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

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(D) Since $\vec{a}$ lies in the plane of $\vec{b}$ and $\vec{c}$,we can write $\vec{a} = x\vec{b} + y\vec{c}$ for some scalars $x$ and $y$.$\vec{a} =

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$\alpha = 1, \beta = 1$

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(D) Since $\vec{a}$ lies in the plane of $\vec{b}$ and $\vec{c}$,we can write $\vec{a} = x\vec{b} + y\vec{c}$ for some scalars $x$ and $y$.$\vec{a} = x(\hat{i} + \hat{j}) + y(\hat{j} + \hat{k}) = x\hat{i} + (x+y)\hat{j} + y\hat{k}$.Comparing this with $\vec{a} = \alpha \hat{i} + 2\hat{j} + \beta \hat{k}$,we get $x = \alpha$,$y = \beta$,and $x+y = 2$,so $\alpha + \beta = 2$.Since $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$,$\vec{a}$ must be proportional to the sum of the unit vectors of $\vec{b}$ and $\vec{c}$.$\hat{b} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$ and $\hat{c} = \frac{\hat{j} + \hat{k}}{\sqrt{2}}$.$\hat{b} + \hat{c} = \frac{1}{\sqrt{2}} (\hat{i} + 2\hat{j} + \hat{k})$.Thus,$\vec{a} = k(\hat{i} + 2\hat{j} + \hat{k})$ for some constant $k$.Comparing coefficients with $\vec{a} = \alpha \hat{i} + 2\hat{j} + \beta \hat{k}$,we get $\alpha = k$,$2 = 2k$,and $\beta = k$.From $2 = 2k$,we find $k = 1$.Therefore,$\alpha = 1$ and $\beta = 1$.

The vector $\vec{a} = \alpha \hat{i} + 2\hat{j} + \beta \hat{k}$ lies in the plane of $\vec{b} = \hat{i} + \hat{j}$ and $\vec{c} = \hat{j} + \hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Find the possible values of $\alpha$ and $\beta$.

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