# Three vectors lie in a plane

• Jul 22nd 2010, 05:48 AM
Three vectors lie in a plane
Can i just find the triple scalar product a.(b x c)

And if it = 0 then it lies in the plane?
• Jul 22nd 2010, 06:41 AM
Plato
Quote:

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.
• Jul 22nd 2010, 07:04 AM
Quote:

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
• Jul 22nd 2010, 07:29 AM
Plato
I don't understand what you are doing.

$\displaystyle v_1\cdot(v_2\times v_3)=72$
• Jul 22nd 2010, 07:45 AM
$\displaystyle v_1\cdot(v_2\times v_3)=72$
If $\displaystyle v_1= <x_1, y_1, z_1>$, $\displaystyle v_2= <x_2, y_2, z_2>$, and [tex]v_3= <x_3, y_3, z_3> then the cross product is the determinant:
$\displaystyle v_1\cdot(v_2\times v_3)= \left|\begin{array}{ccc}x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array}\right|$