Calculating the unit normal of a plane

**posix_memalign** · May 29th 2010, 01:10 PM

Hi,
I've tried to solve a problem and I was just wondering if my answer is correct or not:

Assume three points $p_1, p_2$ and $p_3$ :
$p_1 = (1,2,3)^T$
$p_2 = (2,2,1)^T$
$p_3 = (0,1,3)^T$

The problem is to find the unit normal of this plane.

I take the cross product of $p_1p_2$ and $p_1p_3$ , I get:

$N = -2i + 2j - k$
I find that the length of N is $sqrt((-2)^2 + 2^2 + (-1)^2) = 3$ .

I.e. the answer is: $(-2i + 2j - k) / 3$

Is this correct?

**Plato** · May 29th 2010, 01:58 PM

Correct.

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