91Ó°ÊÓ

The vector $\stackrel{→}{C}=3\stackrel{→}{A}−4\stackrel{→}{B}$.

Consider a Vector quantity$\stackrel{→}{R}=R_x\widehat{i}+R_y\widehat{j}$, Here$R_x,R_y$are the components along x,y, directions respectively and $\widehat{i},\widehat{j}$are the unit vectors along x,y directions respectively. The magnitude of this vector is expressed as,

$\left|\stackrel{→}{R}\right|=\sqrt{{R_x}^2+{R_y}^2}$

Part (a)

$\stackrel{→}{A}$ can be represented as $\stackrel{→}{A}=A_x\widehat{i}+A_y\widehat{j}$. Here A_X and A_y are the components of $\stackrel{→}{A}$ in x and y directions respectively. $\stackrel{→}{i}$and $\stackrel{→}{j}$are the unit vectors in x and y directions respectively.

Using trigonometry, the components of $\stackrel{→}{A}$ can be expressed as,

$\begin{array}{c}{A}_x=\left(3.60\text{″¾}\right)\cos \left(70°\right)\\ =1.23\text{″¾}\\ A_y=\left(3.60\text{″¾}\right)\sin \left(70°\right)\\ =3.38\text{″¾}\end{array}$

Thus, $\stackrel{→}{A}$ can be represented as,

$\begin{array}{l}\stackrel{→}{A}=A_x\widehat{i}+A_y\widehat{j}\\ \stackrel{→}{A}=\left(1.23\text{″¾}\right)\widehat{i}+\left(3.38\text{″¾}\right)\widehat{j}\end{array}$

Similarly, the components of $\stackrel{→}{B}$ can be expressed as,

$\begin{array}{c}{B}_x=\left(−2.40\text{″¾}\right)\cos \left(30\right)\\ =−2.08\text{″¾}\\ B_{\text{y}}=\left(−2.40\text{″¾}\right)\sin \left(30\right)\\ =−1.2\text{″¾}\end{array}$

Thus, $\stackrel{→}{B}$ can be represented as,

$\begin{array}{l}\stackrel{→}{B}=B_x\widehat{i}+B_y\widehat{j}\\ \stackrel{→}{B}=\left(−2.08\text{″¾}\right)\widehat{i}+\left(−1.20\text{″¾}\right)\widehat{j}\end{array}$

Thus, the representation of $\stackrel{→}{A}$ and $\stackrel{→}{B}$ in terms of unit vectors is $\stackrel{→}{A}=\left(1.23\text{″¾}\right)\widehat{i}+\left(3.38\text{″¾}\right)\widehat{j}and\stackrel{→}{B}=\left(−2.08\text{″¾}\right)\widehat{i}+\left(−1.20\text{″¾}\right)\widehat{j}$.

Part (b)

The vector $\stackrel{→}{C}$ can be expressed as,

$\stackrel{→}{C}=3\stackrel{→}{A}−4\stackrel{→}{B}$

Substitute $\left(1.23\text{″¾}\right)\widehat{i}+\left(3.38\text{″¾}\right)\widehat{j}for\stackrel{→}{A}$$and\left(−2.08\text{″¾}\right)\widehat{i}+\left(−1.20\text{″¾}\right)\widehat{j}for\stackrel{→}{B}$

$\begin{array}{c}\stackrel{→}{C}=3\stackrel{→}{A}−4\stackrel{→}{B}\\ =3\left(\left(1.23\text{″¾}\right)\widehat{i}+\left(3.38\text{″¾}\right)\widehat{j}\right)−4\left(\left(−2.08\text{″¾}\right)\widehat{i}+\left(−1.20\text{″¾}\right)\widehat{j}\right)\\ =\left(3.69\text{″¾}\right)\widehat{i}+\left(10.14\text{″¾}\right)\widehat{j}−\left(−8.32\text{″¾}\right)\widehat{i}+\left(−4.8\text{″¾}\right)\widehat{j}\\ =\left(12.01\text{″¾}\right)\widehat{i}+\left(14.94\text{″¾}\right)\widehat{j}\end{array}$

The representation of $\stackrel{→}{C}$in terms of unit vectors is $\stackrel{→}{C}=\left(12.01\text{″¾}\right)\widehat{i}+\left(14.94\text{″¾}\right)\widehat{j}$.

Part (c)

The Vector $\stackrel{→}{C}$ can be expressed as,

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

Substitute 12.01 m for C_x and 14.94m for C_y,

$\stackrel{→}{C}=\left(12.01\text{″¾}\right)\widehat{i}+\left(14.94\text{″¾}\right)\widehat{j}$

The magnitude of vector $\stackrel{→}{C}$ can be expressed as,

$\left|\stackrel{→}{C}\right|=\sqrt{{C_x}^2+{C_y}^2}$

Substitute 12.01 m for C_x , and 14.94 m for C_y ,

$\begin{array}{c}\left|\stackrel{→}{C}\right|=\sqrt{{\left(12.01\text{″¾}\right)}^2+{\left(14.94\text{″¾}\right)}^2}\\ =\sqrt{367.4\text{″¾}}\\ =19.17\text{″¾}\end{array}$

The direction can be calculated as,

$\tan \mathrm{θ}=\frac{C_y}{C_x}$

Substitute 12.01 m for C_x and 14.94m for C_y .

$\begin{array}{c}\mathrm{θ}={\tan }^{−1}\left(\frac{14.94\text{″¾}}{12.01\text{″¾}}\right)\\ =51.2°\end{array}$

Thus, the magnitude of $\stackrel{→}{C}$ is 19.17 m, and its directions is 51.2^o in the first quadrant.