Find the equation of the plane containing the vectors $(1, 0, \sqrt{3})$ and $(1, \sqrt{3}, 0)$. The vectors are in standard position.
I first get the vector between the 2 points
$\vec{v}^{\ } = (1, 0, \sqrt{3}) – (1, \sqrt{3}, 0) = (0, -\sqrt{3}, \sqrt{3})$
Next I determine part of the equation by taking the norm and the dot product of the vector I calculated.
$(x, y, z).(0, -\sqrt{3}, \sqrt{3}) = 0\ -\sqrt{3}y + \sqrt{3}z = 0$
Next I calculate d by plugging one of my original points
$ -\sqrt{3}(0) + \sqrt{3}(\sqrt{3}) = d = 3 $
which gives me this equation
$ -\sqrt{3}y + \sqrt{3}z = 3 $
But I know this is wrong, because when I plug $(1, 0, \sqrt{3})$ into the equation it makes no sense.
Where did I make a mistake?
Best Answer
If you already have two vectors, remember that the plane equation requires a normal vector. This normal vector should be perpendicular to both vectors, which means you should take the cross product of the two vectors you were given. The result is your normal vector.