(a) Write each vector in Fig. E1.39 in terms of the unit vectors i^ and j^ . (b) Use unit vectors to express vector C→ , where C→=3.00A→−4.00B→ (c) Find the magnitude and direction C→ .

AnswerThe representation of A→ and B→ in terms of unit vectors is A→=1.23 mi^+3.38 mj^ and B→=−2.08 mi^+−1.20 mj^ .The representation of C→ in terms of unit vectors is C→=12.01 mi^+14.94 ...

Q39E - University Physics with Modern Physics

Q39E

Question

(a) Write each vector in Fig. E1.39 in terms of the unit vectors $\hat{i}$ and $\hat{j}$ . (b) Use unit vectors to express vector $\vec{C}$ , where $\vec{C} = 3 .00 \vec{A} - 4 .00 \vec{B}$ (c) Find the magnitude and direction $\vec{C}$ .

Step-by-Step Solution

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Answer

Answer

The representation of $\vec{A}$ and $\vec{B}$ in terms of unit vectors is $\vec{A} = (1.23 m) \hat{i} + (3.38 m) \hat{j}$ and $\vec{B} = (- 2.08 m) \hat{i} + (- 1.20 m) \hat{j}$ .
The representation of $\vec{C}$ in terms of unit vectors is $\vec{C} = (12.01 m) \hat{i} + (14.94 m) \hat{j}$ .
The magnitude of $\vec{C}$ is 19.17 m and its directions is 51.2^o in the first quadrant.

1Step 1: Identification of given data

The vector $\vec{C} = 3 \vec{A} - 4 \vec{B}$ .

2Step-2: Vector Quantities and their magnitudes.

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

$|\vec{R}| = \sqrt{{R_{x}}^{2} + {R_{y}}^{2}}$

3Step 3: Determination of vectors from the given figure

Part (a)

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

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

$\begin{matrix} A_{x} = (3.60 m) \cos (70 °) \\ = 1.23 m \\ A_{y} = (3.60 m) \sin (70 °) \\ = 3.38 m \end{matrix}$

Thus, $\vec{A}$ can be represented as,

$\begin{array}{l} \vec{A} = A_{x} \hat{i} + A_{y} \hat{j} \\ \vec{A} = (1.23 m) \hat{i} + (3.38 m) \hat{j} \end{array}$

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

$\begin{matrix} B_{x} = (- 2.40 m) \cos (30) \\ = - 2.08 m \\ B_{y} = (- 2.40 m) \sin (30) \\ = - 1.2 m \end{matrix}$

Thus, $\vec{B}$ can be represented as,

$\begin{array}{l} \vec{B} = B_{x} \hat{i} + B_{y} \hat{j} \\ \vec{B} = (- 2.08 m) \hat{i} + (- 1.20 m) \hat{j} \end{array}$

Thus, the representation of $\vec{A}$ and $\vec{B}$ in terms of unit vectors is $\vec{A} = (1.23 m) \hat{i} + (3.38 m) \hat{j} a n d \vec{B} = (- 2.08 m) \hat{i} + (- 1.20 m) \hat{j}$ .

4Step-4: Representation of Vector C → in terms of unit vectors

Part (b)

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

$\vec{C} = 3 \vec{A} - 4 \vec{B}$

Substitute $(1.23 m) \hat{i} + (3.38 m) \hat{j} f o r \vec{A}$ $a n d (- 2.08 m) \hat{i} + (- 1.20 m) \hat{j} f o r \vec{B}$

$\begin{matrix} \vec{C} = 3 \vec{A} - 4 \vec{B} \\ = 3 ((1.23 m) \hat{i} + (3.38 m) \hat{j}) - 4 ((- 2.08 m) \hat{i} + (- 1.20 m) \hat{j}) \\ = (3.69 m) \hat{i} + (10.14 m) \hat{j} - (- 8.32 m) \hat{i} + (- 4.8 m) \hat{j} \\ = (12.01 m) \hat{i} + (14.94 m) \hat{j} \end{matrix}$