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.
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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.
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