If the vector -i¯+j¯-k¯ bisects the angle between the vector c¯ and the vector 3i¯+4j¯, then the unit vector in the direction of c¯ is

cosθ2=-3+43(5)=153 c^+3i+4j52cosθ/2=-i+j-k3 c^+3i+4j5=2.153 -i+j-k3 c^=215(-i+j-k)-3i+4j5

If the vector -i¯+j¯-k¯ bisects the angle between the vector c¯ and the vector 3i¯+4j¯, then the unit vector in the direction of c¯ is

cosθ2=-3+43(5)=153 c^+3i+4j52cosθ/2=-i+j-k3 c^+3i+4j5=2.153 -i+j-k3 c^=215(-i+j-k)-3i+4j5

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If the vector $- \bar{i} + \bar{j} - \bar{k}$ bisects the angle between the vector $\bar{c}$ and the vector $3 \bar{i} + 4 \bar{j}$ , then the unit vector in the direction of $\bar{c}$ is

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$11 \bar{i} + 10 \bar{j} + 2 \bar{k}$

$- (11 \bar{i} + 10 \bar{j} + 2 \bar{k})$

$- \frac{1}{15} (11 \bar{i} + 10 \bar{j} + 2 \bar{k})$

$\frac{1}{15} (11 \bar{i} + 10 \bar{j} + 2 \bar{k})$

answer is C.

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Detailed Solution

$\cos \frac{θ}{2} = \frac{- 3 + 4}{\sqrt{3} (5)} = \frac{1}{5 \sqrt{3}} \frac{\hat{c} + \frac{3 i + 4 j}{5}}{2 cosθ / 2} = \frac{- i + j - k}{\sqrt{3}} \hat{c} + \frac{3 i + 4 j}{5} = 2 . \frac{1}{5 \sqrt{3}} (\frac{- i + j - k}{\sqrt{3}}) \hat{c} = \frac{2}{15} (- i + j - k) - (\frac{3 i + 4 j}{5})$