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If two planes intersect in a line, explain why the cross product of the normal vectors of the planes is collinear with the direction vector of the line.

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Not only is a normal vector $\hat{n}$ perpendicular to every line in the plane $P$, but the converse is true as well: every line perpendicular to $\hat{n}$ is parallel to $P$.

So, if we take the cross product of $\hat{n}_1$ and $\hat{n}_2$, we get a vector perpendicular to both, which means it is parallel to both $P_1$ and $P_2$.

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  • $\begingroup$ Thank you. Very good explanation! $\endgroup$ – Ol' Reliable Oct 7 '13 at 21:46
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Let $P_i = \{x | \langle d_i , x \rangle = c_i \}$ for $i=1,2$ be the two planes.

Let $L = P_1 \cap P_2$, and let $d = d_1 \times d_2$. Note that by the properties of the cross product, we have $\langle d_i , d \rangle = 0$, for $i=1,2$.

Suppose $x \in L$, then $\langle d_i , x \rangle = c_i$ for $i=1,2$.

Now consider $\phi(\lambda) = x+ \lambda d$, then since $\langle d_i , x+ \lambda d \rangle = \langle d_i , x \rangle + \lambda \langle d_i , d \rangle = \langle d_i , x \rangle = c_i$, and so $\phi(\lambda) = x+ \lambda d \in L$ for all $\lambda$.

Since $L = \{x_0+ t h\}_{t \in \mathbb{R}}$, we have $\phi(0), \phi(1) \in L$, and so $\phi(0) = x = x_0+ t_0 h$, and $\phi(1) = x+d = x_0+ t_1 h$ for some $t_0,t_1$. Hence $d = (t_1-t_0) h$, and so $d,h$ are collinear.

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  • $\begingroup$ Thanks, but I didn't understand it! But it's all good! I get it now! $\endgroup$ – Ol' Reliable Oct 7 '13 at 21:47
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The cross product $v\times w$ yields a vector that is orthogonal to both $v$ and $w$. If $v,w$ are independent (that is, neither is a scalar multiple of the other), then all vectors orthogonal to $v,w$ will be a scalar multiple of $v\times w$. If $v,w$ are normal vectors of two planes intersecting in a line, then $v,w$ are independent, and both orthogonal to the direction vector of the line (by definition of normal vector, since the line lies in both planes), so that the direction vector of the line is a scalar multiple of $v\times w.$

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  • $\begingroup$ That really helped! Thanks! $\endgroup$ – Ol' Reliable Oct 7 '13 at 21:46

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