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Reading a book on Differential Geometry I saw a statement that says the following:

Given two nonparallel planes $ a_{i} x + b_{i} y + c_{i} z + d_{i} = 0 $, $ i = 1, 2 $, their line of intersection may be parametrized as $$ x - x_0 = u_{1} t, y - y_0 = u_{2} t, z - z_0 = u_{3} t $$ where $ (x_0, y_0, z_0) $ belongs to the intersection and $ u = (u_1, u_2, u_3) $ is the vector product $ u = v_ {1} \times v_{2} $, $ v_{i} = (a_{i}, b_{i}, c_{i}) $, $ i = 1, 2 $.

But I'm not sure about this. Could you please explain to me how this is derived?

I like to read this kind of thing, but sometimes I have a hard time understanding it.

Curious
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  • Each plane has a normal vector $v_i=(a_1,b_i,c_i)$, and so far, $u=v_1\times v_2$ is orthogonal to $v_1$ and $v_2$ in the same time, so the line in the intersection of the planes has this vector as a director vector, thus, just take any point $(x_0,y_0,z_0)$ in the intersection and yout line is contructed – DiegoMath Sep 24 '20 at 22:01
  • @DiegoMath I don't understand how to get the parametrization of the intersection of the two planes based on what you said. Could you explain a little more please. How to apply what you mentioned to obtain said parameterization. – Curious Sep 25 '20 at 04:44

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