Question

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

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

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

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

Short answer

(a) $\overrightarrow{D}=7\hat{i}-7\hat{j}.$

(b) The graph is,

(c) The magnitude is $7\sqrt{2}$ and the direction is $-{45}^{\circ }.$

Step-by-step solution

Part (a) Step 1. Given information

It is given that$\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}.$

Step 2. Solution

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

$\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}.$

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

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

$\overrightarrow{D}=7\hat{i}-7\hat{j}.$

Part (b) Step 1. Given information

It is given that $\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}=7\hat{i}-7\hat{j}.$From, part (a) we need to draw the coordinate system and show the vector$\overrightarrow{A},\overrightarrow{B}and\overrightarrow{D}.$

Step 2. Graph

The graph of$\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{D}=7\hat{i}-7\hat{j}$is

Part (c) Step 1. Given information

It is given that $\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}=7\hat{i}-7\hat{j}.$We need to determine the magnitude and the direction of the vector$\overrightarrow{D}.$

Step 2. Calculation

$\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}=7\hat{i}-7\hat{j}.$From part (a).

$D_x=7and{D}_y=-7.$

$\left|D\right|=\sqrt{D_x^2+D_y^2}.$

$=\sqrt{{\left(7\right)}^2+{\left(-7\right)}^2}.$

$=\sqrt{49+49}=\sqrt{98}=7\sqrt{2}.$

$\tan \theta =\frac{D_x}{D_y}=\frac{-7}{7}=-1.$

$\theta ={\tan }^{-1}\left(-1\right)=-{45}^{\circ }.$