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Misc 11 - Find coordinates of point where line through (5, 1, 6) and

Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5
Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6

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Misc 11 (Method 1) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.The equation of a line passing through two points with position vectors 𝑎 ⃗ & 𝑏 ⃗ is 𝒓 ⃗ = 𝒂 ⃗ + 𝜆(𝒃 ⃗ − 𝒂 ⃗) Given, the line passes through (𝑏 ⃗ − 𝑎 ⃗) = (3𝑖 ̂ + 4𝑗 ̂ + 1𝑘 ̂) − (5𝑖 ̂ + 1𝑗 ̂ + 6𝑘 ̂) = (3 −5)𝑖 ̂ + (4 − 1)𝑗 ̂ + (1 − 6)𝑘 ̂ A (5, 1, 6) 𝑎 ⃗ = 5𝑖 ̂ + 1𝑗 ̂ + 6𝑘 ̂ B(3, 4, 1) 𝑏 ⃗ = 3𝑖 ̂ + 4𝑗 ̂ + 1𝑘 ̂ = −2𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂ ∴ 𝒓 ⃗ = (5𝒊 ̂ + 𝒋 ̂ + 6𝒌 ̂) + 𝜆 (−2𝒊 ̂ + 3𝒋 ̂ − 5𝒌 ̂) Let the coordinates of the point where the line crosses the ZX plane be (x, 0, z) So, 𝒓 ⃗ = x𝒊 ̂ + 0𝒋 ̂ + z𝒌 ̂ Since point lies in line, it will satisfy its equation, Putting (2) in (1) x𝑖 ̂ + 0𝑗 ̂ + z𝑘 ̂ = 5𝑖 ̂ + 𝑗 ̂ + 6𝑘 ̂ −2𝜆𝑖 ̂ + 3𝜆𝑗 ̂ − 5𝜆𝑘 ̂ x𝑖 ̂ + 0𝑗 ̂ + z𝑘 ̂ = (5 −2𝜆)𝑖 ̂ + (1 + 3𝜆)𝑗 ̂ + (6 − 5𝜆)𝑘 ̂ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 3𝜆 3𝜆 = −1 ∴ 𝜆 = (−𝟏)/𝟑 Now, x = 5 − 2𝜆 = 5 − 2 × (−1)/3 = 5 + 2/3 = 17/13 z = 6 − 5𝜆 = 6 − 5 × (−1)/3 = 6 + 5/3 = 23/3 Therefore, the coordinate of the required point are (𝟏𝟕/𝟑,𝟎,𝟐𝟑/𝟑) Misc 11 (Method 2) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.The equation of a line passing through two points A(𝑥_1, 𝑦_1, 𝑧_1) and B(𝑥_2, 𝑦_2, 𝑧_2) is (𝒙 − 𝒙_𝟏)/(𝒙_𝟐 − 𝒙_𝟏 ) = (𝒚 − 𝒚_𝟏)/(𝒚_𝟐 − 𝒚_𝟏 ) = (𝒛 − 𝒛_𝟏)/(𝒛_𝟐 − 𝒛_𝟏 ) Given the line passes through the points A (5, 1, 6) ∴ 𝑥_1= 5, 𝑦_1= 1, 𝑧_1= 6 B(3, 4, 1) ∴ 𝑥_2= 3, 𝑦_2= 4, 𝑧_2= 1 So, the equation of line is (𝑥 − 5)/(3 − 5) = (𝑦 − 1)/(4 − 1) = (𝑧 − 6)/(1 − 6) (𝒙 − 𝟓)/(−𝟐) = (𝒚 − 𝟏)/𝟑 = (𝒛 − 𝟔)/(−𝟓) = k So, Since the line crosses the ZX plane at (x, 0, z), y = 0 3k + 1 = 0 3k = −1 ∴ k = (−𝟏)/𝟑 So, x = –2k + 5 = −2 × (−1)/3 + 5 = 2/3 + 5 = 17/3 y = 0 & z = −5k + 6 = −5 × (−1)/3 + 6 = 5/3 + 6 = 23/3 therefore, the coordinate of the required point are (𝟏𝟕/𝟑,𝟎,𝟐𝟑/𝟑)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.