# Equation of a plane and distance from a point to that plane

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• March 1st 2011, 01:11 PM
Rumor
Equation of a plane and distance from a point to that plane
Here's the problem:

"Consider the three points P = (1,7,3), Q = (5,9,2) and R = (2,6,5) that form a triangle.

a) Do these three points form the vertices of a triangle? If so, determine which point is the vertex of the right angle. (Use vector arithmetic and not the Pythagorean Theorem, to check if this is a right triangle.)

b) Find the equation of the plane that contains this triangle using a cross product calculation.

c) Find the distance from the point (2,2,0) to the plane that contains this triangle using a dot product calculation."

My work:

a) PQ = (4,2,-1) and PR = (1,-1,2)
There's no number t where PQ = PR, which means they're non-collinear and hence, a triangle.

PQ (dot) PR = 4-2-2 = 0, which means they're perpendicular at P, and so the right angle is also at P.

b) PQ x PR = [4,2,-1] x [1,-1,2] = 3i+9j-6k

I haven't gotten to part c yet, but does my work so far look good? Although, I did something wrong in part b, because the equation doesn't fit all the points, but I'm not sure what.
• March 1st 2011, 02:15 PM
Soroban
Hello, Rumor!

Quote:

Consider the three points: $P(1,7,3),\;Q(5,9,2),\;R(2,6,5)$

(a) Do these three points form the vertices of a triangle?
If so, determine which point is the vertex of the right angle.

(b) Find the equation of the plane that contains this triangle.

(c) Find the distance from the point (2,2,0) to the plane that contains this triangle/

My work:

$(a)\;\overrightarrow{PQ} = \langle4,2,\text{-}1\rangle,\;\overrightarrow{PR} = \langle1,\text{-}1,2\rangle$
There's no number $\,t$ where $\overrightarrow{PQ} = t\overrightarrow{PR}$
which means they're non-collinear and hence a triangle.

$\overrightarrow{PQ} \cdot\overrightarrow{PR} \:=\: 4-2-2 \:=\:0$
which means they're perpendicular at $\,P$, and so $\angle P = 90^o.$

$(b)\;\overrightarrow{PQ} \times \overrightarrow{PR} \:=\: \langle4,2,-1\rangle \times \langle1,-1,2\rangle \:=\: 3i+9j-6k$
This is correct, but you haven't answered the question yet.

I haven't gotten to part (c) yet, but does my work so far look good?
Although I did something wrong in part (b),
because the equation doesn't fit all the points.

What is the equation you referred to?

In part (b), you found the normal vector to the plane (only):

. . . . $\vec n \:=\:\langle 3,\text{-}9,\text{-}6\rangle \:=\:\langle 1,\text{-}3,\text{-}2\rangle$

The equation of the plane through $P(1,7,3)$
. . with normal $\vec n =\langle 1,\text{-}3,\text{-}2\rangle$ is given by:

. . . . $1(x-1) - 3(y-7) - 2(z-3) \:=\:0$

. . . . . . . . . $x - 3y - 2z + 26 \;=\;0$

• March 1st 2011, 07:57 PM
Rumor
Right, right. So, for part C then, in order to find the distance from (2,2,0) to the plane that contains the triangle, I can use any of the points given to figure it out, right?

For example: (1,7,3) and (2,2,0) = (2-1)i+(2-7)j+(0-3)k = i-5j-3k or (1,-5,-3).
So then, it becomes (1,-5,-3) (dot product) (1,-3,-2)/sqrt(1^2+(-5)^2+(-3)^2), which is 32/sqrt(14), correct? And that's the distance to the plane?