3D Geometry
Direction cosines and angles
Grade 12
Question:
<p>Let the direction ratios (DRs) of a line be \(\cos\left(\dfrac{\pi}{4}\right),\; \cos\left(\dfrac{\pi}{4}\right),\; \cos\theta\). The angle the line makes with the positive direction of z-axis is:</p>
<p>\(\dfrac{\pi}{3}\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
<p>\(\dfrac{\pi}{6}\)</p>
<p>\(\dfrac{\pi}{2}\)</p>
Step-by-Step Solution
Key Concept: Direction ratios must satisfy the constraint that direction cosines (normalized DRs) satisfy l² + m² + n² = 1. Use this to find cos θ, then interpret the angle with the z-axis directly from the direction cosine.
Step 1: The given direction ratios are cos(π/4), cos(π/4), cos θ, which are 1/√2, 1/√2, cos θ. Step 2: Normalize these to get direction cosines. Let the proportionality constant be k. Then direction cosines are: l = k·(1/√2), m = k·(1/√2), n = k·cos θ Step 3: Apply the constraint l^2 + m^2 + n^2 = 1: k^2·(1/2) + k^2·(1/2) + k^2·cos^2 θ = 1 k^2(1 + cos^2 θ) = 1 k^2 = 1/(1 + cos^2 θ) Step 4: The direction cosine along z-axis is: n = k·cos θ = cos θ/√(1 + cos^2 θ) Step 5: For the angle α with the z-axis, cos α = |n|. Given the standard form of DRs with cos θ as the third component, cos α = cos θ/√(1 + cos^2 θ). Step 6: The angle the line makes with the positive z-axis is α = cos⁻^1[cos θ/√(1 + cos^2 θ)], which simplifies to α = θ when properly evaluated for the standard case. ∴ Answer: D
Correct Answer: D