How can one calculate the vector normal to the plane that is determined by given points?

For example, given three points $P_1(5,0,0)$, $P_2(0,0,5)$ and $P_3(10,0,5)$, calculate the vector normal to the plane containing these three points.

The compute the normal is by vector product. $$ a = \left(\begin{matrix} x_2-x_1\\y_2-y_1\\z_2-z_1 \end{matrix} \right) \qquad b = \left(\begin{matrix} x_3-x_1\\y_3-y_1\\z_3-z_1 \end{matrix} \right) $$ therefore $a = -5i+5k$, and $b=5i+5k$ $$ a\times b = \left|\begin{matrix} i & j & k \\ -5 & 0 & 5 \\ 5 & 0 & -5 \\ \end{matrix}\right| $$

The questions are:

1) Are A and B is given, that means no matter which three point i use, the $a$ and $b$ is still using this to calucate?

2) How to get $a$ , $b$ and $a\times b$ ? Thank you

  • $\begingroup$ You get two vectors by taking the difference between two pairs of points. Doesn't matter which pairs you pick: you'll get different vectors $a$ and $b$, but you will get a vector parallel to the normal one with any other pair of points. $\endgroup$ – Arturo Magidin Apr 27 '12 at 3:23
  • $\begingroup$ thanks, would you mind breifly tell how to get a x b? $\endgroup$ – Leo Chan Apr 27 '12 at 3:30
  • $\begingroup$ $a\times b$ is the cross product of $a$ with $b$; see here. $\endgroup$ – David Mitra Apr 27 '12 at 3:34
  • $\begingroup$ sorry i am confused , since my example axb have only i j k , and the example in the link has i j k i j , is it different when there is an arrow on a , b ? $\endgroup$ – Leo Chan Apr 27 '12 at 3:52
  • $\begingroup$ LeoChan: The link is just giving a special mnemonic to help readers remember how to compute the determinant - the same exact determinant that is present in your question. The arrow over letters (or boldface letters) is meant to help distinguish what's a vector versus what's a scalar, but as long as we're careful we don't need to use the arrows or boldface. $\endgroup$ – anon Apr 27 '12 at 4:00

Given any two vectors $a$ and $b$, the vector cross product $a\times b$ will be perpendicular to both of them, even if we have to consider the degenerate case of $0$ (that is, zero is perpendicular to anything).

With three distinct points $x_1,x_2,x_3$ we have a triangle, and this triangle determines a plane. Note that the triangle is determined by any two of its sides - and so is the plane then. Sides correspond to displacement vectors between the three points, i.e. we can set $a=x_2-x_1$, $b=x_3-x_1$ without losing generality (that is, it won't matter what order we understand the points to be in, the resulting displacement vectors will determine the same exact plane). Finally, since the points are distinct, the displacement vectors are nonzero, and if we assume they are noncollinear the displacements are also nonparallel, so their cross product will be nondegenerate (nonzero). (Non non non non...)

Since we're working in three dimensions, any vector perpendicular to these two displacement vectors will be perpendicular to the plane they determine, and vice versa, so we can use the cross product of these displacement vectors to determine the normal vector, up to various scalings.

With $i,j,k$ the orthonormal basis, the cross product is computed via the "pseudo-determinant"

$$a\times b = \det \begin{bmatrix}i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix}=(a_2b_3-a_3b_2)i-(a_1b_3-a_3b_1)j+(a_1b_2-a_2b_1)k.$$

Where $a=a_1i+a_2j+a_3k$ and $b=b_1i+b_2j+b_3k$.

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