Chapter 11.2, Problem 57E

To determine

Thecorrect option for the magnitude of the vector from (A) 4.684 (B) 7.4356 (C) 8.062 (D) 9.369 (E) 11.

Answer to Problem 57E

The correct option is (B).

Explanation of Solution

Given information:

First vector with magnitude 7 and direction angle $40°$ .

Second vector with magnitude 4 and direction angle $140°$ .

Formula Used:

For a vector $\overrightarrow{v}$ with magnitude $m$ angle $\theta$ , the components are:

  $\begin{array}{l}x=m\cos \theta \\ y=m\sin \theta \\ \therefore \left(x,y\right)=\left(m\cos \theta ,m\sin \theta \right)\end{array}$

Pythagoras theorem for a right angle triangle:

  $c=\sqrt{a^2+b^2}$ , $c=\text{hypotenuse},a=\text{adjecent}\text{side},b=\text{opposite}\text{side}$ .

Calculation:

Let $v_1$ be the vector with magnitude 7 and direction angle $40°$ .

Calculate the components of the vector. Use the formula $\left(x,y\right)=\left(m\cos \theta ,m\sin \theta \right)$ .

  $\begin{array}{c}{v}_1=\left(x_1,y_1\right)\\ =\left(7\cos 40°,7\sin 40°\right)\\ =\left(5.362,4.500\right)\end{array}$

Let $v_2$ be the vector with magnitude 4 and direction angle $140°$ .

Calculate the components of the vector. Use the formula $\left(x,y\right)=\left(m\cos \theta ,m\sin \theta \right)$ .

  $\begin{array}{c}{v}_2=\left(x_2,y_2\right)\\ =\left(4\cos 140°,4\sin 140°\right)\\ =\left(-3.064,2.571\right)\end{array}$

Let $v_r$ be the resultant vector of the addition of $v_1$ and $v_2$ .

  $\begin{array}{c}{v}_r=v_1+v_2\\ =\left(5.362,4.500\right)+\left(-3.064,2.571\right)\\ =\left(5.362-3.064,4.500+2.571\right)\\ =\left(2.298,7.071\right)\end{array}$

Now, calculate the magnitude. Use the Pythagoras theorem $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{c}{v}_r=\sqrt{{2.298}^2+{7.071}^2}\\ =7.435\end{array}$

Conclusion:

Hence, the correct option is (B).