7.2 Q-12

Question Statement

Determine which of the following triples can represent the direction angles of a single vector:

(i) $\alpha = 45^\circ$, $\beta = 45^\circ$, $\gamma = 60^\circ$
(ii) $\alpha = 30^\circ$, $\beta = 45^\circ$, $\gamma = 60^\circ$
(iii) $\alpha = 45^\circ$, $\beta = 60^\circ$, $\gamma = 60^\circ$

---

Background and Explanation

The direction angles of a vector $\mathbf{v}$ are the angles between the vector and the coordinate axes (x-axis, y-axis, z-axis). If $\alpha$, $\beta$, and $\gamma$ represent the direction angles, the vector’s direction cosines are given by: $\cos \alpha, \cos \beta, \cos \gamma$

The sum of the squares of the direction cosines must always be equal to 1: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$

This relationship is derived from the Pythagorean theorem and is a key property of direction cosines.

---

Solution

(i) Check if $\alpha = 45^\circ$, $\beta = 45^\circ$, $\gamma = 60^\circ$ can be direction angles

To verify if these angles can be direction angles, we calculate: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$

1. Find the cosine values: $\cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 60^\circ = \frac{1}{2}$

2. Substitute the values: $\cos^2 45^\circ + \cos^2 45^\circ + \cos^2 60^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2$

3. Simplify: $= \frac{1}{2} + \frac{1}{2} + \frac{1}{4} = \frac{2 + 1}{4} = \frac{5}{4}$

Since the result is $\frac{5}{4} \neq 1$, the given angles cannot be direction angles.

(ii) Check if $\alpha = 30^\circ$, $\beta = 45^\circ$, $\gamma = 60^\circ$ can be direction angles

Again, calculate: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$

1. Find the cosine values: $\cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 60^\circ = \frac{1}{2}$

2. Substitute the values: $\cos^2 30^\circ + \cos^2 45^\circ + \cos^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2$

3. Simplify: $= \frac{3}{4} + \frac{1}{2} + \frac{1}{4} = \frac{3 + 2 + 1}{4} = \frac{6}{4} = \frac{3}{2}$

Since the result is $\frac{3}{2} \neq 1$, the given angles cannot be direction angles.

(iii) Check if $\alpha = 45^\circ$, $\beta = 60^\circ$, $\gamma = 60^\circ$ can be direction angles

Now, calculate: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$

1. Find the cosine values: $\cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 60^\circ = \frac{1}{2}$

2. Substitute the values: $\cos^2 45^\circ + \cos^2 60^\circ + \cos^2 60^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2$

3. Simplify: $= \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = \frac{2 + 1 + 1}{4} = \frac{4}{4} = 1$

Since the result is $1$, the given angles can be direction angles.

---

Key Formulas or Methods Used

• Direction Cosines: The cosines of the angles between a vector and the coordinate axes are called the direction cosines.
• Condition for Direction Angles: The direction angles $\alpha$, $\beta$, and $\gamma$ of a vector must satisfy: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$

---

Summary of Steps

1. (i) Calculate $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$ for $\alpha = 45^\circ$, $\beta = 45^\circ$, $\gamma = 60^\circ$ and find that it does not equal 1, so they cannot be direction angles.
2. (ii) Calculate $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$ for $\alpha = 30^\circ$, $\beta = 45^\circ$, $\gamma = 60^\circ$ and find that it does not equal 1, so they cannot be direction angles.
3. (iii) Calculate $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$ for $\alpha = 45^\circ$, $\beta = 60^\circ$, $\gamma = 60^\circ$ and find that it equals 1, so they can be direction angles.