# Find the equation for the plane containing the point

1. May 29, 2010

### number0

1. The problem statement, all variables and given/known data

Find the equation for the plane containing the point (3,-8,7) and the line

l(t)=<1,-8,5>+t<3,-2,5>

2. Relevant equations

Cross product

Equation of a plane: A(x-x0)+B(y-y0)+C(z-z0)+D=0

3. The attempt at a solution

First I set t=0 in l(t)=<1,-8,5>+t<3,-2,5> to obtain the point (1,-8,5)

I then proceed to to subtract the given point (3,-8,7) by (1,-8,5) to get a vector <2,0,2>.

Since I have two vectors, that is, <3,-2,5> and <2,0,2>, I can use the cross product on

the two vectors to obtain the vector for the plane. By <3,-2,5> x <2,0,2>, I obtained the

vector <-4,4,4>. Simplifying it would result in <-1,1,1>.

Thus, using the equation for the plane:

A(x-x0)+B(y-y0)+C(z-z0)+D=0

where <A,B,C> = <-1,1,1> and (x0,y0,z0) = (3,-8,7)

So now, the equation looks like this:

-(x-3)+(y+8)+(z-7)=0

-x+3+y+8+z-7=0

-x+y+z=-4

I am not even sure if I did it right. This question was a question on my last exam (and the

only one that I missed). Sadly, my professor did not even post up the solutions. Can anyone

please see if my results and methods were correct? Thanks! :)

Last edited: May 29, 2010
2. May 29, 2010