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Chapter 11 Class 12 Three Dimensional Geometry
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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Question 6 (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 ⃗) = (3𝑖 ̂ + 4𝑗 ̂ + 1𝑘 ̂) − (5𝑖 ̂ + 1𝑗 ̂ + 6𝑘 ̂) = (3 −5)𝑖 ̂ + (4 − 1)𝑗 ̂ + (1 − 6)𝑘 ̂ = −2𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂ 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 ̂ = −2𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂ B(3, 4, 1) 𝑏 ⃗ = 3𝑖 ̂ + 4𝑗 ̂ + 1𝑘 ̂ ∴ 𝒓 ⃗ = (5𝒊 ̂ + 𝒋 ̂ + 6𝒌 ̂) + 𝜆 (−2𝒊 ̂ + 3𝒋 ̂ − 5𝒌 ̂) Let the coordinates of the point where the line crosses the YZ plane be (0, y, z) So, 𝒓 ⃗ = 0𝒊 ̂ + y𝒋 ̂ + z𝒌 ̂ Since point lies in line, it will satisfy its equation, Putting (2) in (1) 0𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂ = 5𝑖 ̂ + 𝑗 ̂ + 6𝑘 ̂ −2𝜆𝑖 ̂ + 3𝜆𝑗 ̂ − 5𝜆𝑘 ̂ 0𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂ = (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 = 𝟓/𝟐 Now, y = 1 + 3𝜆 = 1 + 3 × 5/2 = 1 + 15/2 = 17/2 & z = 6 − 5𝜆 = 6 − 5 × 5/2 = 6 − 25/2 = (−13)/2 Therefore, the coordinates of the required point is (𝟎,𝟏𝟕/𝟐, (−𝟏𝟑)/𝟐). Question 6 (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(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 So, the equation of line is (𝑥 − 5)/(3 − 5) = (𝑦 − 1)/(4 − 1) = (𝑧 − 6)/(1 − 6) 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 (𝒙 − 𝟓)/(−𝟐) = (𝒚 − 𝟏)/𝟑 = (𝒛 − 𝟔)/(−𝟓) = k So, Since the line crosses the YZ plane at (0, y, z) x = 0 −2k + 5 = 0 2k = 5 k = 𝟓/𝟐 x = –2k + 5 So, x = −2k + 5 = −2 × 5/2 + 5 = − 10/2 + 5 = 0 y = 3k + 1 = 3 × 5/2 + 1 = 15/2 + 1 = 17/2 & z = −5k + 6 = −5 × 5/2 + 6 = (−25)/2 + 6 = (−13)/2 Therefore, the coordinates of the required point are (𝟎,𝟏𝟕/𝟐,(−𝟏𝟑)/𝟐). 