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Expert-verified Found in: Page 150 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # Vector $\stackrel{\mathbf{\to }}{\mathbf{B}}$ has x, y, and z components of 4.00, 6.00, and 3.00 units, respectively. Calculate (a) the magnitude of role="math" localid="1663647737200" $\stackrel{\mathbf{\to }}{\mathbf{B}}$ and (b) the angle that $\stackrel{\mathbf{\to }}{\mathbf{B}}$ makes with each coordinate axis.

(a) The magnitude of $\stackrel{\to }{\mathrm{B}}$ is $|\stackrel{\to }{\mathrm{B}}|=7.81$

(b) The vector's angle with the x-axis is $\alpha =59.2°$ , The vector's angle with the y-axis is $\beta =39.8°$, The vector's angle with the z-axis is $\gamma =67.4°$

See the step by step solution

## Step 1: Definition of magnitude and angle of vector

• In simple terms, magnitude means 'distance or number.' It depicts the absolute or relative size or direction in which an object moves in the feeling of motion.
• It's a term for describing the size or scope of something. In general, magnitude in physics refers to distance or amount.
• The shortest angle at which any of two vectors can be rotated around the other vector so that both vectors have the same direction is called an angle between two vectors.

## Step 2(a): Determine the magnitude

Let us consider,

${\mathrm{B}}_{\mathrm{x}}=4\mathrm{units}\phantom{\rule{0ex}{0ex}}{\mathrm{B}}_{\mathrm{y}}=6\mathrm{units}\phantom{\rule{0ex}{0ex}}{\mathrm{B}}_{\mathrm{z}}=3\mathrm{units}$

In order to determine the magnitude, $\stackrel{\to }{\mathrm{B}}$ , this is the formula we utilise.,

$\begin{array}{rcl}|\stackrel{\to }{\mathrm{B}}|& =& \sqrt{{{\mathrm{B}}_{\mathrm{x}}}^{2}+{{\mathrm{B}}_{\mathrm{y}}}^{2}+{{\mathrm{B}}_{\mathrm{z}}}^{2}}\\ |\stackrel{\to }{\mathrm{B}}|& =& \sqrt{{4}^{2}+{6}^{2}+{3}^{2}}\\ |\stackrel{\to }{\mathrm{B}}|& =& 7.81\end{array}$

Hence, the magnitude of $\stackrel{\to }{\mathrm{B}}$ is $|\stackrel{\to }{\mathrm{B}}|=7.81$

## Step 3(b): Determine the angle that  B→  makes with each coordinate axis

The vector's angle with the x-axis

$\begin{array}{rcl}\mathrm{\alpha }& =& {\mathrm{cos}}^{-1}\frac{{\mathrm{B}}_{\mathrm{x}}}{|\stackrel{\to }{\mathrm{B}}|}\\ & =& {\mathrm{cos}}^{-1}\frac{4}{7.81}\\ & =& 59.2°\end{array}$

The vector's angle with the y-axis

$\begin{array}{rcl}\mathrm{\beta }& =& {\mathrm{cos}}^{-1}\frac{\mathrm{By}}{|\stackrel{\to }{\mathrm{B}}|}\\ & =& {\mathrm{cos}}^{-1}\frac{6}{7.81}\\ & =& 39.8°\end{array}$

The vector's angle with the z-axis

$\begin{array}{rcl}\mathrm{\gamma }& =& {\mathrm{cos}}^{-1}\frac{{\mathrm{B}}_{2}}{|\stackrel{\to }{\mathrm{B}}|}\\ & =& {\mathrm{cos}}^{-1}\frac{3}{7×\mathrm{B}1}\\ & =& 67.4°\end{array}$

Hence, the vector's angle with the x-axis is $\alpha =59.2°$, The vector's angle with the y-axis is $\beta =39.8°$ , The vector's angle with the z-axis is $\gamma =67.4°$ ### Want to see more solutions like these? 