Question

The figure represents a ABCD irregular quadrilateral. Determine the coords of the vector AP knowing:

The points A, B, C, D, E, P belongs to the same plane

AD = [2 2 -2]
BC = [2 4 1]
||AP|| = $\frac{\sqrt{6}}{3}$

AP is ortogonal to BC line

The AP abscissa is negative

Answer 1

There may be an easier way, but this is a way I know will work.

The vectors AD and BC are both in the same plane, so that plane will have a normal that is in the direction of their cross product.

... AD × BC = [10 -6 4], which can be reduced to [5 -3 2]

Since AP ⊥ BC also lies in the same plane, it will be orthogonal to [5 -3 2] and to [2 4 1], so its direction can be found from the cross product of these vectors.

... [5 -3 2] × [2 4 1] = [-11 -1 26]

Now all we need to do is to find the factor this needs to be multiplied by to get the desired magnitude.

||[-11 -1 26]|| = √798, so our multiplying factor is (√6)/3/√798 = (√133)/399

Then the coordinates of P are [-11√133/399 -√133/399 26√133/399]