Write each vector in Fig. E1.24 in terms of the unit vectors $\widehat{\mathbf{i}}$ and $\widehat{j}$

• Length of vector $\overrightarrow{A}$ is 8.00 m.
• Length of vector $\overrightarrow{B}$ is 15.0 m.
• Length of vector $\overrightarrow{C}$ is 12.0 m.
• Length of vector $\overrightarrow{D}$ is 10.0 m.

The representation of $\overrightarrow{A}$ in terms of unit vectors is,

$\overrightarrow{A}=A_x\widehat{i}+A_y\widehat{j}$.

From the given diagram the components,

$\begin{array}{l}{A}_x=0\\ A_y=-8.0\text{\hspace{0.17em}m}\end{array}$

Thus, the representation of $\overrightarrow{A}$ in terms of unit vectors is $\overrightarrow{A}=0\widehat{i}-\left(8.0\text{\hspace{0.17em}m}\right)\widehat{j}$.

The representation of $\overrightarrow{B}$ in terms of unit vectors is,

$\overrightarrow{B}=B_x\widehat{i}+B_y\widehat{j}$.

From the given diagram angle between $\overrightarrow{B}$ and x-axis is,

$\begin{array}{c}\theta =90°-30°\\ =60°\end{array}$

Thus, the components of vector $\overrightarrow{B}$ is,

$\begin{array}{l}{B}_x=B\cos \theta \\ B_y=B\sin \theta \end{array}$

Substitute 15.0 m for B, and 60^o for$\theta$in the above equations,

$\begin{array}{c}{B}_x=\left(15.0\text{\hspace{0.17em}m}\right)\cos \left(60°\right)\\ =7.50\text{\hspace{0.17em}m}\\ B_y=\left(15.0\text{\hspace{0.17em}m}\right)\sin \left(60°\right)\\ =13.0\text{\hspace{0.17em}m}\end{array}$

Thus, the representation of vector $\overrightarrow{B}$in terms of unit vectors is, $\overrightarrow{B}=\left(7.50\text{\hspace{0.17em}m}\right)\widehat{i}+\left(13.0\text{\hspace{0.17em}m}\right)\widehat{j}$.

The representation of $\overrightarrow{C}$ in terms of unit vectors is,

$\overrightarrow{C}=C_x\widehat{i}+C_y\widehat{j}$.

From the given diagram angle between $\overrightarrow{C}$and x-axis is,

$\theta ={25}^o$

Thus, the components of vector $\overrightarrow{C}$ is,

$\begin{array}{l}{C}_x=C\cos \theta \\ C_y=C\sin \theta \end{array}$

Substitute -12.0 m for C and 25^o for$\theta$in the above equations,

$\begin{array}{c}{C}_x=\left(-12.0\text{\hspace{0.17em}m}\right)\cos \left(25°\right)\\ =-10.9\text{\hspace{0.17em}m}\\ C_y=\left(-12.0\text{\hspace{0.17em}m}\right)\sin \left(25°\right)\\ =-5.07\text{\hspace{0.17em}m}\end{array}$

Thus, the representation of vector $\overrightarrow{C}$in terms of unit vectors is, $\overrightarrow{C}=\left(-10.9\text{\hspace{0.17em}m}\right)\widehat{i}+\left(-5.07\text{\hspace{0.17em}m}\right)\widehat{j}$.

The representation of $\overrightarrow{D}$ in terms of unit vectors is,

$\overrightarrow{D}=D_x\widehat{i}+D_y\widehat{j}$

From the given diagram angle between $\overrightarrow{D}$ and x-axis is,

$\begin{array}{c}\theta =90°-53°\\ =37°\end{array}$

Thus, the components of vector $\overrightarrow{D}$ is,

$\begin{array}{l}{D}_x=D\cos \theta \\ D_y=D\sin \theta \end{array}$

Substitute 10.0m for D and 37^o for$\theta$in the above equations,

$\begin{array}{c}{D}_x=\left(10.0\text{\hspace{0.17em}m}\right)\cos \left(37°\right)\\ =-7.99\text{\hspace{0.17em}m}\\ D_y=\left(-12.0\text{\hspace{0.17em}m}\right)\sin \left(37°\right)\\ =6.02\text{\hspace{0.17em}m}\end{array}$

Thus, the representation of vector $\overrightarrow{D}$ in terms of unit vectors is, data-custom-editor="chemistry" $\overrightarrow{D}=\left(-7.99\text{\hspace{0.17em}m}\right)\widehat{i}+\left(6.02\text{\hspace{0.17em}m}\right)\widehat{j}$.