3D Geometry
Direction Cosines and Direction Ratios of a Line
Grade 12

Question:

<p>An angle between the lines whose direction cosines are given by the equations, \(l + 3m + 5n = 0\) and \(5lm - 2mn + 6nl = 0\), is</p>
<p>\(\cos^{-1}\left(\dfrac{1}{3}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{1}{4}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{1}{6}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{1}{8}\right)\)</p>

Step-by-Step Solution

Key Concept: From the linear constraint l + 3m + 5n = 0, express one direction cosine in terms of others, then substitute into the quadratic equation to find the relationship between the remaining direction cosines, yielding two sets of direction ratios whose angle can be computed using the dot product formula.
Step 1: From l + 3m + 5n = 0, express l = -3m - 5n Step 2: Substitute into 5lm - 2mn + 6nl = 0: 5(-3m - 5n)m - 2mn + 6n(-3m - 5n) = 0 -15m^2 - 25mn - 2mn - 18mn - 30n^2 = 0 -15m^2 - 45mn - 30n^2 = 0 m^2 + 3mn + 2n^2 = 0 (m + n)(m + 2n) = 0 Step 3: This gives m = -n or m = -2n Case 1: If m = -n, then l = -3(-n) - 5n = -2n, so direction ratios are (-2, -1, 1) or (2, 1, -1) Case 2: If m = -2n, then l = -3(-2n) - 5n = n, so direction ratios are (1, -2, 1) Step 4: Normalize the direction cosines: For (2, 1, -1): magnitude = √6, so (2/√6, 1/√6, -1/√6) For (1, -2, 1): magnitude = √6, so (1/√6, -2/√6, 1/√6) Step 5: Apply angle formula: cos θ = |(2/√6)(1/√6) + (1/√6)(-2/√6) + (-1/√6)(1/√6)| = |2/6 - 2/6 - 1/6| = |-1/6| = 1/6 ∴ θ = cos⁻^1(1/6)
Correct Answer: C

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