Three vectors lie in a plane

**adam_leeds** · July 22nd 2010, 05:48 AM

Can i just find the triple scalar product a.(b x c)

And if it = 0 then it lies in the plane?

**Plato** · July 22nd 2010, 06:41 AM

Originally Posted by adam_leeds

Can i just find the triple scalar product a.(b x c) = 0 then it lies in the plane?

Yes this is true for three non-zero vectors.

**adam_leeds** · July 22nd 2010, 07:04 AM

Originally Posted by Plato

Yes this is true for three non-zero vectors.

Thanks

So for v1 = (−2, 2, 0), v2 = (6, 1, 4) and v3 = (2, 0,−4)

The cross product is 58

**Plato** · July 22nd 2010, 07:29 AM

I don't understand what you are doing.

$v_1\cdot(v_2\times v_3)=72$

**adam_leeds** · July 22nd 2010, 07:45 AM

Originally Posted by Plato

I don't understand what you are doing.

$v_1\cdot(v_2\times v_3)=72$

Yeah i did something wrong, i used this to help me

thanks

**HallsofIvy** · July 23rd 2010, 02:49 AM

A short cut for the triple product is this:
If $v_1= <x_1, y_1, z_1>$ , $v_2= <x_2, y_2, z_2>$ , and [tex]v_3= <x_3, y_3, z_3> then the cross product is the determinant:
$v_1\cdot(v_2\times v_3)= \left|\begin{array}{ccc}x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array}\right|$

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