Let a be a non-zero vector parallel to the line of intersection of the planes normal to i+…

Question

Let $\vec a$ be a non-zero vector parallel to the line of intersection of the planes normal to $\hat i+\hat j,\ \hat i+\hat k$ and to $\hat i-\hat j,\ \hat i-\hat k$. If $	heta$ is the angle between $\vec a$ and $\vec b=2\hat i-2\hat j+\hat k$, with $\vec a\cdot\vec b=6$, then $(	heta,|\vec a	imes\vec b|)$ is:

Accepted Answer

Normal to first plane: $\vec n_1 = (\hat i+\hat j)\times(\hat i+\hat k) = \hat i-\hat j-\hat k$. Normal to second: $\vec n_2 = (\hat i-\hat j)\times(\hat i-\hat k) = \hat i+\hat j+\hat k$.

$\vec a \parallel \vec n_1\times\vec n_2 = (1,-1,-1)\times(1,1,1) = -2\hat j+2\hat k$, so take $\vec a = -2\hat j+2\hat k$ (checking sign against $\vec a\cdot\vec b=6$ fixes the scale/sign).

$$\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \frac6{2\sqrt2\cdot3} = \frac1{\sqrt2} \implies \theta=\frac\pi4$$

$$|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2 = 8\times9-36 = 36 \implies |\vec a\times\vec b|=6$$

Solutions · 1