Question

Let $\overrightarrow{A}=4\widehat{i}-2\widehat{j}$,$\overrightarrow{B}=-3\widehat{i}+5\widehat{j}$ 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. The vector$\overrightarrow{D}=7\widehat{i}-7\widehat{j}$ .

b. The vector $\overrightarrow{D}$has a magnitude $\left|\overrightarrow{D}\right|=\sqrt{98}$and it is along the direction that makes an angle $-{45}^{°}$with the positive $x$-axis.

Step-by-step solution

Step.1.

VISUALIZE

Each vector is expressed as the addition of two components, one component along the x-direction and the other component in the $y$-direction. So we can use the tip-to-tail rule for vector addition and draw the vectors $\overrightarrow{A}$and $\overrightarrow{B}$using the given components, as shown below.

Step2

SOLVE

Given that,

Vector,$\overrightarrow{A}=4\widehat{i}-2\widehat{j}$

Vector,$\overrightarrow{B}=-3\widehat{i}+5\widehat{j}$

(a) Given that the vector

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

Substitute for $\overrightarrow{A}$and $\overrightarrow{B}$from equations (1) and (2).

$\begin{array}{l}\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}\\ =(4\widehat{i}-2\widehat{j})-(-3\widehat{i}+5\widehat{j})\\ =7\widehat{i}-7\widehat{j}\end{array}$

Thus, the vector, $\overrightarrow{D}=7\widehat{i}-7\widehat{j}$.

Step.3

(b) We are given that $\overrightarrow{D}=7\widehat{i}-7\widehat{j}$. Note that,

$\begin{array}{l}\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}\\ =\overrightarrow{A}+(-\overrightarrow{B})\\ =(4\widehat{i}-2\widehat{j})+(3\widehat{i}-5\widehat{j})\end{array}$

Step.4.

Now use the tip-to-tail rule for the vectors $\overrightarrow{A}$and $-\overrightarrow{B}$to obtain the vector $\overrightarrow{D}$as shown in the below figure.

Step.5

(c) From part (a), the resultant vector,$\overrightarrow{D}=7\widehat{i}-7\widehat{j}$

Hence the magnitude of the vector${\overrightarrow{D}}^{\text{is,}}$

$\begin{array}{l}\left|\overrightarrow{D}\right|=\sqrt{7^2+{(-7)}^2}\\ =\sqrt{49+49}\\ =\sqrt{98}\end{array}$

And the direction of the vector $\overrightarrow{D}$is given by,

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{-7}{7}\right)\\ ={\tan }^{-1}(-1)\\ =-45\end{array}$

Thus, the vector $\overrightarrow{D}$has a magnitude $\left|\overrightarrow{D}\right|=\sqrt{98}$and it is along the direction that makes an angle $-{45}^{°}$with the positive $x$-axis.