Find point C, if it lies on L2 and AC and BC are perpendicular?
I understand a and c and I thought I understood c until I checked the mark scheme.
So, here's my workings:
AC.BC (dot product) = 0 so,
x^(2) -4x+y^)2) -2y+z^(2)-14z=0
Then I completed the square for each x, y and z:
So, I would assume my answer is
However, I know we haven't proved that this lies on l2 but is this not right so far?
Also, to prove that ( 1 3 0) is a possibility:
I just plugged into AC.BC = 0 and it worked.
Be the first to answer this question.