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The vector $\bar{a} = α \hat{i} + 2 \hat{j} + β \hat{k}$ lies in the plane of vectors $\bar{b} = \hat{i} + \hat{j} a n d \bar{c} = \hat{j} + \hat{k}$ and bisects the angle between $\bar{b} & \bar{c}$ . Then which of the following gives possible values of $α a n d β$

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$α = 2; β = 1$

$α = 1; β = 1$

$α = 2; β = 2$

$α = 1; β = 2$

answer is B.

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Detailed Solution

$\bar{a} = t (\frac{\bar{b}}{| \bar{b} |} + \frac{\bar{c}}{| \bar{c} |})$

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