Chapter 1, Problem 40SP

Find (a) $\stackrel{\to }{A}+\stackrel{\to }{B}+\stackrel{\to }{C}$, (b) $\stackrel{\to }{A}-\stackrel{\to }{B}$, and (c) $\stackrel{\to }{A}-\stackrel{\to }{C}$ if $\stackrel{\to }{A}=7\hat{i}-6\hat{j},\stackrel{\to }{B}=3\hat{i}+12\hat{j}$, and

$\stackrel{\to }{C}=4\hat{i}-4\hat{j}$.

To determine

The value of $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$ if $\overrightarrow{A}=7\widehat{i}-6\widehat{j}$, $\overrightarrow{B}=-3\widehat{i}+12\widehat{j}$, and $\overrightarrow{C}=4\widehat{i}-4\widehat{j}$.

Answer to Problem 40SP

Solution:

$8\widehat{i}+2\widehat{j}$

Explanation of Solution

Given data:

The value of $\overrightarrow{A}$ is $7\widehat{i}-6\widehat{j}$.

The value of $\overrightarrow{B}$ is $-3\widehat{i}+12\widehat{j}$.

The value of $\overrightarrow{C}$ is $4\widehat{i}-4\widehat{j}$.

Formula used:

Consider two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ such that $\overrightarrow{P}=a_1\widehat{i}+a_2\widehat{j}$ and $\overrightarrow{Q}=b_1\widehat{i}+b_2\widehat{j}$. The sum of vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is calculated as

$\begin{array}{c}\overrightarrow{P}+\overrightarrow{Q}=a_1\widehat{i}+a_2\widehat{j}+b_1\widehat{i}+b_2\widehat{j}\\ =\left(a_1+b_1\right)\widehat{i}+\left(a_2+b_2\right)\widehat{j}\end{array}$

Explanation:

Consider the sum of vectors $\overrightarrow{A}$, $\overrightarrow{B}$, and $\overrightarrow{C}$ as

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

Here, $\overrightarrow{S}$ represents the sum of vectors $\overrightarrow{A}$, $\overrightarrow{B}$, and $\overrightarrow{C}$.

Substitute $7\widehat{i}-6\widehat{j}$ for $\overrightarrow{A}$, $-3\widehat{i}+12\widehat{j}$ for $\overrightarrow{B}$, and $4\widehat{i}-4\widehat{j}$ for $\overrightarrow{C}$

$\begin{array}{c}\overrightarrow{S}=\left(7\widehat{i}-6\widehat{j}\right)+\left(-3\widehat{i}+12\widehat{j}\right)+\left(4\widehat{i}-4\widehat{j}\right)\\ =\left(7-3+4\right)\widehat{i}+\left(-6+12-4\right)\widehat{j}\\ =8\widehat{i}+2\widehat{j}\end{array}$

Conclusion:

The value of $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$ is $8\widehat{i}+2\widehat{j}$.

To determine

The value of $\overrightarrow{A}-\overrightarrow{B}$ if $\overrightarrow{A}=7\widehat{i}-6\widehat{j}$, $\overrightarrow{B}=-3\widehat{i}+12\widehat{j}$, and $\overrightarrow{C}=4\widehat{i}-4\widehat{j}$.

Answer to Problem 40SP

Solution:

$10\widehat{i}-18\widehat{j}$

Explanation of Solution

Given data:

The value of $\overrightarrow{A}$ is $7\widehat{i}-6\widehat{j}$.

The value of $\overrightarrow{B}$ is $-3\widehat{i}+12\widehat{j}$.

The value of $\overrightarrow{C}$ is $4\widehat{i}-4\widehat{j}$.

Formula used:

Consider two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ such that $\overrightarrow{P}=a_1\widehat{i}+a_2\widehat{j}$ and $\overrightarrow{Q}=b_1\widehat{i}+b_2\widehat{j}$. The difference of vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is calculated as

$\begin{array}{c}\overrightarrow{P}-\overrightarrow{Q}=\left(a_1\widehat{i}+a_2\widehat{j}\right)-\left(b_1\widehat{i}+b_2\widehat{j}\right)\\ =\left(a_1-b_1\right)\widehat{i}+\left(a_2-b_2\right)\widehat{j}\end{array}$

Explanation:

Consider the expression for $\overrightarrow{A}-\overrightarrow{B}$ as

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

Here, $\overrightarrow{D}$ represents the resultant vector equal to $\overrightarrow{A}-\overrightarrow{B}$.

Substitute $7\widehat{i}-6\widehat{j}$ for $\overrightarrow{A}$ and $-3\widehat{i}+12\widehat{j}$ for $\overrightarrow{B}$

$\begin{array}{c}\overrightarrow{D}=\left(7\widehat{i}-6\widehat{j}\right)-\left(-3\widehat{i}+12\widehat{j}\right)\\ =\left(7+3\right)\widehat{i}+\left(-6-12\right)\widehat{j}\\ =10\widehat{i}-18\widehat{j}\end{array}$

Conclusion:

The value of $\overrightarrow{A}-\overrightarrow{B}$ is $10\widehat{i}-18\widehat{j}$.

To determine

The value of $\overrightarrow{A}-\overrightarrow{C}$ if $\overrightarrow{A}=7\widehat{i}-6\widehat{j}$, $\overrightarrow{B}=-3\widehat{i}+12\widehat{j}$, and $\overrightarrow{C}=4\widehat{i}-4\widehat{j}$.

Answer to Problem 40SP

Solution:

$3\widehat{i}-2\widehat{j}$

Explanation of Solution

Given data:

The value of $\overrightarrow{A}$ is $7\widehat{i}-6\widehat{j}$.

The value of $\overrightarrow{B}$ is $-3\widehat{i}+12\widehat{j}$.

The value of $\overrightarrow{C}$ is $4\widehat{i}-4\widehat{j}$.

Formula used:

Consider two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ such that $\overrightarrow{P}=a_1\widehat{i}+a_2\widehat{j}$ and $\overrightarrow{Q}=b_1\widehat{i}+b_2\widehat{j}$. The difference between vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is calculated as

$\begin{array}{c}\overrightarrow{P}-\overrightarrow{Q}=\left(a_1\widehat{i}+a_2\widehat{j}\right)-\left(b_1\widehat{i}+b_2\widehat{j}\right)\\ =\left(a_1-b_1\right)\widehat{i}+\left(a_2-b_2\right)\widehat{j}\end{array}$

Explanation:

Consider the expression for $\overrightarrow{A}-\overrightarrow{C}$ as

$\overrightarrow{F}=\overrightarrow{A}-\overrightarrow{C}$

Here, $\overrightarrow{F}$ represents the resultant vector equal to $\overrightarrow{A}-\overrightarrow{C}$.

Substitute $7\widehat{i}-6\widehat{j}$ for $\overrightarrow{A}$ and $4\widehat{i}-4\widehat{j}$ for $\overrightarrow{C}$

$\begin{array}{c}\overrightarrow{F}=\left(7\widehat{i}-6\widehat{j}\right)-\left(4\widehat{i}-4\widehat{j}\right)\\ =\left(7-4\right)\widehat{i}+\left(-6+4\right)\widehat{j}\\ =3\widehat{i}-2\widehat{j}\end{array}$

Conclusion:

The value of $\overrightarrow{A}-\overrightarrow{C}$ is $3\widehat{i}-2\widehat{j}$.