Equation of a plane (not sure if I got this)

Question

When I was doing my calculus midterm, I came across a question that I didn't really know how to solve, I think I skipped over these problems in my studies. The question is:

Find an equation of the plane through $(1,2,-2)$ that contains the line:
$x=2t$ $y=3-t$ $z=1+3t$

Since I didn't know how to solve it the "correct" way, what I tried doing was finding two points on the line by plugging in $t=0$ and $t=1$ and I got the points $(0,3,1)$ and $(2,2,2)$.

So now I needed to find the equation of a plane containing $A(1,2,-2),B(0,3,1),C(2,2,4)$.

I found AB and AC and took the cross product and then just plugged in the values into:

$a(x_0-x)+b(y_0-y)+c(z_0-z) = 0$

Does this make sense? Did I salvage at least some partial credit for this answer? Thanks

Your method is correct. – Mhenni Benghorbal Dec 04 '13 at 02:02 — Mhenni Benghorbal, Dec 04 '13 at 02:02
A related problem. – Mhenni Benghorbal Dec 04 '13 at 02:09 — Mhenni Benghorbal, Dec 04 '13 at 02:09

score 1 · Accepted Answer · answered Dec 04 '13 at 02:06

1

If $(a, b, c)$ are the components of your normal vector (from the cross product), and $(x_0, y_0, z_0)$ is one of the points in the plane, that looks good!

The way I've seen it before was with terms like $(x-x_0)$ instead of $(x_0 - x)$ but that shouldn't make a difference. If $\vec N$ is a normal vector, so is $-\vec N$.

answered Dec 04 '13 at 02:06

John

26,319

wooooooppsss yes I meant $(x-x_0)$ etc... – Sam Creamer Dec 04 '13 at 02:17
thanks for confirming this for me – Sam Creamer Dec 04 '13 at 02:20

Equation of a plane (not sure if I got this)

1 Answers1