The projection of a vector $\vec{r} = 3\hat{i} + \hat{j} + 2\hat{k}$ on the $xy$-plane has magnitude:

If two forces of $5 \, N$ each are acting along $X$ and $Y$ axes,then the magnitude and direction of the resultant is

$A$ vector in the $x-y$ plane makes an angle of $30^{\circ}$ with the $y$-axis. The magnitude of the $y$-component of the vector is $2 \sqrt{3}$. The magnitude of the $x$-component of the vector will be:

Two forces act on point $A$ along the sides $AC$ and $AB$ of a right-angled triangle $ABC$ (where $\angle A = 90^{\circ}$). The magnitudes of these forces are inversely proportional to the lengths of the sides $AC$ and $AB$ respectively. If $F_1 = 1/AC$ and $F_2 = 1/AB$,then the resultant of these forces is proportional to:

$A$ person walks $25.0^{\circ}$ north of east for $3.18 \,km$. How far would she have to walk due north and then due east to arrive at the same location?

Find the angle the vector $\sqrt{3} \hat{i} - \hat{j}$ makes with the positive $x$-axis. (in $^{\circ}$)