A vectors question?

[ichigo_ice9]

Please, can anyone help me? I got this question and i don't know what to do.

Vector OA is i+2j-2k
Vector OB is 2i-3j+6k

The point P on AB is such that AP : PB = x : 1-x
Find the value of x for which angles AOP and POB are equal.

[krtxmrtz]

Welcome to the forums!

Refer to the attached figure. I have called v₁, v₂ and w the vectors OA, OB and OP.

You can write w as:

w = v₁ + x(v₂ - v₁)

The angles are (writing 'a' for alpha):

cos a₁ = v₁ . w / v₁ w
cos a₂ = v₂ . w / v₂ w

where non-bold characters mean modulus.

If the angles are to be equal:

cos a₂ = cos a₁ => v₁ . w / v₁ w = v₂ . w / v₂ w

v₂ v₁ . w = v₁ v₂ . w

(v₂ v₁ - v₁ v₂) . w = 0

Now you can substitute the values of the known vectors:

v₁ = (1,2,-2)
v₂ = (2,-3,6)
v₁ = Sqr(1² + 2² + (-2)²) = 3
v₂ = Sqr(2² + (-3)² + 6²) = 7

and you arrive at:

(1,23,-32) . w = 0

Using the above expression for w:

0 = (1,23,-32)[(1,2,-2) + x(1,-5,8)] = 111 - 370 x

Finally:

x = 111 / 370