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Q34P

Expert-verifiedFound in: Page 150

Book edition
9th Edition

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

Pages
1624 pages

ISBN
9781133947271

**Vector ** $\overrightarrow{\mathbf{B}}$** 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" $\overrightarrow{\mathbf{B}}$

(a) The magnitude of $\overrightarrow{\mathrm{B}}$ is $\left|\overrightarrow{\mathrm{B}}\right|=7.81$

(b) The vector's angle with the *x*-axis is $\alpha =59.2\xb0$ , The vector's angle with the* y*-axis is $\beta =39.8\xb0$, The vector's angle with the *z*-axis is $\gamma =67.4\xb0$

- 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.

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, $\overrightarrow{\mathrm{B}}$ , this is the formula we utilise.,

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

Hence, the magnitude of $\overrightarrow{\mathrm{B}}$ is $\left|\overrightarrow{\mathrm{B}}\right|=7.81$

The vector's angle with the *x*-axis

$\begin{array}{rcl}\mathrm{\alpha}& =& {\mathrm{cos}}^{-1}\frac{{\mathrm{B}}_{\mathrm{x}}}{\left|\overrightarrow{\mathrm{B}}\right|}\\ & =& {\mathrm{cos}}^{-1}\frac{4}{7.81}\\ & =& 59.2\xb0\end{array}$

The vector's angle with the *y*-axis

$\begin{array}{rcl}\mathrm{\beta}& =& {\mathrm{cos}}^{-1}\frac{\mathrm{By}}{\left|\overrightarrow{\mathrm{B}}\right|}\\ & =& {\mathrm{cos}}^{-1}\frac{6}{7.81}\\ & =& 39.8\xb0\end{array}$

The vector's angle with the *z*-axis

$\begin{array}{rcl}\mathrm{\gamma}& =& {\mathrm{cos}}^{-1}\frac{{\mathrm{B}}_{2}}{\left|\overrightarrow{\mathrm{B}}\right|}\\ & =& {\mathrm{cos}}^{-1}\frac{3}{7\times \mathrm{B}1}\\ & =& 67.4\xb0\end{array}$

Hence, the vector's angle with the *x*-axis is $\alpha =59.2\xb0$, The vector's angle with the* y*-axis is $\beta =39.8\xb0$ , The vector's angle with the *z*-axis is $\gamma =67.4\xb0$

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