Question 34 (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

The equations of motion of a rocket are: 𝑥 = 2𝑡, 𝑦 = −4𝑡, 𝑧 = 4𝑡, where the time t is given in seconds, and the coordinates of a moving point in km. What is the path of the rocket? At what distances will the rocket be from the starting point O(0,0,0) and from the following line in 10 seconds?

r = 20i ̂-10j ̂+4k ̂+μ(10i ̂-20j ̂+10k ̂ )?

Transcript

Question 34 (Choice 2) The equations of motion of a rocket are: 𝑥 = 2𝑡, 𝑦 = −4𝑡, 𝑧 = 4𝑡, where the time t is given in seconds, and the coordinates of a moving point in km. What is the path of the rocket? At what distances will the rocket be from the starting point O(0, 0, 0) and from the following line in 10 seconds 𝑟 ⃗ = 20𝑖 ̂−10𝑗 ̂+40𝑘 ̂+𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ )?Given equation of motion of rocket
𝑥 = 2𝑡, 𝑦 = −4𝑡, 𝑧 = 4𝑡,
We note that if we join the points, we get a line.
At t = 0
x = 0, y = 0, z = 0
So, point at t = 0 is (0, 0, 0)
At t = 1
x = 2, y = −4, z = 4
So, point at t = 0 is (2, −4, 4)
So, path of rocket is a line
Equation of Path of Rocket
Since rocket passes through points (0, 0, 0) and (2, −4, 4), equation of line is
(𝑥 − 𝑥_1)/(𝑥_2 − 𝑥_1 ) = (𝑦 − 𝑦_1)/(𝑦_2 − 𝑦_1 ) = (𝑧 − 𝑧_1)/(𝑧_2 − 𝑧_1 )
(𝑥 − 0)/(2 − 0) = (𝑦 − 0)/( −4 − 0) = (𝑧 − 0)/(4 − 0)
𝒙/𝟐 = 𝒚/( −𝟒) = 𝒛/𝟒
Now, we need to find
At what distances will the rocket be from the starting point O(0, 0, 0) and from the following line in 10 seconds 𝑟 ⃗ = 20𝑖 ̂−10𝑗 ̂+40𝑘 ̂+𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ )?
At t = 10 seconds, point is
𝑥 = 2𝑡, 𝑦 = −4𝑡, 𝑧 = 4𝑡,
𝑥 = 20, 𝑦 = −40, 𝑧 = 40
So,
At t = 0, point is (0, 0, 0)
At t = 10, point is (20, −40, 40)
Thus,
We need to find distance of point (20, −40, 40) from point (0, 0, 0) And
Distance of point (20, −40, 40) from line 𝒓 ⃗ = 20𝑖 ̂−10𝑗 ̂+4𝑘 ̂+𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ )?
Thus,
We need to find distance of point (20, −40, 40) from point (0, 0, 0) And
Distance of point (20, −40, 40) from line 𝒓 ⃗ = 20𝑖 ̂−10𝑗 ̂+40𝑘 ̂+𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ )?
Line
𝒓 ⃗ = 20𝑖 ̂−10𝑗 ̂+4𝑘 ̂ +𝜇(10𝑖 ̂−20𝑗 ̂+10𝑘 ̂ )
Distance of point (20, −40, 40) from point (0, 0, 0)
D = √((20−0)^2+(−40−0)^2+(40−0)^2 )
= √(〖𝟐𝟎〗^𝟐+(−𝟒𝟎)^𝟐+〖𝟒𝟎〗^𝟐 )
= √(400+1600+1600)
= √3600
= √((60)^2 )
= 60 kmDistance of point (20, −40, 40) from line 𝒓 ⃗ = 20𝒊 ̂−𝟏𝟎𝒋 ̂+𝟒𝟎𝒌 ̂+𝝁(𝟏𝟎𝒊 ̂−𝟐𝟎𝒋 ̂+𝟏𝟎𝒌 ̂ )
In three dimensions, the perpendicular distance, 𝐷,between a point 𝑃(𝑥, 𝑦, 𝑧) (position vector 𝑝 ⃗) and line 𝒓 ⃗ = a ⃗ + λb ⃗ is
𝐷=|(𝑝 ⃗ − 𝑎 ⃗ ) × 𝑏 ⃗ |/|𝑏 ⃗ |
Putting values
𝐷=|([20𝑖 ̂ − 40𝑗 ̂ + 40𝑘 ̂ ]−[20𝑖 ̂ −10𝑗 ̂ + 4𝑘 ̂ ]) × (𝟏𝟎𝒊 ̂ − 𝟐𝟎𝒋 ̂ + 𝟏𝟎𝒌 ̂)|/|𝟏𝟎𝒊 ̂ − 𝟐𝟎𝒋 ̂ + 𝟏𝟎𝒌 ̂ |
𝐷=|(30𝒋 ̂ ) × (𝟏𝟎𝒊 ̂ − 𝟐𝟎𝒋 ̂ + 𝟏𝟎𝒌 ̂)|/√(〖10〗^2 + (−20)^2 + 〖10〗^2 )
𝑫=|■8(𝒊 ̂&𝒋 ̂&𝒌 ̂@𝟎&𝟑𝟎&𝟎@𝟏𝟎&−𝟐𝟎&𝟏𝟎)|/√(𝟏𝟎𝟎 + 𝟒𝟎𝟎 + 𝟏𝟎𝟎)
𝐷=|𝑖 ̂(300 − 0) − 𝑗 ̂(0 − 0) + 𝑘 ̂(0 − 300)|/√600
𝑫=|𝟑𝟎𝟎𝒊 ̂ − 𝟑𝟎𝟎𝒌 ̂ |/√𝟔𝟎𝟎
𝐷=√((300)^2+(300)^2 )/√(6 × 100)
𝐷=√(2 × (300)^2 )/√(6 × 100)
𝑫=(𝟑𝟎𝟎√𝟐)/(𝟏𝟎√𝟔)
𝐷=30/√3
𝐷=30/√3 ×√3/√3
𝐷=(30√3)/3
D = 10 √𝟑 km

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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