How do you find the unit vector which bisects the angle AOB given A and B are position vectors a=2i-2j-k and b=3i+4k?

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Answer 1 · 2016-12-08T22:56:43

Please see the explanation.

Given: $bara = 2hati - 2hatj - hatk and barb = 3hati + 4hatk$

Here is a reference regarding how to find an Angle Bisector Vector , $barc$ :

$barc = ||bara||barb + ||barb||bara$

Compute the magnitude of $bara$ :

$||bara|| = sqrt(2^2 + (-2)^2 + (-1)^2) = sqrt(9) = 3$

Compute the magnitude of $barb$ :

$||barb|| = sqrt(3^2 + 4^2) = sqrt(25) = 5$

$||bara||barb = 3(3hati + 4hatk) = 9hati + 12hatk$

$||barb||bara = 5(2hati - 2hatj - hatk) = 10hati - 10hatj - 5hatk$

$barc = 9hati + 12hatk + 10hati - 10hatj - 5hatk$

$barc = 19hati - 10hatj + 7hatk$

However, this is not the unit vector, $hatc$ . To make is at unit vector we must divide vector $barc$ by its magnitude:

||barc|| = sqrt(19^2 + (-10)^2 + 7^2) = sqrt(510)

$hatc = 19/sqrt(510)hati - 10/sqrt(510)hatj + 7/sqrt(510)hatk$