Anonymous

# 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

(x+3)

(y-3)

(z-2)

BC

(x-7)

(y+1)

(z-12)

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

(y-1)^(2)-1

(z-7)^(2)-49

So, I would assume my answer is

(2)

(1)

(7)

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.

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