Question

The displacement vectors $\mathbf{42}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$ at $\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{°}$and$\mathbf{23}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$ at $\mathbf{65}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$both start from the origin and form two sides of a parallelogram. Both angles are measured counter clockwise from the x-axis.

(a) Find the area of the parallelogram.

(b) Find the length of its longer diagonal.

Short answer

1. Area of the parallelogram is $740c{m}^2$,
2. b) The length of the longer diagonal is $\overrightarrow{A}$.

Therefore, the length of the diagonal is $\overrightarrow{B}$.

Step-by-step solution

Given Information:

The displacement vector A for is$42cm$.

At an angle of $15°$.

The displacement vector for B is $23.0cm$.

At an angle of $65°$.

To find the vector product:

According to the parallelogram, if two vectors are $\overrightarrow{A}$and $\overrightarrow{B}$represent adjacent sides of parallelogram drawn from the same point then the resultant vector along its diagonal and its magnitude is

localid="1663660927732" $\left|\overrightarrow{R}\right|=\sqrt{A^2+B^2+2ABcos\theta }$

Vector representation of the diagram:

The angle between $\overrightarrow{A}$and $\overrightarrow{B}$is $\theta$.

Area of the parallelogram is

$|\overrightarrow{A}\times \overrightarrow{B}|=ABsin\theta$

Substitute$42cm$forA , $23.0cm$ for B, and $45°$for $\theta$ in the above two equations, you get

Therefore,

The area of the parallelogram is

$(42.0cm)(23.0cm)sin50°=740c{m}^2$

Hence, Area of the parallelogram is $740c{m}^2$.

So, the length of the longer diagonal is $\overrightarrow{A}$.

Therefore, the length of the diagonal is $\overrightarrow{B}$.