Question

A straight vector line makes equal acute angles \( \alpha = \beta = \gamma \) with all three primary coordinate axes. Find the absolute value of \( \cos\alpha \).

• A. \( \frac{1}{\sqrt{3}} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( 1 \)

Hint:

Since \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \), remember the related identity for sine terms by substituting \( \cos^2\theta = 1 - \sin^2\theta \), which gives: \( \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2 \).

Answer 1

The Correct Option is A

Concept: If a line forms directional orientation angles \( \alpha \), \( \beta \), and \( \gamma \) with the positive \(x\), \(y\), and \(z\) axes respectively, its direction cosines are defined as \( l = \cos\alpha \), \( m = \cos\beta \), and \( n = \cos\gamma \). These parameters must satisfy the fundamental geometric identity: \[ l^2 + m^2 + n^2 = 1 \quad \Rightarrow \quad \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \]

Step 1: Apply the equal angle condition to the direction cosine identity.
Since the problem states that the line is oriented at equal angles to all three axes, substitute \( \alpha = \beta = \gamma \) into the equation: \[ \cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \] Combine the matching terms: \[ 3\cos^2\alpha = 1 \]

Step 2: Isolate the value of \( \cos\alpha \).
Divide both sides by the scalar factor 3: \[ \cos^2\alpha = \frac{1}{3} \] Take the square root of both sides. Because the problem specifies an acute angle (\( \alpha \le 90^\circ \)), the cosine value must be positive: \[ \cos\alpha = \frac{1}{\sqrt{3}} \]