Question

Vector $\mathbf{r}$ is inclined at equal angles to the three axes $x$, $y$, and $z$. If the magnitude of $\mathbf{r}$ is $5\sqrt{3}$ units, then find $\mathbf{r}$.

Hint:

When a vector is inclined at equal angles to the axes, its components are all equal, and the magnitude can be used to determine the components.

Answer 1

Since the vector $\mathbf{r}$ is inclined at equal angles to the $x$, $y$, and $z$ axes, the direction cosines of $\mathbf{r}$ with respect to the $x$, $y$, and $z$ axes are equal. Let the direction cosines be $\cos \theta$ for all three axes. Then, the components of the vector $\mathbf{r}$ are: \[ \mathbf{r} = (r \cos \theta, r \cos \theta, r \cos \theta). \] The magnitude of $\mathbf{r}$ is given by: \[ |\mathbf{r}| = \sqrt{(r \cos \theta)^2 + (r \cos \theta)^2 + (r \cos \theta)^2} = \sqrt{3r^2 \cos^2 \theta}. \] Since the magnitude of $\mathbf{r}$ is $5\sqrt{3}$, we have: \[ 5\sqrt{3} = \sqrt{3r^2 \cos^2 \theta}. \] Squaring both sides: \[ 75 = 3r^2 \cos^2 \theta. \] Solving for $r^2 \cos^2 \theta$: \[ r^2 \cos^2 \theta = 25. \] Thus, the vector $\mathbf{r}$ is: \[ \mathbf{r} = (5, 5, 5). \]