- #1

CrosisBH

- 27

- 4

## Homework Statement

*P*(3, 0, 3),

*Q*(−2, 1, 5),

*R*(6, 2, 7)

(a) Find a nonzero vector orthogonal to the plane through the points P, Q,

and

*R*.

(b) Find the area of the triangle

*PQR*

## Homework Equations

[tex] A = \frac{1}{2}|\vec{AB}\times\vec{AC}| [/tex]

Source: http://www.maths.usyd.edu.au/u/MOW/vectors/vectors-11/v-11-7.html

## The Attempt at a Solution

The first part was straight forward and my homework marked it right:

[tex]P(3,0,3)\\

Q(-2,1,5)\\

R(6,2,7)\\

\vec{PR} = \langle 3,2,4 \rangle\\

\vec{PQ} = \langle 5,1,2 \rangle[/tex]

To find a vector orthogonal to two vectors, you do a cross product:

[tex] \vec{PR} \times \vec {PQ} = \langle 0,14,-7 \rangle [/tex]

My homework marked this part right for this answer.

Part two:

First taking the magnitude of the cross product:

[tex]|\vec{PR} \times \vec {PQ}| = \sqrt{245} = 7\sqrt{5} [/tex]

Dividing it by two:

[tex]A = \frac{7}{2}\sqrt{5}[/tex]

Which is my final answer, and the website marks it as wrong, but I can't see why it's wrong. The magnitude is right which I triple checked, by my own hands and Wolfram Alpha, and the order of the cross product is irrelevant in this situation.

I found a calculator that calculates the area from three 3D points and it spit out the nonexact answer of 14.534. My answer is approximately 7.826, a little more than half. I can't see my error unless I'm completely calculating this area wrong. Thanks for the help.