Question

For the three vectors shown in FIGURE EX3.20, $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}=1\hat{j}.$What is vector $\overrightarrow{B}?$

a. Write $\overrightarrow{B}$in component form.

b. Write $\overrightarrow{B}$as a magnitude and a direction.

Short answer

(a) $\overrightarrow{B}=-3\hat{i}+5\hat{j}.$

(a) The magnitude of$\overrightarrow{B}$is$\sqrt{37}$and the direction is$59.{03}^{\circ }.$

Step-by-step solution

Part (a) Step 1. Given information

The given figure is,

and$\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}=1\hat{j}.$We need to determine the vector$\overrightarrow{B}$in component form.

Step 2. Calculation

From the figure,

$\overrightarrow{A}=3\hat{i},\overrightarrow{C}=-4\hat{j}.$

$\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}=1\hat{j}.$

$3\hat{i}+\overrightarrow{B}-4\hat{j}=1\hat{j}.$

$\overrightarrow{B}=-3\hat{i}+5\hat{j}.$

Part (b) Step 1. Given information

The given figure is,

and$\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}=1\hat{j}.$We need to write$\overrightarrow{B}$as a magnitude and direction.

Step 2. Calculation

From part (a), $\overrightarrow{B}=-3\hat{i}+5\hat{j}.$

role="math" $\left|\overrightarrow{B}\right|=\sqrt{{\left(-3\right)}^2+{\left(5\right)}^2}.$

$=\sqrt{9+25}.$

$=\sqrt{34}.$

$\tan \theta =\frac{B_y}{B_x}.$

$\tan \theta =\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$-3$}\right..$

$\theta ={\tan }^{-1}\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$-3$}\right.\right).$

$=59.{03}^{\circ }.$