How Do You Determine the Equation of a Plane Given Points and a Parallel Line?

[mathmari]

Hey!

Find an equation for the plane that passes through $(3, 2, -1)$ and $(1, -1, 2)$ and that is parallel to the line $\overrightarrow{v}=(1, -1, 0)+t(3, 2, -2)$.

The general formula of the equation on the plane is $$Ax+By+Cz+D=0$$

where $(A, B, C)$ is a perpendicular.

Since the plane passes through $(3, 2, -1)$ and $(1, -1, 2)$ we have that
$$3A+2B-C+D=0 \\ A-B+2C+D=0 $$

Also $$(A, B, C) \cdot (3, 2, -2)=0 \Rightarrow 3A+2B-2C=0$$

Is this correct so far??

 

[Prove It]

Two vectors in the plane are (3, 2, -2) and (2, 3, -3), so take their cross product to get a normal vector to the plane. This will be the coefficients of the plane. You can find the last parameter by substituting in the point (1, -1, 0) which you know lies on the plane.

 

[mathmari]

Two vectors in the plane are (3, 2, -2) and (2, 3, -3), so take their cross product to get a normal vector to the plane. This will be the coefficients of the plane. You can find the last parameter by substituting in the point (1, -1, 0) which you know lies on the plane.

How did you find the two vectors of the plan (3, 2, -2) and (2, 3, -3) ?? (Wondering)

 

[Dethrone]

How did you find the two vectors of the plan (3, 2, -2) and (2, 3, -3) ?? (Wondering)

He found $(3,2,-2)$ by realizing that the line is parallel to it, and the second vector is the vector produced by the two given points. $(3,2,-1) - (1,-1,2) = (2,3,-3)$. It suffices to take any two linearly independent vectors that are parallel to the plane to find the normal.