Question

A ray makes angles \(\frac{\pi}{3}\) and \(\frac{\pi}{4}\) with \(Y\)-axis and \(Z\)-axis respectively. Then the value of the sine of the angle made by the ray with \(X\)-axis is

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

Hint:

If a line makes angles \(\alpha,\beta,\gamma\) with the coordinate axes, then always use \[ \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1. \]

Answer 1

The Correct Option is A

Step 1: Use the direction cosine relation.
Let the angles made by the ray with \(X\)-axis, \(Y\)-axis, and \(Z\)-axis be \[ \alpha,\quad \beta,\quad \gamma \] Then the direction cosines are \[ \cos \alpha,\quad \cos \beta,\quad \cos \gamma \] For any ray in three-dimensional geometry, \[ \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 \]

Step 2: Substitute the given angles.
Given, \[ \beta=\frac{\pi}{3} \] and \[ \gamma=\frac{\pi}{4} \] Therefore, \[ \cos\beta=\cos\frac{\pi}{3}=\frac{1}{2} \] and \[ \cos\gamma=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}} \]

Step 3: Find \(\cos^2\alpha\).
Using \[ \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1, \] we get \[ \cos^2\alpha+\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2=1 \] \[ \cos^2\alpha+\frac{1}{4}+\frac{1}{2}=1 \] \[ \cos^2\alpha+\frac{3}{4}=1 \] \[ \cos^2\alpha=\frac{1}{4} \] Thus, \[ \sin^2\alpha=1-\cos^2\alpha \] \[ \sin^2\alpha=1-\frac{1}{4} \] \[ \sin^2\alpha=\frac{3}{4} \] Therefore, \[ \sin\alpha=\frac{\sqrt{3}}{2} \]

Step 4: Final conclusion.
Hence, the sine of the angle made by the ray with \(X\)-axis is \[ \boxed{\frac{\sqrt{3}}{2}} \]