Question

The rectangle shown in Figure P3.56 has sides parallel to the xand yaxes. The position vectors of two corners are $\overrightarrow{\mathbf{A}}=10.0\text{m}$at $50.0°$and$\overrightarrow{\mathbf{B}}=12.0\text{m}$at$30.0°$

(a) Find the perimeter of the rectangle.

(b) Find the magnitude and direction of the vector from the origin to the upper-right corner of the rectangle.

Short answer

(a)The perimeter of the rectangle is$11.3\text{m}$

(b)The magnitude and the direction vector for the origin is$\left|\overrightarrow{\mathbf{C}}\right|=12.9\text{m}$ and $\theta =36.4°$.

Step-by-step solution

Definition of Perimeter of Rectangle.

• The entire distance included via way of means of a rectangle's edges or facets is referred to as its perimeter. Because a rectangle has 4 sides, the perimeter of the rectangle is same to the sum of the 4 facets.
• Because the perimeter is a linear measurement, the unit of the rectangle's perimeter might be metres, centimetres, inches, feet, and so on.

• $P=2(l+h)$
  

Determine the perimeter of the rectangle.

(a)

Given that, the position vector of upper left corner of the rectangle is$\overrightarrow{\mathbf{A}}=10.0\text{m}$ at $50.0°$and the position vector of lower right corner of the rectangle is$\overrightarrow{\mathbf{B}}=12.0\text{m}$at$30.0°$

The position vector$\overrightarrow{A}$ in Cartesian coordinate from can be written as

$$\begin{gathered} \overrightarrow{A}=\left(10m\right)cos\left(50.0^{°}\right)\widehat{x}+\left(10m\right)sin\left(50.0^{°}\right)\widehat{y} \\ \overrightarrow{A}=\left(6.43m\right)\widehat{x}+\left(7.67m\right)\widehat{x} \\ \overrightarrow{A}=A_x\widehat{x}+A_y\widehat{y} \end{gathered}$$

Hence, $A_x=6.43$and $A_y=7.67$

The position vector $\overrightarrow{B}$in Cartesian coordinate from can be written as

$$\begin{gathered} \overrightarrow{B}=\left(12m\right)cos\left(30.0^{°}\right)\widehat{x}+\left(12m\right)sin\left(30.0^{°}\right)\widehat{y} \\ \overrightarrow{B}=\left(10.4m\right)\widehat{x}+\left(6.00m\right)\widehat{x} \\ \overrightarrow{B}=B_x\widehat{x}+B_y\widehat{y} \end{gathered}$$

Hence,$B_x=10.4$anddata-custom-editor="chemistry" $B_y=6.00$

So, the length of the rectangle is

$$\begin{gathered} l=\left|{\overrightarrow{B}}_x\right|-\left|{\overrightarrow{A}}_x\right| \\ l=(10.4-6.43)\text{m} \\ l=(3.97)\text{m} \end{gathered}$$

The breath of the rectangle can be calculated as

$$\begin{gathered} h=\left|{\overrightarrow{A}}_y\right|-\left|{\overrightarrow{B}}_y\right| \\ h=(7.66-6.00)\text{m} \\ h=(1.66)\text{m} \end{gathered}$$

Thus, the perimeter of the rectangle is

$$\begin{gathered} p=2(l+h)\backslash \mathrm{hfill} \\ p=2(3.97\text{m}+1.66\text{m}) \\ p=11.3\text{m} \end{gathered}$$

Therefore, the perimeter of the rectangle is$11.3m$

To determine the magnitude and the direction vector from the origin.

(b)

From the above figure, the coordinates of the right upper corner of the rectangle is$\left(B_x,A_y\right).$

So, the position vector from the origin$(0,0)$to the upper right corner of the rectangle is

$$\begin{gathered} \overrightarrow{\mathbf{C}}=\left({\overrightarrow{B}}_x-0\right)+\left({\overrightarrow{A}}_y-0\right) \\ ={\overrightarrow{B}}_x+{\overrightarrow{A}}_y\backslash \mathrm{hfill} \\ =(10.4\hat{\mathbf{i}})\text{m}+(7.66\hat{\mathbf{j}})\text{m} \\ =(10.4\hat{\mathbf{i}}+7.66\hat{\mathbf{j}})\text{m} \end{gathered}$$

The magnitude of the position vector from the origin to the upper right corner of the rectangle is

$$\begin{gathered} \left|\overrightarrow{\mathbf{C}}\right|=\sqrt{{(10.4\text{m})}^2+{(7.66\text{m})}^2} \\ \left|\overrightarrow{\mathbf{C}}\right|=12.9\text{m} \end{gathered}$$

The direction of the position vector from the origin to the upper right corner of the rectangle is

Therefore, the magnitude and the direction vector for the origin is $\left|\overrightarrow{\mathbf{C}}\right|=12.9\text{m}$and$\theta =36.4^{°}$.