Question

What is the length PC? How did you solve this?

Answer 1

See below.

Explanation:

Calling

`vec v_1, vec v_2, vec v_3` the unit vectors with the direction of `A-P, B-P` and `C-P` we have

`A = P + a vec v_1`
`B = P + b vec v_2`
`C = P + c vec v_3`

Here

`a = 10`
`b = 6`

then

`B-A = b vec v_2-a vec v_1`
`C-B = c vec v_3 - b vec v_2`

but `B-A` and `C-B` are perpendicular then

`<< B-A, C-B >> = << b vec v_2-a vec v_1, c vec v_3 - b vec v_2 >> = 0`

`b c << vec v_2, vec v_3 >> - b^2 norm(vec v_2)^2-ac << vec v_1, vec v_3 >>+a b << vec v_1, vec v_2 >> =0`

and then

`c = (b^2-ab << vec v_1, vec v_2>>)/(b << vec v_2, vec v_3 >> - a << vec v_1, vec v_3>>) = 33`

with

`<< vec v_1, vec v_2 >> =-1/2`
`<< vec v_2, vec v_3 >> =-1/2`
`<< vec v_1, vec v_3 >> = -1/2`

NOTE

`<< cdot, cdot >>` represents the scalar product of two vectors.
`norm(cdot)` represents the norm operation.