Question

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

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

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

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

Short answer

(a) $\overrightarrow{C}=\hat{i}+3\hat{j}.$

(b) The graph is,

(c) The magnitude is $\sqrt{10}$and the direction is$71.5^{\circ }.$

Step-by-step solution

Part (a) Step 1. Given information

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

Step 2. Solution

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

$\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}.$

$=4\hat{i}-2\hat{j}-3\hat{i}+5\hat{j}.$

$=\hat{i}+3\hat{j}.$

Part (b) Step 1. Given information

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

Step 2. Graph

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

Part (c) Step 1. Given information

It is given that$\overrightarrow{A}=4\hat{i}-2\hat{j},\overrightarrow{B}=-3\hat{i}+5\hat{j}and\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}.$From part (a) we need to determine the magnitude and direction of the vector$\overrightarrow{C}.$

Step 2. Calculation

$\overrightarrow{C}=\hat{i}+3\hat{j}.$

$C_x=1,and{C}_y=3.$

$\left|C\right|=\sqrt{C_x^2+{C_y}^2}=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}.$

$\tan \theta =\frac{C_x}{C_y}.$

$\tan \theta =\frac{3}{1}.$

$\theta ={\tan }^{-1}3=71.5^{\circ }.$