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7.2 Q-12

Question Statement

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

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


Background and Explanation

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

The sum of the squares of the direction cosines must always be equal to 1: cos2α+cos2β+cos2γ=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 α=45\alpha = 45^\circ, β=45\beta = 45^\circ, γ=60\gamma = 60^\circ can be direction angles

To verify if these angles can be direction angles, we calculate: cos2α+cos2β+cos2γ\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma

  1. Find the cosine values: cos45=12,cos60=12\cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 60^\circ = \frac{1}{2}

  2. Substitute the values: cos245+cos245+cos260=(12)2+(12)2+(12)2\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: =12+12+14=2+14=54= \frac{1}{2} + \frac{1}{2} + \frac{1}{4} = \frac{2 + 1}{4} = \frac{5}{4}

Since the result is 541\frac{5}{4} \neq 1, the given angles cannot be direction angles.

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

Again, calculate: cos2α+cos2β+cos2γ\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma

  1. Find the cosine values: cos30=32,cos45=12,cos60=12\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: cos230+cos245+cos260=(32)2+(12)2+(12)2\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: =34+12+14=3+2+14=64=32= \frac{3}{4} + \frac{1}{2} + \frac{1}{4} = \frac{3 + 2 + 1}{4} = \frac{6}{4} = \frac{3}{2}

Since the result is 321\frac{3}{2} \neq 1, the given angles cannot be direction angles.

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

Now, calculate: cos2α+cos2β+cos2γ\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma

  1. Find the cosine values: cos45=12,cos60=12\cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 60^\circ = \frac{1}{2}

  2. Substitute the values: cos245+cos260+cos260=(12)2+(12)2+(12)2\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: =12+14+14=2+1+14=44=1= \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = \frac{2 + 1 + 1}{4} = \frac{4}{4} = 1

Since the result is 11, 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 α\alpha, β\beta, and γ\gamma of a vector must satisfy: cos2α+cos2β+cos2γ=1\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1

Summary of Steps

  1. (i) Calculate cos2α+cos2β+cos2γ\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma for α=45\alpha = 45^\circ, β=45\beta = 45^\circ, γ=60\gamma = 60^\circ and find that it does not equal 1, so they cannot be direction angles.
  2. (ii) Calculate cos2α+cos2β+cos2γ\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma for α=30\alpha = 30^\circ, β=45\beta = 45^\circ, γ=60\gamma = 60^\circ and find that it does not equal 1, so they cannot be direction angles.
  3. (iii) Calculate cos2α+cos2β+cos2γ\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma for α=45\alpha = 45^\circ, β=60\beta = 60^\circ, γ=60\gamma = 60^\circ and find that it equals 1, so they can be direction angles.