Question 10 - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

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 )

Transcript

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

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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