# vector orthogonal to plane, find triangle area.

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• Feb 23rd 2009, 11:01 PM
rcmango
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.
• Feb 24th 2009, 06:38 AM
earboth
Quote:

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:

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

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

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

2. The area of the triangle PQR is

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

3. With the given vectors you'll get:

$a = \dfrac12 \cdot |(-1,2,0) \times (-1, 0, 3)| = \dfrac12 \cdot |(6,3,2)| = \dfrac12 \cdot \sqrt{49} = 3.5$
• Feb 24th 2009, 10:32 AM
rcmango
okay, ya definately needed some help, thankyou.