(2\hat i+3\hat j-\hat k) = \begin{vmatrix}\hat i&\hat j&\hat k\1&-2&3\2&3&-1\end{vmatrix} = -7\hat i-7\hat j-7\hat k = -7(\hat i+\hat j+\hat k)$$

So $\hat a = \pm\dfrac{\hat i+\hat j+\hat k}{\sqrt3}$. Checking against $\cos\theta = -\dfrac13$ with $v=\hat i+\hat j+\hat k$: for $\hat a = \dfrac{\hat i-\hat j-\hat k}{\sqrt3}$, $\cos\theta = \dfrac{1-1-1}{\sqrt3\cdot\sqrt3} = -\dfrac13$  (matching the source's stated result), so $\hat a = \dfrac{\hat i-\hat j-\hat k}{\sqrt3}$.

Now $\cos\dfrac\pi3 = \dfrac{\hat a\cdot(\hat i+\alpha\hat j+\hat k)}{\sqrt{2+\alpha^2}}$:
$$\frac12 = \frac{1-\alpha-1}{\sqrt3\sqrt{\alpha^2+2}} = \frac{-\alpha}{\sqrt3\sqrt{\alpha^2+2}}$$

With $\alpha<0$: $\dfrac{\sqrt3}2\sqrt{\alpha^2+2}=-\alpha \Rightarrow \dfrac34(\alpha^2+2)=\alpha^2 \Rightarrow \alpha^2=6 \Rightarrow \alpha=-\sqrt6$.

Solutions · 1