Question

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

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 vector$\overrightarrow{C}?$

Short answer

(a) $\overrightarrow{C}=4\hat{i}-\hat{j}.$

(b) The graph is,

(c) The magnitude of $\overrightarrow{C}$is$\sqrt{17}$and the direction is$-{14}^{\circ }.$

Step-by-step solution

Part (a) Step 1. Given information

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

Step 2. Solution

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

role="math" $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}.$

role="math" localid="1649230879168" $=2\hat{i}+3\hat{j}+2\hat{i}-4\hat{j}.$

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

$=4\hat{i}-\hat{j}.$

Part (b) Step 1. Given information

It is given that,$\overrightarrow{A}=2\hat{i}+3\hat{j},\overrightarrow{B}=2\hat{i}-4\hat{j}and\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}.$We need to draw the coordinate system.

Step 2. Graph of the A→,B→ and C→

The graph of$\overrightarrow{A},\overrightarrow{B}and\overrightarrow{C}$is,

Part (c) Step 1. Given information

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

Step 2. Calculation

From part (a) $\overrightarrow{C}=4\hat{i}-\hat{j}.$

Thus, $C_x=4,C_y=-1.$

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

$=\sqrt{{\left(4\right)}^2+{\left(-1\right)}^2}.$

$=\sqrt{16+1}.$

$=\sqrt{17},$

and $\tan \theta =\raisebox{1ex}{$C_y$}\!\left/ \!\raisebox{-1ex}{$C_x$}\right..$

$\theta ={\tan }^{-1}\left(\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)={14}^{\circ }.$