91影视

The vector $\stackrel{鈫�}{C}=3\stackrel{鈫�}{A}鈭�4\stackrel{鈫�}{B}$.

Consider a Vector quantity$\stackrel{鈫�}{R}=R_x\widehat{i}+R_y\widehat{j}$, 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,

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

Part (a)

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

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

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

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

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

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

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

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

$\begin{array}{l}\stackrel{鈫�}{B}=B_x\widehat{i}+B_y\widehat{j}\\ \stackrel{鈫�}{B}=\left(鈭�2.08\text{鈥尘}\right)\widehat{i}+\left(鈭�1.20\text{鈥尘}\right)\widehat{j}\end{array}$

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

Part (b)

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

$\stackrel{鈫�}{C}=3\stackrel{鈫�}{A}鈭�4\stackrel{鈫�}{B}$

Substitute $\left(1.23\text{鈥尘}\right)\widehat{i}+\left(3.38\text{鈥尘}\right)\widehat{j}for\stackrel{鈫�}{A}$$and\left(鈭�2.08\text{鈥尘}\right)\widehat{i}+\left(鈭�1.20\text{鈥尘}\right)\widehat{j}for\stackrel{鈫�}{B}$

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

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

Part (c)

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

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

Substitute 12.01 m for C_x and 14.94m for C_y,

$\stackrel{鈫�}{C}=\left(12.01\text{鈥尘}\right)\widehat{i}+\left(14.94\text{鈥尘}\right)\widehat{j}$

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

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

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

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

The direction can be calculated as,

$\tan 胃=\frac{C_y}{C_x}$

Substitute 12.01 m for C_x and 14.94m for C_y .

$\begin{array}{c}胃={\tan }^{鈭�1}\left(\frac{14.94\text{鈥尘}}{12.01\text{鈥尘}}\right)\\ =51.2掳\end{array}$

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