Intersection of 2 planes concept question

Question

In my calc 3 class, I had to find the equation of line formed when 2 planes intersect. We were given the equations of the 2 planes. My idea was to equate those 2 equations and the equation that I would now receive should be able to give me the direction vector of the line that I need. But it does not. I don't understand where I am going wrong. Any help will be appreciated.

Answer 1

Assuming that what you mean is that you had a pair of equations of the form $ax+by+cz+d=0$ and equated their left-hand sides, here’s what went wrong. Doing that is equivalent to subtracting one of the equations from the other, which doesn’t really get you any closer to working out their line of intersection: Every nonzero linear combination of those two equations just produces another equation of a plane that also contains the line. The normal to this plane is of course perpendicular to the line, as it is for any plane that contains the line.

To find the line of intersection, you must either proceed as gimusi suggests in his answer, which takes advantage of the special properties of $\mathbb R^3$, or buckle down and compute the solution set of the system of equations in the usual way. If all goes well, this solution will be in the form of a parametric equation for the line, which you can then convert into whatever form is required.

Answer 2

Another way to do this is the consider the two equations, $a_1x+ b_1y+ c_1z+ d_1= 0$ and $a_2x+ b_2y+ c_2z+ d_2= 0$, in the three unknowns x, y, and z. We can solve for two of the unknowns in terms of the third, say, solve for x and y in terms of z. Multiply the first equation by $a_2$,to get $a_1a_2x+ b_1a_2y+ c_1a_2z+ d_1a_2= 0$, and the second equation by $a_1$, to get $a_1a_2x+ a_1b_2y+ a_1c_2z+ a_1d2= 0$. Subtract the second equation from the first to eliminate x: $(b_1a_2- a_1b_2)y+ (c_1a_2- a_1c_2)z+ (d_1a_2- d_2a_1)= 0$. Then $y= -\frac{(c_1a_2- a_1c_2)z+ (d_1a_2- d_2a_1)}{b_1a_2- a_1b_2}$. Similarly we can multiply the first equation by $b_2$, to get $a_1b_2x+ b_1b_2y+ c_1b_2z+ d_1b_2= 0$, and multiply the equation by $b_1$ to get $b_1a_2x+ b_1b_2y+ b_1c_2z+ b_1d_2= 0$. Now, subtract to eliminate y: $(a_1b_2- b_1a_2)x+ (c_1b_2- b_1c_2)z+ (d_1b_2- b_2d_1)= 0. Then $x= -\frac{(c_1b_2- b_1c_2)z+ (d_1b_2- b_2d_1)}{(a_1b_2- b_1a_2}. Now we can take the parameter t= z and have parametric equations for x, y, and z.