# Find the equation of the plane given point and line (3D)

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1. Apr 19, 2017

### alexandra2727

∴1. The problem statement, all variables and given/known data
Find an equation for the plane that contains the point (1,-1,2) and the line x=t, y=t+1, z=-3+2t?
2. Relevant equations

3. The attempt at a solution

I dont know what Im doing wrong but everytime I try to do these plane questions my amswer is always the opposite sign of the real answer.

P1(7,0,1)
P2(7,3,0)

P1P2 (0,3,-1)
n(7,-1,2)

P1P2 crossproducted with n = (5,-7,-21)

∴ 5(x-7)-7y-21(z-1)=0

2. Apr 19, 2017

### gv3

Where are your P1 and P2 coming from?
Are you crossing P1P2 vector with the normal vector?

3. Apr 19, 2017

### Staff: Mentor

If you have a point $P_0(x_0, y_0, z_0)$ and a normal to the plane N = <A, B, C>, the equation of the plane is $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$. This equation can be obtained by taking the dot product of N and a vector $\vec{P_0P}$, where P(x, y, z) is an arbitrary point in the plane.

You're given a point, so you have some of the information you need. To get a normal to the plane, find two vectors that lie in the plane. For the first vector, find two points on the line, and construct a displacement vector between the two points. Call this $\vec u$. To get another vector, construct a displacement vector between any point on the line and the given point. Call this $\vec u$.
Take the cross product of $\vec u$ and $\vec v$, which will give you a vector perpendicular to both $\vec u$ and $\vec v$ (and therefore be a normal to the plane).