#### Find the position vector of a point A in space, so that $\overline{OA}$ is inclined at  60⁰ to $\overline{OX}$ and 45⁰ to $\overline{OY}$ and $\left |\overline{OA} \right |$= 10 units.

Given, $\overline{OA}$ is inclined at 600 to and at  $\overline{OX}$ 450 to $\overline{OY}$

$\overline{OA}$ = 10 units.

We want to find the position vector of point A in space, which is nothing but  $\overline{OA}$

We know, there are three axes in space: X, Y, and Z.

Let OA be inclined with OZ at an angle α.

We know, directions cosines are associated by the relation:

l² + m² + n² = 1 ….(i)

In this question, direction cosines are the cosines of the angles inclined by  on  $\overline{OA}$$\overline{OX}$$\overline{OY}$ and  $\overline{OZ}$

So,$l=\cos 60^{\circ},m=\cos 45^{\circ},n=\cos \alpha$

Substituting the values of l, m, and n in equation (i),

$\left (\cos 60^{\circ} \right )^{2}+\left (m=\cos 45^{\circ} \right )^{2}+\left (n=\cos \alpha \right )^{2}=1$

We know the values of $\cos 60^{\circ}$and $\cos 45^{\circ}$, i.e. 1/2 and 1/√2 respectively.

Therefore, we get

$\left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{\sqrt{2}} \right )^{2}+\cos^{2}\alpha =1$

$\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos^{2}\alpha =1$

$\Rightarrow \cos^{2}\alpha =1-\frac{1}{4}-\frac{1}{2}$

$\Rightarrow \cos^{2}\alpha =\frac{4-1-2}{4}$

$\Rightarrow \cos^{2}\alpha =\frac{1}{4}$

$\Rightarrow \cos \alpha =\pm \sqrt{\frac{1}{4}}$

$\Rightarrow \cos \alpha =\pm \frac{1}{2}$

So $\overrightarrow{OA}$ is given as

$\overrightarrow{OA}=\overrightarrow{OA}\left ( l\hat{i}+m\hat{j}+n\hat{k} \right )$..........(ii)

We have,

$l = \cos 60^{\circ} = \frac{1}{2}\\ m =\ cos 45^{\circ} = \frac{1}{\sqrt{2}}\\ n = \cos \alpha = \pm \frac{1}{2}\\$

Inserting these values of l, m and n in equation (ii),

$\overrightarrow{OA}=\left |\overrightarrow{OA} \right |\left ( \frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k} \right )$

Also ,Put $| \overrightarrow{OA}| =10$

$\Rightarrow \overrightarrow{OA}=10\left ( \frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k} \right )$

$\Rightarrow \overrightarrow{OA}=10\times \frac{1}{2} \hat{i}+10\times\frac{1}{\sqrt{2}}\hat{j}+10\times\frac{1}{2}\hat{k}$

$\Rightarrow \overrightarrow{OA}=5i+10\times\frac{\sqrt{2}}{\sqrt{2}}\times+10\times\frac{1}{\sqrt{2}}\hat{j}\pm 5\hat{k}$

$\Rightarrow \overrightarrow{OA}=5i+\frac{10\times\sqrt{2}}{2}\hat{j}\pm 5\hat{k}$

$\Rightarrow \overrightarrow{OA}=5i+5\sqrt{2}\hat{j}\pm 5\hat{k}$

Thus, the position vector of A in space $=5i+5\sqrt{2}\hat{j}\pm 5\hat{k}$