Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 10 (Method 1) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-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) = (3 𝑖 + 4 𝑗 + 1 𝑘) − (5 𝑖 + 1 𝑗 + 6 𝑘) = (3 −5) 𝑖 + (4 − 1) 𝑗 + (1 − 6) 𝑘 = −2 𝑖 + 3 𝑗 − 5 𝑘 ∴ 𝒓 = (5 𝒊 + 𝒋 + 6 𝒌) + 𝜆 (−2 𝒊 + 3 𝒋 − 5 𝒌) Let the coordinates of the point where the line crosses the YZ plane be (0, y, z) So, 𝒓 = 0 𝒊 + y 𝒋 + z 𝒌 Since point lies in line, it will satisfy its equation, Putting (2) in (1) 0 𝑖 + y 𝑗 + z 𝑘 = 5 𝑖 + 𝑗 + 6 𝑘 −2𝜆 𝑖 + 3𝜆 𝑗 − 5𝜆 𝑘 0 𝑖 + y 𝑗 + z 𝑘 = (5 −2𝜆) 𝑖 + (1 + 3𝜆) 𝑗 + (6 − 5𝜆) 𝑘 Two vectors are equal if their corresponding components are equal So, Solving 0 = 5 − 2𝜆 5 = 2𝜆 ∴ 𝜆 = 𝟓𝟐 Now, y = 1 + 3𝜆 = 1 + 3 × 52 = 1 + 152 = 172 & z = 6 − 5𝜆 = 6 − 5 × 52 = 6 − 252 = −132 Therefore, the coordinates of the required point is 𝟎, 𝟏𝟕𝟐, −𝟏𝟑𝟐. Misc 10 (Method 2) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane. The equation of a line passing through two points A( 𝑥1, 𝑦1, 𝑧1) and B( 𝑥2, 𝑦2, 𝑧2) is 𝒙 − 𝒙𝟏 𝒙𝟐 − 𝒙𝟏 = 𝒚 − 𝒚𝟏 𝒚𝟐 − 𝒚𝟏 = 𝒛 − 𝒛𝟏 𝒛𝟐 − 𝒛𝟏 Given the line passes through the points So, the equation of line is 𝑥 − 53 − 5 = 𝑦 − 14 − 1 = 𝑧 − 611 − 6 𝒙 − 𝟓−𝟐 = 𝒚 − 𝟏𝟑 = 𝒛 − 𝟔−𝟓 = k So, Since the line crosses the YZ plane at (0, y, z), x = 0 −2k + 5 = 0 2k = 5 k = 𝟓𝟐 So, x = 2k + 5 = −2 × 52 + 5 = − 102 + 5 = 0 y = 3k + 1 = 3 × 52 + 1 = 152 + 1 = 172 & z = −5k + 6 = −5 × 52 + 6 = −252 + 6 = −132 Therefore, the coordinates of the required point are 𝟎, 𝟏𝟕𝟐, −𝟏𝟑𝟐.

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .