Anonymous asked in Science & MathematicsMathematics · 8 months ago

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:

(x-2)^(2) -4



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.

Attachment image
There are no answers yet.
Be the first to answer this question.