Example 30 - Find coordinates of point where line through A (3, 4, 1)

Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5

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Question 20 (Method 1) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-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 the points (𝑏 ⃗ − 𝑎 ⃗) = (5𝑖 ̂ + 1𝑗 ̂ + 6𝑘 ̂) − (3𝑖 ̂ + 4𝑗 ̂ + 1𝑘 ̂) = 2𝑖 ̂ − 3𝑗 ̂ + 5𝑘 ̂ A (3, 4, 1) 𝑎 ⃗ = 3𝑖 ̂ + 4𝑗 ̂ + 𝑘 ̂ B (5, 1, 6) 𝑏 ⃗ = 5𝑖 ̂ + 1𝑗 ̂ + 6𝑘 ̂ ∴ 𝑟 ⃗ = (3𝑖 ̂ + 4𝑗 ̂ + 1𝑘 ̂) + 𝜆 (2𝑖 ̂ − 3𝑗 ̂ + 5𝑘 ̂) Let the coordinates of the point where the line crosses the XY plane be (x, y, 0). So, 𝑟 ⃗ = x𝑖 ̂ + y𝑗 ̂ + 0𝑘 ̂ Since point crosses the plane, it will satisfy its equation Putting (2) in (1) x𝑖 ̂ + y𝑗 ̂ + 0𝑘 ̂ = 3𝑖 ̂ + 4𝑗 ̂ + 1𝑘 ̂ + 2𝜆𝑖 ̂ − 3𝜆𝑗 ̂ + 5𝜆𝑘 ̂ x𝑖 ̂ + y𝑗 ̂ + 0𝑘 ̂ = (3 + 2𝜆)𝑖 ̂ + (4 − 3𝜆)𝑗 ̂ + (1 + 5𝜆)𝑘 ̂ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 5𝜆 ∴ 𝜆 = (−𝟏)/𝟓 So, x = 3 + 𝜆 = 3 + 2 × (−1)/5 = 3 − 2/5 = 13/5 & y = 4 − 3𝜆 = 4 − 3 × (−1)/5 = 4 + 3/5 = 23/5 Therefore, the required coordinates are (𝟏𝟑/𝟓,𝟐𝟑/𝟓,𝟎) Question 20 (Method 2) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-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 − 3)/(5 − 3) = (𝑦 − 4)/(1 − 4) = (𝑧 − 1)/(6 − 1) A (3, 4, 1) ∴ 𝑥_1= 3, 𝑦_1= 4, 𝑧_1= 1 B(5, 1, 6) ∴ 𝑥_2 = 5, 𝑦_2= 1, 𝑧_2= 6 (𝑥 − 3)/2 = (𝑦 − 4)/(−3) = (𝑧 − 1)/5 = k So, Since, the line crosses the XY plane at (x, y, 0) z = 0 5k + 1 = 0 5k = −1 ∴ k = (−𝟏)/𝟓 So, x = 2k + 3 = 2 × (−1)/5 + 3 = 3 − 2/5 = 13/5 y = −3 × (−1)/5 + 4 = 4 + 3/5 = 23/5 Therefore, the coordinates of the required point are (𝟏𝟑/𝟓, 𝟐𝟑/𝟓, 𝟎).

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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 and Computer Science at Teachoo