Question

1. a vector is given by \\( \vec{r} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k} \\). find (a) the magnitude, (b) the magnitudes of the x, y, and z components; (c) the angles between \\( \vec{r} \\) and the x, y, and z axes.

Explanation:

Step1: Identify vector components

The vector is $\vec{R} = 2\hat{i} + \hat{j} + 3\hat{k}$, so:
$R_x = 2$, $R_y = 1$, $R_z = 3$

Step2: Calculate magnitude of $\vec{R}$

Use 3D vector magnitude formula:
$$|\vec{R}| = \sqrt{R_x^2 + R_y^2 + R_z^2} = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \approx 3.74$$

Step3: Find angles with axes

Use direction cosine formula $\cos\theta = \frac{R_i}{|\vec{R}|}$:

• Angle with x-axis ($\theta_x$):

$\cos\theta_x = \frac{R_x}{|\vec{R}|} = \frac{2}{\sqrt{14}} \approx 0.5345$, so $\theta_x = \arccos(0.5345) \approx 57.7^\circ$

• Angle with y-axis ($\theta_y$):

$\cos\theta_y = \frac{R_y}{|\vec{R}|} = \frac{1}{\sqrt{14}} \approx 0.2673$, so $\theta_y = \arccos(0.2673) \approx 74.5^\circ$

• Angle with z-axis ($\theta_z$):

$\cos\theta_z = \frac{R_z}{|\vec{R}|} = \frac{3}{\sqrt{14}} \approx 0.8018$, so $\theta_z = \arccos(0.8018) \approx 36.7^\circ$

Answer:

(a) Magnitudes of components: $R_x = 2$, $R_y = 1$, $R_z = 3$
(b) Magnitude of $\vec{R}$: $\sqrt{14} \approx 3.74$
(c) Angles with axes: $\theta_x \approx 57.7^\circ$, $\theta_y \approx 74.5^\circ$, $\theta_z \approx 36.7^\circ$