Question 10 The equation of the line in vector form passing through the point (−1, 3, 5) and parallel to line (𝑥 − 3)/2 = (𝑦 − 4)/3, z = 2. is (a) 𝑟 ⃗ = (−𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂) + 𝜆 (2𝑖 ̂ + 3𝑗 ̂ + 𝑘 ̂) (b) 𝑟 ⃗ = (−𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂) + 𝜆 (2𝑖 ̂ + 3𝑗 ̂) (c) 𝑟 ⃗ = (2𝑖 ̂ + 3𝑗 ̂ − 2𝑘 ̂) + 𝜆 (−𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂) (d) 𝑟 ⃗ = (2𝑖 ̂ + 3𝑗 ̂) + 𝜆 (−𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂)
Equation of a line passing through a point with position vector 𝑎 ⃗ and parallel to vector 𝑏 ⃗ is
𝑟 ⃗ = 𝑎 ⃗ + 𝜆𝑏 ⃗
Here,
Point is (–1, 3, 5)
So, 𝑎 ⃗ = –𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂
And it is parallel to line
(𝑥 − 3)/2 = (𝑦 − 4)/3, z = 2
This means that z-component of equation is 0
So, line is (𝑥 − 3)/2 = (𝑦 − 4)/3 = (𝑧 − 2)/0
So, 𝑏 ⃗ = 2𝑖 ̂ + 3𝑗 ̂ + 0𝑘 ̂ = 2𝑖 ̂ + 3𝑗 ̂
Thus, Equation of line is
𝒓 ⃗ = (−𝒊 ̂ + 3𝒋 ̂ + 5𝒌 ̂) + 𝝀 (2𝒊 ̂ + 3𝒋 ̂)
So, (b) is the correct answer
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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