If OP = 8 and overrightarrow{OP} makes angles | vedclass

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(B) Let the direction angles of $\overrightarrow{OP}$ be $\alpha = 45^\circ$,$\beta = 60^\circ$,and $\gamma$. We know that $\cos^2 \alpha + \cos^2 \be

Vedclass · Accepted Answer

$4(\sqrt{2}\hat{i} + \hat{j} \pm \hat{k})$

Vedclass · Answer

(B) Let the direction angles of $\overrightarrow{OP}$ be $\alpha = 45^\circ$,$\beta = 60^\circ$,and $\gamma$. We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Substituting the values: $\cos^2 45^\circ + \cos^2 60^\circ + \cos^2 \gamma = 1$. $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 + \cos^2 \gamma = 1$. $\frac{1}{2} + \frac{1}{4} + \cos^2 \gamma = 1 \implies \frac{3}{4} + \cos^2 \gamma = 1 \implies \cos^2 \gamma = \frac{1}{4}$. Thus,$\cos \gamma = \pm \frac{1}{2}$. The vector $\overrightarrow{OP}$ is given by $OP(\cos \alpha \hat{i} + \cos \beta \hat{j} + \cos \gamma \hat{k})$. $\overrightarrow{OP} = 8(\cos 45^\circ \hat{i} + \cos 60^\circ \hat{j} \pm \cos 60^\circ \hat{k})$. $\overrightarrow{OP} = 8(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} \pm \frac{1}{2} \hat{k})$. $\overrightarrow{OP} = 4(\sqrt{2} \hat{i} + \hat{j} \pm \hat{k})$.

If $OP = 8$ and $\overrightarrow{OP}$ makes angles $45^\circ$ and $60^\circ$ with the $OX$-axis and $OY$-axis respectively,then $\overrightarrow{OP} = $

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