91Ó°ÊÓ

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

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

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

From the given diagram the components,

$\begin{array}{l}{A}_x=0\\ A_y=−8.0\text{″¾}\end{array}$

Thus, the representation of $\stackrel{→}{A}$ in terms of unit vectors is $\stackrel{→}{A}=0\widehat{i}−\left(8.0\text{″¾}\right)\widehat{j}$.

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

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

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

$\begin{array}{c}\mathrm{θ}=90°−30°\\ =60°\end{array}$

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

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

Substitute 15.0 m for B, and 60^o for$\mathrm{θ}$in the above equations,

$\begin{array}{c}{B}_x=\left(15.0\text{″¾}\right)\cos \left(60°\right)\\ =7.50\text{″¾}\\ B_y=\left(15.0\text{″¾}\right)\sin \left(60°\right)\\ =13.0\text{″¾}\end{array}$

Thus, the representation of vector $\stackrel{→}{B}$in terms of unit vectors is, $\stackrel{→}{B}=\left(7.50\text{​″¾}\right)\widehat{i}+\left(13.0\text{″¾}\right)\widehat{j}$.

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

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

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

$\mathrm{θ}={25}^o$

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

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

Substitute -12.0 m for C and 25^o for$\mathrm{θ}$in the above equations,

$\begin{array}{c}{C}_x=\left(−12.0\text{″¾}\right)\cos \left(25°\right)\\ =−10.9\text{″¾}\\ C_y=\left(−12.0\text{″¾}\right)\sin \left(25°\right)\\ =−5.07\text{″¾}\end{array}$

Thus, the representation of vector $\stackrel{→}{C}$in terms of unit vectors is, $\stackrel{→}{C}=\left(−10.9\text{​″¾}\right)\widehat{i}+\left(−5.07\text{″¾}\right)\widehat{j}$.

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

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

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

$\begin{array}{c}\mathrm{θ}=90°−53°\\ =37°\end{array}$

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

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

Substitute 10.0m for D and 37^o for$\mathrm{θ}$in the above equations,

$\begin{array}{c}{D}_x=\left(10.0\text{″¾}\right)\cos \left(37°\right)\\ =−7.99\text{″¾}\\ D_y=\left(−12.0\text{″¾}\right)\sin \left(37°\right)\\ =6.02\text{″¾}\end{array}$

Thus, the representation of vector $\stackrel{→}{D}$ in terms of unit vectors is, data-custom-editor="chemistry" $\stackrel{→}{D}=\left(−7.99\text{​″¾}\right)\widehat{i}+\left(6.02\text{″¾}\right)\widehat{j}$.