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Q 26P

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Found in: Page 73

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

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

a) The graph is drawn below.

b) $\stackrel{\to }{C}=5\stackrel{^}{i}+4\stackrel{^}{j}and\stackrel{\to }{D}=-1\stackrel{^}{i}+8\stackrel{^}{j}$

c) $|\stackrel{\to }{D}|=8.06and{\theta }_{D}=97.1°$

See the step by step solution

## Step 1: Define vector, polar coordinates, and unit vector

• 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.

## Step 2: (a) Plot the graph of the vector sum and vector difference

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

## Step 3: (b) Determine the vector total and difference

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

$\stackrel{\to }{C}=\stackrel{\to }{A}+\stackrel{\to }{B}\phantom{\rule{0ex}{0ex}}=\left(2\stackrel{^}{i}+6\stackrel{^}{j}\right)+\left(3\stackrel{^}{i}-2\stackrel{^}{j}\right)\phantom{\rule{0ex}{0ex}}=5\stackrel{^}{i}+4\stackrel{^}{j}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{D}=\stackrel{\to }{A}-\stackrel{\to }{B}\phantom{\rule{0ex}{0ex}}=\left(2\stackrel{^}{i}+6\stackrel{^}{j}\right)-\left(3\stackrel{^}{i}-2\stackrel{^}{j}\right)\phantom{\rule{0ex}{0ex}}=-1\stackrel{^}{i}+8\stackrel{^}{j}$

## Step 4: (c) Determine the polar coordinates

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

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

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