Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the direction cosines of the vector $\overrightarrow{OP}$ be $l,m,n$.

Given that the vector is inclined at $45^{\circ }$ to the X-axis and $60^{\circ }$ to the Y-axis.

$l=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$

$m=\cos 60^{\circ }=\frac{1}{2}$

Using the relation $l^2+m^2+n^2=1$:

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

$\frac{1}{2}+\frac{1}{4}+n^2=1$

$n^2=1-\frac{3}{4}=\frac{1}{4}$

$n=\pm \frac{1}{2}$

The position vector $\vec{r}$ is given by $\vec{r}=|\overrightarrow{OP}|(l\hat{i}+m\hat{j}+n\hat{k})$.

Substituting $|\overrightarrow{OP}|=12$ and the values of $l,m,n$:

$\vec{r}=12\left(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{2}\hat{j}\pm \frac{1}{2}\hat{k}\right)$

$\vec{r}=6\sqrt{2}\hat{i}+6\hat{j}\pm 6\hat{k}$

Answer: $\vec{r}=6\sqrt{2}\hat{i}+6\hat{j}\pm 6\hat{k}$