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## Q. 1.7

Referring to Fig. (a), determine (1) the rectangular representation of the position vector A; and (2) the angles between A and each of the positive coordinate axes.

## Verified Solution

Part 1
We first resolve $A$ into two components as shown in Fig. (b): $A_z$ along the $z$-axis and $A_{xy}$ in the $xy$-plane. (Once again we see that a carefully drawn sketch is an essential aid in performing vector resolution.) Because $A$, $A_{z}$, and $A_{xy}$ lie in the same plane (a diagonal plane of the parallelepiped), we obtain by trigonometry

$A_{z}=A\cos30^\circ=12\cos30^\circ=10.392\:$ m

$A_{xy}=A\sin30^\circ=12\sin30^\circ=6\:$ m

The next step, illustrated in Fig. (c), is to resolve $A_{xy}$ into the components along the coordinate axes:

$A_{x}=A_{xy}\cos40^\circ=6\cos40^\circ=4.596\:$ m

$A_{y}=A_{xy}\sin40^\circ=6\sin40^\circ=3.857\:$ m

Therefore, the rectangular representation of $A$ is $A=A_{x}i+A_{y}j+A_{z}k=4.60i+3.86j+10.39k\:$ m

Part 2
The angles between A and the coordinate axes can be computed from Eqs.(1.6):

$A_x = A\cos θ_x \: A_y = A\cosθ_y \: A_z = A\cosθ_z$                 (1.6)

$θ_{x}=\cos^{−1}\frac{A_{x}}{A}=\cos^{−1}\frac{4.596}{12}=67.5^\circ$

$θ_{y}=\cos^{−1}\frac{A_{y}}{A}=\cos^{−1}\frac{3.857}{12}=71.3^\circ$

$θ_{z}=\cos^{−1}\frac{A_{z}}{A} =\cos^{−1}\frac{10.392}{12}=30.0^\circ$

These angles are shown in Fig. (d). Note that it was not necessary to compute $θ_{z}$, because it was already given in  Fig. (a).