problem 2.2 - find a) the position vector r_ab from point a to b, b) the unit vector u_ab in the same…

Explanation:

Step1: Find coordinates of points A and B

Point A: $x_A = 4 + 3=7$ m, $y_A = 2$ m, $z_A = 0$ m. For point B, $x_B=8\sin30^{\circ}\cos50^{\circ}\approx 2.57$ m, $y_B = 8\cos30^{\circ}\approx6.93$ m, $z_B=- 8\sin30^{\circ}\sin50^{\circ}\approx - 3.06$ m.

Step2: Calculate position - vector $\vec{r}_{AB}$

$\vec{r}_{AB}=(x_B - x_A)\vec{i}+(y_B - y_A)\vec{j}+(z_B - z_A)\vec{k}$ $=(2.57 - 7)\vec{i}+(6.93 - 2)\vec{j}+(-3.06 - 0)\vec{k}$ $=-4.43\vec{i}+4.93\vec{j}-3.06\vec{k}$ m.

Step3: Calculate magnitude of $\vec{r}_{AB}$

$|\vec{r}_{AB}|=\sqrt{(-4.43)^2+(4.93)^2+(-3.06)^2}$ $=\sqrt{19.62 + 24.30+9.36}$ $=\sqrt{53.28}\approx7.3$ m.

Step4: Calculate unit - vector $\vec{u}_{AB}$

$\vec{u}{AB}=\frac{\vec{r}{AB}}{|\vec{r}_{AB}|}=\frac{-4.43\vec{i}+4.93\vec{j}-3.06\vec{k}}{7.3}$ $=-0.607\vec{i}+0.675\vec{j}-0.42\vec{k}$.

Step5: Calculate direction cosines

The direction cosines are the components of the unit - vector. So, $\cos\alpha=-0.607$, $\cos\beta = 0.675$, $\cos\gamma=-0.42$.

Answer:

a. $\vec{r}{AB}=-4.43\vec{i}+4.93\vec{j}-3.06\vec{k}$ m b. $\vec{u}{AB}=-0.607\vec{i}+0.675\vec{j}-0.42\vec{k}$ c. $\cos\alpha=-0.607$, $\cos\beta = 0.675$, $\cos\gamma=-0.42$