91影视

Consider the distance between the original position and the base camp as vector $\stackrel{鈫�}{A}$and the distance between the new position and original position as vector $\stackrel{鈫�}{B}$. The distance between the new position and the base camp can be computed by subtracting vector B from vector A. The magnitude of the vector which would be the distance between the new location and the base camp is to be calculated by the general formula of magnitude of the vector. Similarly, angle between components of the vector, can be found as well.

Formulae

$$\begin{gathered} \stackrel{鈫�}{C}=\stackrel{鈫�}{A}鈥�\stackrel{鈫�}{B} \\ C=\left|\stackrel{鈫�}{C}\right|=\sqrt{{\left(\stackrel{鈫�}{C_x}\right)}^2+{\left(\stackrel{鈫�}{C_y}\right)}^2} \\ 胃={\tan }^{-1}\left(\frac{\stackrel{鈫�}{C_y}}{\stackrel{鈫�}{C_x}}\right) \end{gathered}$$

Given

$$\begin{gathered} \stackrel{鈫�}{A}=\left(5.6\mathrm{km}\right)\hat{\mathrm{j}},90掳 \\ \stackrel{鈫�}{\mathrm{B}}=\left(7.8\mathrm{km}\right),50掳 \end{gathered}$$

$\stackrel{鈫�}{A}=\left(5.6\mathrm{km}\right)\hat{\mathrm{j}},90掳$or

$\stackrel{鈫�}{A}=\left(5.6\mathrm{km}\right)\cos \left(90\right)\hat{\mathrm{i}}+\left(5.6\mathrm{km}\right)\sin \left(90\right)\hat{\mathrm{j}}$

$\stackrel{鈫�}{B}=\left(7.8\mathrm{km}\right),50掳$ or

$\stackrel{鈫�}{B}=\left(7.8\mathrm{km}\right)\cos \left(50\right)\hat{\mathrm{i}}+\left(10\mathrm{m}\right)\sin \left(50\right)\hat{\mathrm{j}}$

The displacement vector $\stackrel{鈫�}{C}$is

$$\begin{gathered} \stackrel{鈫�}{C}=\stackrel{鈫�}{A}-\stackrel{鈫�}{B} \\ =\left[\left(5.6\mathrm{km}\right)\cos \left(90\right)\hat{\mathrm{i}}+\left(5.6\mathrm{km}\right)\sin \left(90\right)\hat{\mathrm{j}}\right]-\left[\left(7.8\mathrm{km}\right)\cos \left(50\right)\hat{\mathrm{i}}+\left(10\mathrm{m}\right)\sin \left(50\right)\hat{\mathrm{j}}\right] \\ \stackrel{鈫�}{C}=\left(-5.01\mathrm{km}\right)\hat{\mathrm{i}}-\left(0.38\mathrm{km}\right)\hat{\mathrm{j}} \end{gathered}$$

The magnitude of $\stackrel{鈫�}{C}$is

$\sqrt{{\left(-5.01\right)}^2+{\left(-0.38\right)}^2}=5.0\mathrm{km}$

Therefore, magnitude of the displacement to reach base camp is$5.0\mathrm{km}$

The angle is

${\tan }^{-1}\left(\frac{-0.38}{-5.01}\right)=4.3掳$ South of due west

Therefore, angle of the displacement to reach base camp $4.3掳$south of west.