Chapter 3, Problem 38PQ

(a)

To determine

The magnitude of the scalar components of the vector $\overrightarrow{A}$.

Answer to Problem 38PQ

The magnitude of the scalar components of the vector $\overrightarrow{A}$ are $\underset{\_}{A_x=3.00,A_y=2.00,A_z=5.00}$.

Explanation of Solution

Given vector is $\overrightarrow{A}=3.00\widehat{i}+2.00\widehat{j}+5.00\widehat{k}$.

Write the general form of a vector (say $\overrightarrow{A}$) in terms of the magnitude of its scalar components and the corresponding unit vectors.

  $\overrightarrow{A}=A_x\widehat{i}+A_y\widehat{j}+A_z\widehat{k}$

Here, $A_x$, $A_y$, and $A_z$ are the scalar components along $x$, $y$, and $z$ directions respectively.

Compare the given vector with the general form to deduce its scalar components.

  $A_x=3.00$

  $A_y=2.00$

  $A_z=5.00$

Conclusion:

Therefore, the magnitude of the scalar components of the vector $\overrightarrow{A}$ are $\underset{\_}{A_x=3.00,A_y=2.00,A_z=5.00}$.

(b)

To determine

The magnitude of the vector $\overrightarrow{A}$.

Answer to Problem 38PQ

The magnitude of the vector $\overrightarrow{A}$ is $\underset{\_}{6.16}$.

Explanation of Solution

Given vector is $\overrightarrow{A}=3.00\widehat{i}+2.00\widehat{j}+5.00\widehat{k}$.

Write the expression for the magnitude of the vector $\overrightarrow{A}$.

  $\left|\overrightarrow{A}\right|=\sqrt{A_x^2+A_y^2+A_z^2}$

Conclusion:

Substitute $3.00$ for $A_x$, $2.00$ for $A_y$, and $5.00$ for $A_z$ in the above equation to find $\left|\overrightarrow{A}\right|$.

  $\begin{array}{c}\left|\overrightarrow{A}\right|=\sqrt{{3.00}^2+{2.00}^2+{5.00}^2}\\ =6.16\end{array}$

Therefore, the magnitude of the vector $\overrightarrow{A}$ is $\underset{\_}{6.16}$.

(c)

To determine

The angles that the vector $\overrightarrow{A}$ makes with the $x$, $y$, and $z$ axes.

Answer to Problem 38PQ

The angles that the vector $\overrightarrow{A}$ makes with the $x$, $y$, and $z$ axes are $\underset{\_}{{\theta }_x=60.9°\text{from }+x}$, $\underset{\_}{{\theta }_y=71.1°\text{from }+y}$, and $\underset{\_}{{\theta }_z=35.8°\text{from }+z}$ respectively.

Explanation of Solution

Given vector is $\overrightarrow{A}=3.00\widehat{i}+2.00\widehat{j}+5.00\widehat{k}$.

In order to find the angle that the vector make with the coordinate axes, a right triangle can be formed with vector $\overrightarrow{A}$ as hypotenuse and making an angle $\theta$ with the coordinate axis about with the angle has to be measured. For such a construction, trigonometric relation can be deduced corresponding to each coordinate axes.

  $\cos {\theta }_x=\frac{A_x}{\left|\overrightarrow{A}\right|}$ (I)

Here, ${\theta }_x$ is the angle that the vector $\overrightarrow{A}$ makes with the $x$ axis.

  $\cos {\theta }_y=\frac{A_y}{\left|\overrightarrow{A}\right|}$ (II)

Here, ${\theta }_y$ is the angle that the vector $\overrightarrow{A}$ makes with the $y$ axis.

  $\cos {\theta }_z=\frac{A_z}{\left|\overrightarrow{A}\right|}$ (III)

Here, ${\theta }_z$ is the angle that the vector $\overrightarrow{A}$ makes with the $z$ axis.

Solve equation (I) for ${\theta }_x$, equation (II) for ${\theta }_y$, and equation (III) for ${\theta }_z$.

  ${\theta }_x={\cos }^{-1}\left(\frac{A_x}{\left|\overrightarrow{A}\right|}\right)$ (IV)

  ${\theta }_y={\cos }^{-1}\left(\frac{A_y}{\left|\overrightarrow{A}\right|}\right)$ (V)

  ${\theta }_z={\cos }^{-1}\left(\frac{A_z}{\left|\overrightarrow{A}\right|}\right)$ (VI)

Conclusion:

Substitute $3.00$ for $A_x$, and $6.16$ for $\left|\overrightarrow{A}\right|$ in equation (IV) to find ${\theta }_x$.

  $\begin{array}{c}{\theta }_x={\cos }^{-1}\left(\frac{3.00}{6.16}\right)\\ =60.9°\text{from }+x\end{array}$

Substitute $2.00$ for $A_y$, and $6.16$ for $\left|\overrightarrow{A}\right|$ in equation (V) to find ${\theta }_y$.

  $\begin{array}{c}{\theta }_y={\cos }^{-1}\left(\frac{2.00}{6.16}\right)\\ =71.1°\text{from }+y\end{array}$

Substitute $5.00$ for $A_z$, and $6.16$ for $\left|\overrightarrow{A}\right|$ in equation (VI) to find ${\theta }_z$.

  $\begin{array}{c}{\theta }_z={\cos }^{-1}\left(\frac{5.00}{6.16}\right)\\ =35.8°\text{from }+z\end{array}$

Therefore, the angles that the vector $\overrightarrow{A}$ makes with the $x$, $y$, and $z$ axes are $\underset{\_}{{\theta }_x=60.9°\text{from }+x}$, $\underset{\_}{{\theta }_y=71.1°\text{from }+y}$, and $\underset{\_}{{\theta }_z=35.8°\text{from }+z}$ respectively.