Question

Let $\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{F}=\overrightarrow{A}-4\overrightarrow{B}.$

a. Write vector $\overrightarrow{F}$in component form.

b. Draw a coordinate system and on it show vectors $\overrightarrow{A},\overrightarrow{B}and\overrightarrow{F}.$

c. What are the magnitude and direction of vector$\overrightarrow{F}?$

Short answer

(a) $\overrightarrow{F}=20\hat{i}-22\hat{j}.$

(b) The graph is,

(c) The magnitude is$2\sqrt{221}$and the direction is$-47.7^{\circ }.$

Step-by-step solution

Part (a) Step 1. Given information

It is given that,$\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{F}=\overrightarrow{A}-4\overrightarrow{B}.$We need to write the vector$\overrightarrow{F}$in component form.

Step 2. Solution

$\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}.$

$\overrightarrow{F}=\overrightarrow{A}-4\overrightarrow{B}.$

$=\left(4\hat{i}-2\hat{j}\right)-4\left(-3\hat{i}+5\hat{j}\right).$

$=4\hat{i}-2\hat{j}+12\hat{i}-20\hat{j}.$

$=20\hat{i}-22\hat{j}.$

Part (b) Step 1. Given information

It is given that$\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{F}=\overrightarrow{A}-4\overrightarrow{B}=20\hat{i}-22\hat{j}.$We need to draw the coordinate system and on it show vector$\overrightarrow{A},\overrightarrow{B}and\overrightarrow{F}.$

Step 2. Graph

The graph of$\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{F}=20\hat{i}-22\hat{j}$is,

Part (c) Step 1. Given information

It is given that$\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{F}=\overrightarrow{A}-4\overrightarrow{B}=20\hat{i}-22\hat{j}.$We need to determine the magnitude and direction of$\overrightarrow{F}.$

Step 2. Calculation

$\overrightarrow{F}=20\hat{i}-22\hat{j}.$

$F_x=20and{F}_y=-22.$

$\left|\overrightarrow{F}\right|=\sqrt{{\left(F_x\right)}^2+{\left(F_y\right)}^2}.$

$=\sqrt{{\left(20\right)}^2+{\left(-22\right)}^2.}$

$=\sqrt{400+484}.$

$=\sqrt{884}.$

$=2\sqrt{221}.$

$\tan \theta =\frac{F_y}{F_x}=\frac{-22}{20}=-\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$10$}\right..$

$\theta ={\tan }^{-1}\left(\raisebox{1ex}{$-11$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right).$

$=-47.7^{\circ }.$