The vector overline{a}=alpha hat{i}+2 hat{j}+ | vedclass

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(A) Since $\overline{a}$ bisects the angle between $\overline{b}$ and $\overline{c}$,it must be proportional to the sum of the unit vectors along $\ov

Vedclass · Accepted Answer

$\alpha=1, \beta=1$

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(A) Since $\overline{a}$ bisects the angle between $\overline{b}$ and $\overline{c}$,it must be proportional to the sum of the unit vectors along $\overline{b}$ and $\overline{c}$.Let $\hat{b} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$ and $\hat{c} = \frac{\hat{j}+\hat{k}}{\sqrt{2}}$.The vector $\overline{a}$ is given by $\overline{a} = k(\hat{b} + \hat{c})$ for some scalar $k$.$\overline{a} = k \left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} + \frac{\hat{j}+\hat{k}}{\sqrt{2}} \right) = \frac{k}{\sqrt{2}} (\hat{i} + 2\hat{j} + \hat{k})$.Given $\overline{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k}$.Comparing the components,we have $\alpha = \frac{k}{\sqrt{2}}$,$2 = \frac{2k}{\sqrt{2}}$,and $\beta = \frac{k}{\sqrt{2}}$.From $2 = \frac{2k}{\sqrt{2}}$,we get $k = \sqrt{2}$.Substituting $k = \sqrt{2}$ into the expressions for $\alpha$ and $\beta$,we get $\alpha = \frac{\sqrt{2}}{\sqrt{2}} = 1$ and $\beta = \frac{\sqrt{2}}{\sqrt{2}} = 1$.Thus,$\alpha=1$ and $\beta=1$.

The vector $\overline{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}$ lies in the plane of the vectors $\bar{b}=\hat{i}+\hat{j}$ and $\bar{c}=\hat{j}+\hat{k}$ and bisects the angle between $\bar{b}$ and $\bar{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$?

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