# Thread: vector orthogonal to plane, find triangle area.

1. ## vector orthogonal to plane, find triangle area.

Find vector orthogonal to plane through these points P, Q and R.

P(1, 0, 0)
Q(0, 2, 0)
R(0,0,3)

then find the area of the triangle PQR.

i tried using cross product of:

|1 0 0|
|0 2 0|
|0 0 3|

i got 1*(6-0) - 0(0-0) + 0(0-0)

...thats not correct because it should be <6, 3, 2>
i'm not sure i set this up correctly, haven't really tried a cross product with three points, usually just two.

then need the area of triangle.

thanks for any help here.

2. Originally Posted by rcmango
Find vector orthogonal to plane through these points P, Q and R.

P(1, 0, 0)
Q(0, 2, 0)
R(0,0,3)

then find the area of the triangle PQR.

i tried using cross product of:

|1 0 0|
|0 2 0|
|0 0 3|

i got 1*(6-0) - 0(0-0) + 0(0-0)

...thats not correct because it should be <6, 3, 2>
i'm not sure i set this up correctly, haven't really tried a cross product with three points, usually just two.

then need the area of triangle.

thanks for any help here.
1. The sides of the triangle are calculated as differences of the stationary vectors determining the vertices of the triangle:

$\displaystyle \vec a=\overrightarrow{PQ}=(-1,2,0)$

$\displaystyle \vec b=\overrightarrow{PR}=(-1,0,3)$

$\displaystyle \vec c=\overrightarrow{QR}=(0, -2, 3)$

2. The area of the triangle PQR is

$\displaystyle a = \dfrac12 \cdot |\overrightarrow{PQ} \times \overrightarrow{PR}|$

3. With the given vectors you'll get:

$\displaystyle a = \dfrac12 \cdot |(-1,2,0) \times (-1, 0, 3)| = \dfrac12 \cdot |(6,3,2)| = \dfrac12 \cdot \sqrt{49} = 3.5$

3. okay, ya definately needed some help, thankyou.