A vector overrightarrow{a} = alpha hat{i} + 2 | vedclass

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(D) Since $\overrightarrow{a}$ bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$,$\overrightarrow{a}$ must be parallel to the su

Vedclass · Accepted Answer

$\overrightarrow{a} \cdot \hat{k} - 4 = 0$

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(D) Since $\overrightarrow{a}$ bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$,$\overrightarrow{a}$ must be parallel to the sum of the unit vectors along $\overrightarrow{b}$ and $\overrightarrow{c}$.First,find the unit vectors:$\hat{b} = \frac{\overrightarrow{b}}{|\overrightarrow{b}|} = \frac{\hat{i} + \hat{j}}{\sqrt{1^2 + 1^2}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$$\hat{c} = \frac{\overrightarrow{c}}{|\overrightarrow{c}|} = \frac{\hat{i} - \hat{j} + 4 \hat{k}}{\sqrt{1^2 + (-1)^2 + 4^2}} = \frac{\hat{i} - \hat{j} + 4 \hat{k}}{\sqrt{18}} = \frac{\hat{i} - \hat{j} + 4 \hat{k}}{3 \sqrt{2}}$Thus,$\overrightarrow{a} = \lambda (\hat{b} + \hat{c}) = \lambda \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} + \frac{\hat{i} - \hat{j} + 4 \hat{k}}{3 \sqrt{2}} ight) = \frac{\lambda}{3 \sqrt{2}} [3(\hat{i} + \hat{j}) + (\hat{i} - \hat{j} + 4 \hat{k})] = \frac{\lambda}{3 \sqrt{2}} (4 \hat{i} + 2 \hat{j} + 4 \hat{k})$.Comparing this with $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k}$,we have the $y$-component as $2$. So,$\frac{\lambda}{3 \sqrt{2}} 	imes 2 = 2$,which gives $\frac{\lambda}{3 \sqrt{2}} = 1$.Therefore,$\overrightarrow{a} = 4 \hat{i} + 2 \hat{j} + 4 \hat{k}$.Checking the options:$\overrightarrow{a} \cdot \hat{k} = 4$. Thus,$\overrightarrow{a} \cdot \hat{k} - 4 = 0$.

Similar Questions

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If $a = i + 2j - 3k$ and $b = 3i - j + 2k$,find the angle between the vectors $a + b$ and $a - b$ in degrees.

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