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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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.

Explanation:

Given: $¯ a = 2 ˆ i - 2 ˆ j - ˆ k and ¯ b = 3 ˆ i + 4 ˆ k$ $bara = 2hati - 2hatj - hatk and barb = 3hati + 4hatk$

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

$¯ c = | | ¯ a | | ¯ b + ∣ ∣ ∣ ∣ ¯ b ∣ ∣ ∣ ∣ ¯ a$ $barc = ||bara||barb + ||barb||bara$

Compute the magnitude of $¯ a$ $bara$ :

$| | ¯ a | | = \sqrt{2^{2} + {(- 2)}^{2} + {(- 1)}^{2}} = \sqrt{9} = 3$ $||bara|| = sqrt(2^2 + (-2)^2 + (-1)^2) = sqrt(9) = 3$

Compute the magnitude of $¯ b$ $barb$ :

$∣ ∣ ∣ ∣ ¯ b ∣ ∣ ∣ ∣ = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$ $||barb|| = sqrt(3^2 + 4^2) = sqrt(25) = 5$

$| | ¯ a | | ¯ b = 3 (3 ˆ i + 4 ˆ k) = 9 ˆ i + 12 ˆ k$ $||bara||barb = 3(3hati + 4hatk) = 9hati + 12hatk$

$∣ ∣ ∣ ∣ ¯ b ∣ ∣ ∣ ∣ ¯ a = 5 (2 ˆ i - 2 ˆ j - ˆ k) = 10 ˆ i - 10 ˆ j - 5 ˆ k$ $||barb||bara = 5(2hati - 2hatj - hatk) = 10hati - 10hatj - 5hatk$

$¯ c = 9 ˆ i + 12 ˆ k + 10 ˆ i - 10 ˆ j - 5 ˆ k$ $barc = 9hati + 12hatk + 10hati - 10hatj - 5hatk$

$¯ c = 19 ˆ i - 10 ˆ j + 7 ˆ k$ $barc = 19hati - 10hatj + 7hatk$

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

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

$ˆ c = \frac{19}{\sqrt{510}} ˆ i - \frac{10}{\sqrt{510}} ˆ j + \frac{7}{\sqrt{510}} ˆ k$ $hatc = 19/sqrt(510)hati - 10/sqrt(510)hatj + 7/sqrt(510)hatk$

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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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