Algebraic Vectors in r cubed plane

Question

A vector $\vec p$ with magnitude |$\vec P|=9$ in $R^3$ makes an equal direction angles with the coordinate axes. Find the coordinates of vector $\vec p$

Answer 1

Write the vector as $\vec{p}=(x,y,z).$ Since the angle with any of the coordinate axes is the same you have:

$$\cos \alpha = \frac{|(x,y,z)\cdot (1,0,0)|}{9}=\frac{|(x,y,z)\cdot (0,1,0)|}{9}=\frac{|(x,y,z)\cdot (0,0,z)|}{9},$$ where $\alpha$ is the common angle. So you have that $|x|= |y|=|z|,$ and since $\sqrt{x^2+y^2+z^2}=9,$ you have that $\sqrt{3}|x|=\sqrt{3}|y|=\sqrt{3}|z|=9,$ that is, $|x|=|y|=|z|= \frac{9}{\sqrt{3}}=3\sqrt{3}.$