Question

Find and equation for the plane that contains the line x = -1+3t, y = 5+2t, z = 2-t and is perpendicular to the plane 2x-4y+2z=9

Explain your reasonings behind the solution, thanks!

Answer 1

you are given 2 vectors; one of them is in the line equation itself as the coeffs of the variable t. <3,2,-1>

the other vector in the plane is the normal attached to the plane equation that is "perp" to this one; <2,-4,2>

we are given a point to work off of from the line equation as well: the constants: (-1,5,2)

Answer 2

we can cross the vectors to get a normal to our plane and then dot it to our point

Answer 3

<3,2,-1>
<2,-4,2>
---------
nx = 2(2)-(-4)(-1) = 0
ny = 2(-1)-3(2) = -8
nz = 3(-4) - 2(2) = -16

<0,-8,-16> looks to be our normal; or <0,8,16>

0(x-Px)+8(y-Py)+16(z-Pz) = 0

8(y-5)+16(z-2) = 0

8y-40+16z-32 = 0

7y+16z-72 = 0

maybe

Answer 4

7y meant to be 8y

Answer 5

Thank you, great answer!

Answer 6

youre welcome, i hope its right; its been awhile since i did these

Answer 7

It's correct, the only tricky part really, was to realise you can actually use the normal from the plane to cross with the vector from the line. It was there somewhere, but I obviously didnt remember it :D

Answer 8

:) yay!!