# Thread: Vector equation of line perpendicular to plane

1. ## Vector equation of line perpendicular to plane

Consider three points A,B,C with coordinates (2,2,3) (1,1,0) and (1,0,1) respectively. Find a vector equation of the line through point A that is perpendicular to the plane that passes through all three points A,B,C. [Hint: Use point-normal equation]

I am completely lost; the point normal equation bit throws me off the most.

Point normal equation: A(x-x0) + B(y-y0) + C(z-z0) = 0

2. Originally Posted by axdv
Consider three points A,B,C with coordinates (2,2,3) (1,1,0) and (1,0,1) respectively. Find a vector equation of the line through point A that is perpendicular to the plane that passes through all three points A,B,C. [Hint: Use point-normal equation]

I am completely lost; the point normal equation bit throws me off the most.

Point normal equation: A(x-x0) + B(y-y0) + C(z-z0) = 0
I don't understand that hint. the "point normal equation" is the equation of a plane and you want to find a line, not a plane.

Instead, use the fact that the vector equation of a line through $(x_0, y_0, z_0)$ and in the direction of vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ is $\vec{r}(t)= (At+ x_0)\vec{i}+ (Bt+ y_0)\vec{j}+ (Ct+ z_0)\vec{k}$.
You already know that $x_0= 2$, $y_0= 2$, $z_0= 3$ so you just need to find a vector perpendicular to the plane. And you can do that by taking the cross product of the vectors $(2- 1)\vec{i}+ (2- 1)\vec{j}+ (3- 0)\vec{k}$ (from B to A) and $(1- 1)\vec{i}+ (0- 1)\vec{j}+ (1- 0)\vec{k}$ (from B to C).

By the way, this is a "Calculus" problem. It doesn't belong in "Linear and Abstract Algebra".

3. Thanks for the help
And sorry about posting in the wrong place, I have difficulty telling different topics of maths appart.

4. ## Im confused

Can someone please explain that to me. Im a little lost.

5. Originally Posted by Surreptitous_Smiles
Can someone please explain that to me. Im a little lost.
Explain what to you?