Question

A vector is given by$\overrightarrow{\mathbf{R}}=2\hat{\mathbf{i}}+\hat{\mathbf{j}}+3\hat{\mathbf{k}}$. Find

(a) The magnitudes of the$x.y$, and$z$components;

(b) The magnitude of$\overrightarrow{\mathbf{R}}$; and

(c) The angles between$\overrightarrow{\mathbf{R}}$and $x.y$theand$z$axes.

Short answer

a)The magnitudes are$R_x=2,R_y=1,R_z=3.$

(b)The magnitude of$\left|\overrightarrow{\mathbf{R}}\right|=3.74\text{units}$.

(c)The angle between is$\overrightarrow{R}$and$x,y,z$islocalid="1663674629114" ${\theta }_x=57.7^{°}$,${\theta }_y=75.5^{°},{\theta }_z=36.7^{°}$.

Step-by-step solution

Definition of Pythagorean theorem and expression used in solution.

• Pythagoras' theorem is a fundamental relationship between the three sides of a right triangle in Euclidean geometry.
• The area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides, according to this formula.
• The Pythagorean equation, which relates the lengths of the legs a, b, and the hypotenuse c, can be stated as a theorem.
• Formula used in solution is
• $R=\left|\overrightarrow{\mathbf{R}}\right|=\sqrt{R_x^2+R_y^2+R_z^2}$
  

Determine the magnitudes of the x,y and z

We know that any vector$\overrightarrow{R}$ can be expressed in unit-vector form as follows:

$\overrightarrow{\mathbf{R}}=R_x\hat{\mathbf{i}}+R_y\hat{\mathbf{j}}+R_z\hat{\mathbf{k}}$

By comparing with the vector we have,

$\overrightarrow{\mathbf{R}}=2\hat{\mathbf{i}}+\hat{\mathbf{j}}+3\hat{\mathbf{k}}$

We find that:

The $X-$component of the vector$\overrightarrow{R}$ is$R_x=2$

The $y-$component of the vector $\overrightarrow{R}$is $R_y=1$

The $z-$component of the vector $\overrightarrow{R}$is $R_z=3$

Therefore, the magnitudes are$R_x=2,R_y=1,R_z=3.$

Determine the magnitude of R→

(b)

We can get the magnitude of the using Pythagorean theorem as follows:

$$\begin{gathered} R=\left|\overrightarrow{\mathbf{R}}\right|=\sqrt{R_x^2+R_y^2+R_z^2} \\ R=\sqrt{{\left(2\right)}^2+{\left(1\right)}^2+{\left(3\right)}^2}=3.4 \\ R=3.74\text{units} \end{gathered}$$

Hence, the magnitude of$\left|\overrightarrow{\mathbf{R}}\right|=3.74\text{units}$ .

Determine the angle between R→ and x,y,z

(c)

We know that the angle ${\theta }_x$that$\overrightarrow{R}$makes with the $X-$axis is found from the expression:

$$\begin{gathered} \cos {\theta }_x=\frac{R_x}{R} \\ \theta ={\cos }^{-1}\frac{R_x}{R} \end{gathered}$$

Similarly for the angles with the $y$and $Z$axis, so we have:

$$\begin{gathered} {\theta }_y={\cos }^{-1}\frac{R_y}{R} \\ {\theta }_z={\cos }^{-1}\frac{R_z}{R} \end{gathered}$$

Substituting with the values we have, we get:

Hence, the angle between is $\overrightarrow{R}$and $x,y,z$is${\theta }_x=57.7^{°},$${\theta }_y=75.5^{°},{\theta }_z=36.7^{°}.$