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Q 26P

Expert-verifiedFound in: Page 73

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Given the vectors ** $\overrightarrow{A}=2.00\hat{i}+6.00\hat{j}$** and **$\overrightarrow{B}=3.00\hat{i}-2.00\hat{j}$**, (a) draw the vector sum **$\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ **and the vector difference $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}.$ ****(b) Calculate **$\overrightarrow{C}$** and $\overrightarrow{D}$ ** **, in terms of unit vectors. (c) Calculate $\overrightarrow{C}$** **and $\overrightarrow{D}$** **in terms of polar coordinates, with angles measured with respect to the positive x axis.**

a) The graph is drawn below.

b) $\overrightarrow{C}=5\hat{i}+4\hat{j}and\overrightarrow{D}=-1\hat{i}+8\hat{j}$

c) $\left|\overrightarrow{D}\right|=8.06and{\theta}_{D}=97.1\xb0$

- A quantity that has both magnitude and direction is called a vector. It's usually represented by an arrow with the same direction as the amount and a length proportionate to the magnitude of the quantity.
- A vector does not have a position, even though it has magnitude and direction.
- Polar coordinates are two integers that indicate the position of a point on a plane: a distance and an angle.
- A unit vector is a vector with the same magnitude as the magnitude of the magnitude.

The vector sum $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ and the vector difference $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}.$

We just add or subtract the corresponding components to determine the vector total or difference.

$\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}\phantom{\rule{0ex}{0ex}}=(2\hat{i}+6\hat{j})+(3\hat{i}-2\hat{j})\phantom{\rule{0ex}{0ex}}=5\hat{i}+4\hat{j}\phantom{\rule{0ex}{0ex}}\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}\phantom{\rule{0ex}{0ex}}=(2\hat{i}+6\hat{j})-(3\hat{i}-2\hat{j})\phantom{\rule{0ex}{0ex}}=-1\hat{i}+8\hat{j}$

In order to find $\overrightarrow{C}$ and $\overrightarrow{D}$ in polar coordinates, we must first determine their magnitudes and angles with the *+x*-axis.

$\left|\overrightarrow{C}\right|=\sqrt{{5}^{2}+{4}^{2}}=6.40\phantom{\rule{0ex}{0ex}}{\theta}_{c}=ta{n}^{-1}\frac{4}{5}\phantom{\rule{0ex}{0ex}}=38.7\xb0\phantom{\rule{0ex}{0ex}}\left|\overrightarrow{D}\right|=\sqrt{{(-1)}^{2}+{8}^{2}}=8.06$

${\theta}_{D}=ta{n}^{-1}\frac{8}{-1}\phantom{\rule{0ex}{0ex}}=-82.9+180\xb0\phantom{\rule{0ex}{0ex}}=97.1\xb0\phantom{\rule{0ex}{0ex}}Hence,\left|\overrightarrow{D}\right|=8.06and{\theta}_{D}=97.1\xb0\phantom{\rule{0ex}{0ex}}$

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