Question

Consider the following three displacement vectors: Vector A has a magnitude of 3.70 km and a direction that makes an angle of 37.0° north of west, vector B has a magnitude of 7.50 km and a direction that makes an angle of 30.0° south of west, vector has a magnitude of 3.20 km and a direction that makes an angle of 42.0° north of east. Determine the magnitude of the vector Z=B-C+A

Answer 1

Answer:

7.1 units

Explanation:

We are given that

Magnitude of vector A=3.7 km

$\theta_1=37^{\circ}$ N of W

Magnitude of vector B=7.5 km

$\theta_2=$30 degree S of W

Magnitude of vector C=3.2 km

$\theta_3=42^{\circ}$

We have to determine the magnitude of vector

Z=B-C+A

$\vec{A}=-3.7cos 37i+3.7sin 37j=-2.95i+2.23 j$

$\vec{B}=-7.5cos30i-7.5sin 30j=-6.50 i-3.75 j$

$\vec{C}=3.2cos42i+3.2sin42 j=2.38i+2.14 j$

Substitute the values

$Z=-2.95i+2.23 j-6.5i-3.75j+2.38i+2.14j$

$Z=-7.07 i+0.62 j$

$\mid Z\mid=\sqrt{(-7.07)^2+(0.62)^2}=7.1$ units.