Question

Consider the three displacement vectors$\overrightarrow{A}=$ $(3\hat{i}-3\hat{j})m,\overrightarrow{B}=(\hat{i}-4\hat{j})m$, and$\overrightarrow{C}=(-2\hat{i}+5\hat{j})m$ . Use the component method to determine (a) the magnitude and direction of$\overrightarrow{D}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$ and (b) the magnitude and direction of$\overrightarrow{E}=-\overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}.$

Short answer

(a) $2.83m$and $315°$

(b) $13.4m$and $117°$.

Step-by-step solution

Define magnitude and direction

• In simple terms, magnitude means' distance or number.' It describes the absolute or relative size or direction in which an object moves in the feeling of motion. It's a term for describing the size or scope of something. In general, magnitude in physics refers to distance or amount.
• When we talk about direction, we're talking about a section of straight line that connects two places in space. The direction is from the beginning (first) to the end (second).

(a): Determine the magnitude and the direction with the positive axis in a counterclockwise orientation.

The component method is a sort of vector addition in which the same type of components are added.

Add the vectors together to get the total.

$$\begin{gathered} \overrightarrow{A}\times \overrightarrow{B}and\overrightarrow{C}. \\ \overrightarrow{D}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C} \end{gathered}$$

Substitute $(3\hat{i}-3\hat{j})m$for $\overrightarrow{A},(\hat{i}-4\hat{j})m$for B and $(-2\hat{i}+5\hat{j})m$for $\overrightarrow{C}.$

$\overrightarrow{D}=(3\hat{i}-3\hat{j})m+(\hat{i}-4\hat{j})m+(-2\hat{i}+5\hat{j})m$

Consider using the component approach of vector addition to simplify the equation.

$$\begin{gathered} \overrightarrow{D}=(3\hat{i}-3\hat{j})m+(\hat{i}-4\hat{j})m+(-2\hat{i}+5\hat{j})m \\ =(2\hat{i}-2\hat{j})m \end{gathered}$$

Determine the size of the problem D.

$$\begin{gathered} \left|\overrightarrow{D}\right|=\sqrt{2^2+{(-2)}^2} \\ \left|\overrightarrow{D}\right|=2.83m \end{gathered}$$

As a result, the magnitude of the vector data-custom-editor="chemistry" $\overrightarrow{D}$is $2.83m$.

Determine the direction of $\overrightarrow{D}.$

$\alpha =ta{n}^{-1}\left(\frac{y}{x}\right)$

Substitute -2 for y-component and 2 for x-component.

$$\begin{gathered} \alpha =ta{n}^{-1}\left(\frac{-2}{2}\right) \\ =-45 \end{gathered}$$

Here, $\alpha$is the angle formed by vector $\overrightarrow{D}$ in counterclockwise direction with positive x-axis. The negative sign indicates that the angle formed by $\overrightarrow{D}$is in a counterclockwise direction. As a result, the vector $\overrightarrow{D}$made a $360°-45°=315°$angle with the positive axis in a counterclockwise orientation.

Hence, the magnitude and the direction is 2.83m and$315°$

(b): Determine the magnitude and the direction with the positive axis in a counterclockwise orientation.

Add the two numbers together $-\overrightarrow{A},-\overrightarrow{B}$ and $\overrightarrow{C}.$

$\overrightarrow{E}=(-\overrightarrow{A})+(-\overrightarrow{B})+\overrightarrow{C}$

Substitute $(3\hat{i}-3\hat{j})m$ for $\overrightarrow{A},(\hat{i}-4\hat{j})m$for $\overrightarrow{B}$ and $(-2\hat{i}+5\hat{j})m$ for $\overrightarrow{C}$ .

$\overrightarrow{E}=-(3\hat{i}-3\hat{j})m-(\hat{i}-4\hat{j})m+(-2\hat{i}+5\hat{j})m$

Consider using the component approach of vector addition to simplify the equation.

$$\begin{gathered} \overrightarrow{E}=-(3\hat{i}-3\hat{j})m-(\hat{i}-4\hat{j})m+(-2\hat{i}+5\hat{j})m \\ =(-6\hat{i}+12\hat{j})m \end{gathered}$$

Determine the size of the problem .

$$\begin{gathered} \left|\overrightarrow{E}\right|=\sqrt{{(-6)}^2+{12}^2} \\ \left|\overrightarrow{E}\right|=\sqrt{36+144} \\ \left|\overrightarrow{E}\right|=13.4m \end{gathered}$$

As a result, the magnitude of the vector $\overrightarrow{E}$ is $13.41m.$

Determine the direction of $\overrightarrow{E}$

$\alpha =ta{n}^{-1}\left(\frac{y}{x}\right)$

Substitute 12 for y-component and -6 for x-component.

$$\begin{gathered} \alpha ={\tan }^{-1}\left(\frac{12}{-6}\right) \\ =-63° \end{gathered}$$

Here, $\alpha$is the angle formed by vector $\overrightarrow{E}$ in counterclockwise direction with the positive x-axis. The negative sign indicates that the angle formed by $\overrightarrow{E}$is counterclockwise.

As a result, the vector $\overrightarrow{E}$formed an angle of $180°-63.43°=117°$with the positive axis in a clockwise orientation.

Hence, the magnitude and the direction is $13.4m$and$117°$ .