A vector makes 45^{circ} and 60^{circ} angles | vedclass

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(B) The direction cosines of a vector satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,where $\alpha, \beta, \gamma$ are the an

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{2} \hat{k}$

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(B) The direction cosines of a vector satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,where $\alpha, \beta, \gamma$ are the angles made with the $x, y, z$ axes respectively.Given $\alpha = 45^{\circ}$ and $\beta = 60^{\circ}$.Substituting these values: $\cos^2 45^{\circ} + \cos^2 60^{\circ} + \cos^2 \gamma = 1$.$(\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 + \cos^2 \gamma = 1$.$\frac{1}{2} + \frac{1}{4} + \cos^2 \gamma = 1$.$\frac{3}{4} + \cos^2 \gamma = 1 \Rightarrow \cos^2 \gamma = \frac{1}{4} \Rightarrow \cos \gamma = \frac{1}{2}$.The unit vector $\hat{n}$ is given by $\hat{n} = \cos \alpha \hat{i} + \cos \beta \hat{j} + \cos \gamma \hat{k}$.Therefore,$\hat{n} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{2} \hat{k}$.

$A$ vector makes $45^{\circ}$ and $60^{\circ}$ angles with the $x$ and $y$ axes respectively. The unit vector along the given vector is:

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