Question

Find the equation of the plane that passes through the point (6,0,-2) and contains the line x=4-2t, y=3+5t, z=7+4t... I can't figure this out, perhaps I just need a formula? Thanks

Answer 1

Try Gaussian elimination?

Answer 2

It seems this won't work. I have tried taking the dot product of V (where V=<4-2t,3+5t,7+4t>) and an unknown vector, say N, and setting them equal to 0 (so as to say they are perpendicular. But....I end up getting 4x+3y+7z+t(-2x+5y+4z)=0. I don't know what to do with it from there. I can't use Gaussian elimination because I have unknown a, b, and c (which is in essence what I'm solving for

Answer 3

Since your plane has to contain that point, and that vector, you know that it must contain the vector from the point (6, 0, -2) to the beginning of the other vector. In other words, the plane must contain the vectors \[\left(\begin{matrix}4\\3\\7\end{matrix}\right)+t\left(\begin{matrix}-2\\5\\4\end{matrix}\right)\]and \[\left(\begin{matrix}6\\0\\-2\end{matrix}\right)+t\left(\begin{matrix}-2\\3\\9\end{matrix}\right)\]

Answer 4

Now take the cross product of these vectors to get a vector that is perpendicular to both of the vectors, and therefore must be perpendicular to the plane. The values in that vector will correspond directly to \(a, b, c\) in the equation \[ax+by+cz+d=0\]Then you just have to solve for \(d\) based on your point (6, 0, -2)

Answer 5

Basically, I think you should just take the cross product of \[<-2, 5, 4>\text{and}<-2, 3, 9>\]You'll get a vector of the form \[<a, b, c>\]where you can simply use the same \(a, b, c\) in the equation \(ax+by+cz+d=0\) and then solve for \(d\).

Answer 6

It keeps saying [Math processing error] in red print so I can't really see what you have said =\

Answer 7

oops. Try opening the page in a new tab. See if that helps. Otherwise, I'll just write it normally.

Answer 8

ok 1 sec, thanks a bunch

Answer 9

ok, I can see your work now, let me check it over and see if what I can learn =)

Answer 10

On the reply where you told me to take the cross products, where did u get the vector <-2,3,9>?

Answer 11

That's the vector from (6, 0, -2) to (4, 3, 7).

Answer 12

Oh for when t=0?

Answer 13

When t=1 I think. But the t doesn't really matter too much.

Answer 14

Aha! But the moral of the story is one has to get another point on the plane to get a vector that one can use....Thanks I'll crunch the numbers real quick!

Answer 15

@mathguyz what's your plane equation?