A vector r of magnitude 3*sqrt(6) is directed along the bisector of the angle between vectors a = 7i - 4j - 4k and b = -2i - j + 2k. Find r.

i - 7j + 2k
i + 7j - 2k
-i + 7j - 2k
i - 7j - 2k

Correct answer: i - 7j + 2k

Solution

Unit vector of a: (7i-4j-4k)/9. Unit vector of b: (-2i-j+2k)/3. Bisector direction = (7i-4j-4k)/9 + (-2i-j+2k)/3 = (7i-4j-4k+(-6i-3j+6k))/9 = (i-7j+2k)/9. Magnitude of (i-7j+2k) = sqrt(1+49+4) = sqrt(54) = 3*sqrt(6). So r = (i-7j+2k)/9 * 3*sqrt(6) = but wait, |r| must be 3*sqrt(6). r = (i-7j+2k) has magnitude 3*sqrt(6) directly. So r = i - 7j + 2k.

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