Find the position vector of a point A in spac | vedclass

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

Vedclass · Accepted Answer

$5\hat{i} + 5\sqrt{2}\hat{j} + 5\hat{k}$

Vedclass · Answer

(A) Let the direction angles of $\overrightarrow{OA}$ be $\alpha = 60^{\circ}$,$\beta = 45^{\circ}$,and $\gamma$. We know that $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$. Substituting the values,we get $\cos^{2} 60^{\circ} + \cos^{2} 45^{\circ} + \cos^{2} \gamma = 1$. $\Rightarrow (\frac{1}{2})^{2} + (\frac{1}{\sqrt{2}})^{2} + \cos^{2} \gamma = 1$. $\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^{2} \gamma = 1$. $\Rightarrow \frac{3}{4} + \cos^{2} \gamma = 1$. $\Rightarrow \cos^{2} \gamma = 1 - \frac{3}{4} = \frac{1}{4}$. $\Rightarrow \cos \gamma = \pm \frac{1}{2}$. Assuming $\gamma$ is acute,$\cos \gamma = \frac{1}{2}$. The direction cosines are $l = \cos 60^{\circ} = \frac{1}{2}$,$m = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,$n = \cos 60^{\circ} = \frac{1}{2}$. The position vector $\overrightarrow{OA} = |\overrightarrow{OA}| (l\hat{i} + m\hat{j} + n\hat{k})$. $\overrightarrow{OA} = 10 (\frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k})$. $\overrightarrow{OA} = 5\hat{i} + 5\sqrt{2}\hat{j} + 5\hat{k}$.

Find the position vector of a point $A$ in space such that $\overrightarrow{OA}$ is inclined at $60^{\circ}$ to $OX$ and at $45^{\circ}$ to $OY$ and $|\overrightarrow{OA}|=10$ units.

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