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

Given that the vector $\mathbf{r}$ is inclined at equal angles to the $x$, $y$, and $z$ axes, its direction cosines with respect to these axes are identical. Let this common direction cosine be $\cos \theta$. Consequently, the components of vector $\mathbf{r}$ are expressed as: \[ \mathbf{r} = (r \cos \theta, r \cos \theta, r \cos \theta). \] The magnitude of $\mathbf{r}$ is calculated as: \[ |\mathbf{r}| = \sqrt{(r \cos \theta)^2 + (r \cos \theta)^2 + (r \cos \theta)^2} = \sqrt{3r^2 \cos^2 \theta}. \] As the magnitude of $\mathbf{r}$ is given as $5\sqrt{3}$, we equate this to the derived magnitude: \[ 5\sqrt{3} = \sqrt{3r^2 \cos^2 \theta}. \] Squaring both sides yields: \[ 75 = 3r^2 \cos^2 \theta. \] Solving for $r^2 \cos^2 \theta$: \[ r^2 \cos^2 \theta = 25. \] Therefore, the vector $\mathbf{r}$ is determined to be: \[ \mathbf{r} = (5, 5, 5). \]