Hi guys,

I've got this exercise where I'm required to describe the equation of the plane determined by the fact that the following

three points lie in it: $\displaystyle 1, 2, -4), (2, 3, 7), (4, -1, 3)$

So we find two vectors that lie in the plane

$\displaystyle v_1= \underbrace{(1,2,-4)}_{Head}-\underbrace{(4,-1,3)}_{tail}= -3 \vec i, 3 \vec j, -7 \vec k$

and

$\displaystyle v_2= \underbrace{(2,3,7)}_{Head}-\underbrace{(4,-1,3)}_{tail}=-2 \vec i, 4 \vec j , 4 \vec k$

To find a normal vector we need to take the cross product

$\displaystyle \begin{vmatrix}

\vec i && \vec j && \vec k \\

-3 && 3 && -7 \\

-2 && 4 && 4 \\

\end{vmatrix} = 40 \vec i +26 \vec j- 6 \vec k

$

$\displaystyle 40(x-4) + 26(y-1) -6(z-3) = 0 $

The equation can now be rewritten to the form

$\displaystyle ax+by+cz = d $

$\displaystyle 40x+26y-6z = d $

I'm not sure how to reorganize that equation to evaluate the d mentioned above. Any help /push in the right direction would be great!

So we distribute out the left hand side.

$\displaystyle 40x-160 + 26y-26 -6z+18 = 0 ? $

If any mistakes can be pointed out and any help towards the completion of this exercise would be greatly appreciated.