Chapter 14.5, Problem 4MRE

To determine

To find: the angle $\text{γ}$ that vector $\text{w}$ makes with the positive x-axis.

Answer to Problem 4MRE

  $\text{γ=329}{\text{.03}}^{\text{0}}$ .

Explanation of Solution

Given:

  $\text{u=3i+j}$ and $\text{v=2i-4j}$ .

Concept used:

From Pythagoras theorem:

  ${\text{r}}^{\text{2}}{\text{=x}}^{\text{2}}{\text{+y}}^{\text{2}}$ .

Where here $\text{r}$ represents the absolute value or modulus and the angle $\text{θ}$ is called the argument of the complex number.

By using the basic trigonometric ratios:

  $\text{cosθ=}\frac{\text{x}}{\text{r}}$ and $\text{sinθ=}\frac{\text{y}}{\text{r}}$ or $\text{x=rcosθ and y=rsinθ}$ .

Argument:

  ${\text{θ=tan}}^{\text{-1}}\frac{\text{y}}{\text{x}}$ .

Calculation:

  $\text{u=3i+j}$ and $\text{v=2i-4j}$ .

  $\text{w=u+v}$ is

  $\text{w=3i+j+2i-4j}$ .

Same unit vector will add up.

  $\text{w=5i-3j}$ .

Polar form of the equation:

  $\text{x=rcosθ}$ and $\text{y=rsinθ}$ .

  $\text{tanθ=}\frac{\text{y}}{\text{x}}=-\frac{3}{5}$ .

  ${\text{θ=tan}}^{\text{-1}}\left(\text{-}\frac{\text{3}}{\text{5}}\right)$ .

  $\text{θ=30}\text{.96}$ .

Since the angle $\text{γ}$ that $\text{w}$ makes with the positive x-axis is:

  $\text{γ=360-30}\text{.96}$ .

  $\text{γ=329}{\text{.03}}^{\text{0}}$ .

Hence, $\text{γ=329}{\text{.03}}^{\text{0}}$ .