Find a plane containing the point (5,4,-6) and the line of intersection of the planes -x+2y+8z=-57 and 5x-4y+7z=-27
The intersection of 2 planes ( if there is) is a line, what is it? ( I mean the equation of that line?)
The question does not given any other information
sure, but we have a way to find it out, :) you are supposed to know it.
Would I need to find the normal vector? So the cross product of those two equations?
@sourwing please help
ok, he left. I don't have time, but let me try.
Thank you!
Since I tried solving the cross product and got 46i+47j-6k
let say P1 , P2 are plane 1, 2, so normal vector of P1 is n1 = < -1,2,8> and n2 = <5,-4,7> ok?
Okay
then n1 x n2 =?
I got 46i+47j-6k
Then do I need to make x=0 and solve for y and z?
That is the intersection line, and you need 2 points in that line, right? but if you want to go on your way, you let x =0 for the equations of 2 planes, not this line I mean -x+2y+8z=-57 and 5x -4y +7z = 27 let x =0 for both , solve for y and z
what do you have?
Hold on...
2y + 8z = -57 -4y +7z = 27 so, y = -615/46 z= -87/23
I am sorry, I don't have time, cannot wait for you :)
No worries!
so, you have a point (0, -615/46, -87/23)
And that would be my (x0,y0,z0) point?
And (46,47,-6) would be my (a,b,c) points?
therefore, the equation of the intersection line is x = 46t y =( -615/46) + 47t z = (-87/23) -6t now, just let t =0 to get point A (...,...,..) and then t =1 to get point B (...,...,...) accompanied with the given point (5,4,-6) 3points , so you can write down the equation of the required plane, right?
Got it. I'll try to go through that process then.
ok, good luck, sorry for not letting you work on your problem, I need my time to write my lab report. try to help because I started but sourwing didn't want to get over it. :)
Thank you so much for helping!
;)
I was in class at the time :DD
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