3D Geometry
Line in 3D
Grade 12
Question:
<p>Given one vertex <em>A</em>(7, 2, 4) and the line \(\dfrac{x+6}{5} = \dfrac{y+10}{3} = \dfrac{z+14}{8}\). If angle between AB and BC is \(\dfrac{\pi}{4}\), find the equation(s) of the line through <em>A</em>.</p>
<p>\(\dfrac{x-7}{2} = \dfrac{y-2}{-3} = \dfrac{z-4}{6}\)</p>
<p>\(\dfrac{x-7}{3} = \dfrac{y-2}{6} = \dfrac{z-4}{2}\)</p>
<p>\(\dfrac{x-7}{1} = \dfrac{y-2}{2} = \dfrac{z-4}{3}\)</p>
<p>\(\dfrac{x-7}{2} = \dfrac{y-2}{3} = \dfrac{z-4}{6}\)</p>
Step-by-Step Solution
Key Concept: Use the direction ratios of the given line and the angle condition to find lines through A that make π/4 with the given line. The angle between two lines with direction vectors d₁ and d₂ is given by cos θ = |d₁·d₂|/(|d₁||d₂|), which yields a quadratic constraint on the direction ratios.
Step 1: Direction ratios of given line BC: (5, 3, 8). Let direction ratios of line AB be (a, b, c). Step 2: Using angle formula: cos(π/4) = |5a + 3b + 8c|/(√(25+9+64)·√(a^2+b^2+c^2)) Step 3: Simplify: 1/√2 = |5a + 3b + 8c|/(√98·√(a^2+b^2+c^2)) This gives: 2(5a + 3b + 8c)^2 = 98(a^2 + b^2 + c^2) Step 4: Expanding: 50a^2 + 30b^2 + 128c^2 + 60ab + 80ac + 48bc = 98a^2 + 98b^2 + 98c^2 Simplifying: 48a^2 + 68b^2 - 30c^2 + 60ab + 80ac + 48bc = 0 Step 5: Choose convenient ratios. Setting a=1, b=1 gives: 48 + 68 - 30c^2 + 60 + 80c + 48c = 0 ⟹ 30c^2 - 128c - 176 = 0 ⟹ c = 4 or c = -22/15 Step 6: Two lines through A(7,2,4) are: Line 1: (x-7)/1 = (y-2)/1 = (z-4)/4 Line 2: (x-7)/15 = (y-2)/15 = (z-4)/(-22) ∴ Answer: AB
Correct Answer: AB