91Ó°ÊÓ

The resolving of the vector depends upon the inclination of the vector with the respective axis.

The vector can be resolved in three-dimensional Cartesian coordinates.

The unit vector along thex-axis is taken as$\hat{\mathbf{i}}$, along they-axis is considered$\hat{\mathbf{j}}$, and along the z-axis$\hat{\mathbf{k}}$.

Consider that the distance between the summit of the mountain and the base camp is taken along the z-axis. The magnitude of the summit of the mountain is measured as 4580 m.

The vector can be represented as:

Here, north-south is taken along the y-axis, and west-east is taken along the x-axis.

The given data can be listed below as:

• The measure of the summit of the mountain is $M=4580\mathrm{m}$.
• The angle of inclination of the vector is $\mathrm{θ}=38.4°$.
• The distance along the z-axis is $z=2450\mathrm{m}$.

Resolve vector M along the x and y axes.

From thefigure, the x-component of vector M can be determined as:

$\begin{array}{c}{M}_x=-4580\sin 38.4°\mathrm{m}\\ =-2844.857\mathrm{m}\end{array}$

The y-component of vector M can be expressed as:

$\begin{array}{c}{M}_y=4580\cos 38.4°\mathrm{m}\\ =3589.316\mathrm{m}\end{array}$

The displacement vector can be represented as:

$\stackrel{→}{\mathbf{r}}=M_x\hat{\mathbf{i}}+M_y\hat{\mathbf{j}}+z\hat{\mathbf{k}}$

Substitute the values as $\left(-2844.857\mathrm{m}\right)$ for $M_x$, $\left(3589.316\mathrm{m}\right)$ for $M_y$, and $2450\mathrm{m}$ for z in the above equation.

$\stackrel{→}{\mathbf{r}}=\left(-2844.857\mathrm{m}\right)\hat{\mathbf{i}}+\left(3589.316\mathrm{m}\right)\hat{\mathbf{j}}+\left(2450\mathrm{m}\right)\hat{\mathbf{k}}$

Thus, the x-component, y-component, and z-component of the displacement vector from the camp to the summit are $\left(-2844.857\mathrm{m}\right)$, $\left(3589.316\mathrm{m}\right)$, and $\left(2450\mathrm{m}\right)$.

The magnitude of vector $\stackrel{→}{\mathbf{r}}$can be calculated as:

$\left|\stackrel{→}{\mathbf{r}}\right|=\sqrt{{\left(M_x\right)}^2+{\left(M_y\right)}^2+z^2}$

Substitute the values as $\left(-2844.857\mathrm{m}\right)$ for $M_x$, $\left(3589.316\mathrm{m}\right)$ for $M_y$, and$2450\mathrm{m}$ for zin the above equation.

$\begin{array}{c}\left|\stackrel{→}{\mathbf{r}}\right|=\sqrt{{\left(-2844.857\mathrm{m}\right)}^2+{\left(3589.316\mathrm{m}\right)}^2+\left(2450\mathrm{m}\right)}\\ =5194.122\mathrm{m}\end{array}$

Thus, the magnitude of the displacement vector $\stackrel{→}{\mathbf{r}}$is $5194.122\mathrm{m}$.