Let the vectors $\vec a, \vec b, \vec c$ and $\vec d$ be such that $(\vec a × \vec b) × (\vec c× \vec d) =\vec 0$. Let $P_1$ and $P_2$ be the planes determined by the pairs of vectors $\vec a, \vec b$ and $\vec c, \vec d$ respectively, then the angle between $P_1$ and $P_2$ is
Answer & explanation
Correct answer: option 1
Let $\vec{n_1}$ and $\vec{n_2}$ be the vectors normal to the planes, $P_1$ and $P_2$ respectively. Then, $\vec{n_1} = \vec a × \vec b$ and $\vec{n_2} =\vec c× \vec d$
$∴\vec{n_1}×\vec{n_2}=(\vec a × \vec b) × (\vec c× \vec d)$
$⇒\vec{n_1}×\vec{n_2}=\vec 0$ [∵$(\vec a × \vec b) × (\vec c× \vec d) =\vec 0$ (Given)]
$⇒\vec{n_1}||\vec{n_2}⇒θ=0$