# Thread: Finding equation of plane?

1. I would agree that you've found two points on the line. Now what should you do?

2. find the equation by using the two points:

i get:

$6i+6j-k+t(7i+4j-k)$

3. What is that the equation of? The line? You already have that. What does the equation of a plane look like?

4. ok, let me start again i have the three points (6,6,-1) (13,10,-2) (1,3,2)

$n=OB-OA x OC-OA$
$n= 7i+4j-k x -5i-3j+3k$

find the cross product

$i(12+3)-j(21-5)+k(-21+20)$
$15i-16j-k$

sub points (1,3,2)

i get $15(1)-16(3)-(2)$
$15-48+2=-35$

$15x-16y-z=-35$

this is incorrect, however shouldn't it be still correct even if the points are different

5. Name the three points: $P6,6,-1),~Q13,10,-2),~\&~R1,2,3)" alt="P6,6,-1),~Q13,10,-2),~\&~R1,2,3)" />.

Now find the normal: $N = \overrightarrow {PQ} \times \overrightarrow {PR}$.

Then write the equation of the plane determined by the three points:
$N \cdot < x - 1,y - 2,z - 3 > = 0$

BTW: $$\times$$ gives $\times$ in LaTeX.

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