Find the projection of vec m=\sqrt2,-3,-5\ onto an axis that…

Find the projection of $\vec m=\{\sqrt2,-3,-5\}$ onto an axis that makes $\alpha=45^\circ$ with $Ox$, $\gamma=60^\circ$ with $Oz$ and an obtuse angle $\beta$ with $Oy$.

Explanation of Method

The projection of m onto an axis with given direction angles.

Step 1: Direction cosines

\[ \cos^2\beta=1-\cos^245^\circ-\cos^260^\circ=1-\tfrac12-\tfrac14=\tfrac14,\qquad \cos\beta=-\tfrac12\ \ (\beta\text{ obtuse}), \] so $\beta=120^\circ$ and the unit vector of the axis is

\[ \vec e=\left\{\tfrac{\sqrt2}{2},\ -\tfrac12,\ \tfrac12\right\}. \]

Step 2: Projection

\[ \mathrm{Pr}_{\vec e}\vec m=\vec m\cdot\vec e=\sqrt2\cdot\tfrac{\sqrt2}{2}+(-3)\left(-\tfrac12\right)+(-5)\cdot\tfrac12=1+\tfrac32-\tfrac52=0. \]

\[\mathrm{Pr}_{\vec e}\vec m=0\qquad(\vec m\perp\text{axis})\]

$O$ $\vec m$ $\vec e$ $\text{axis}$

$\vec e=\left\{\tfrac{\sqrt2}{2},-\tfrac12,\tfrac12\right\}$
$\vec m=\{\sqrt2,-3,-5\}$

$\vec m$ is perpendicular to the axis, so its projection onto the axis is $0$.

Result

\[\boxed{\,\mathrm{Pr}_{\vec e}\vec m=0\qquad(\vec m\perp\text{axis})\,}\]