Misc 18 - Find distance of point (-1, -5, -10) from point of intersect

Misc 18 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 18 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3 Misc 18 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4 Misc 18 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5 Misc 18 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6 Misc 18 - Chapter 11 Class 12 Three Dimensional Geometry - Part 7 Misc 18 - Chapter 11 Class 12 Three Dimensional Geometry - Part 8

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Question 14 (Method 1) Find the distance of the point (–1, –5, –10) from the point of intersection of the line 𝑟 ⃗ = 2𝑖 ̂ – 𝑗 ̂ + 2𝑘 ̂ + 𝜆 (3𝑖 ̂ + 4𝑗 ̂ + 2𝑘 ̂) and the plane 𝑟 ⃗ . (𝑖 ̂ – 𝑗 ̂ + 𝑘 ̂) = 5 .Given, the equation of line is 𝑟 ⃗ = (2𝑖 ̂ − 𝑗 ̂ + 2𝑘 ̂) + 𝜆 (3𝑖 ̂ + 4𝑗 ̂ + 2𝑘 ̂) and the equation of the plane is 𝑟 ⃗.(𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) = 5 To find point of intersection of line and plane, Putting value of 𝒓 ⃗ from equation of line into equation of plane. ["(2" 𝑖 ̂" − " 𝑗 ̂" + 2" 𝑘 ̂") + 𝜆 (3" 𝑖 ̂" + 4" 𝑗 ̂" + 2" 𝑘 ̂")" ] . (𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) = 5 ["(2" 𝑖 ̂" − 1" 𝑗 ̂" + 2" 𝑘 ̂+3"𝜆" 𝑖 ̂" + 4𝜆" 𝑗 ̂+2"𝜆" 𝑘 ̂")" ] . (1𝑖 ̂ − 1𝑗 ̂ + 1𝑘 ̂) = 5 ["(2 + 3𝜆) " 𝑖 ̂" + (" −1" + 4𝜆) " 𝑗 ̂+(2+2"𝜆" )𝑘 ̂ ] . (1𝑖 ̂ − 1𝑗 ̂ + 1𝑘 ̂) = 5 (2 + 3𝜆) × 1 + (−1 + 4𝜆) × (−1) + (2 + 2𝜆) × 1 = 5 2 + 3𝜆 + 1 − 4𝜆 + 2 + 2𝜆 = 5 𝜆 + 5 = 5 𝜆 = 5 − 5 𝜆 = 0 So, the equation of line is 𝑟 ⃗ = (2𝑖 ̂ − 𝑗 ̂ + 2𝑘 ̂) + 𝜆 (3𝑖 ̂ + 4𝑗 ̂ + 2𝑘 ̂) 𝒓 ⃗ = 2𝒊 ̂ − 𝒋 ̂ + 2𝒌 ̂ Let the point of intersection be (x, y, z) So, 𝑟 ⃗ = x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂ x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂ = 2𝑖 ̂ − 𝑗 ̂ + 2𝑘 ̂ Hence, x = 2 , y = −1, z = 2 Therefore, the point of intersection is (2, −1, 2) Now, the distance between two points (𝑥_1, 𝑦_1, 𝑧_1) and (𝑥_2, 𝑦_2, 𝑧_2) is √((𝑥_2−𝑥_1 )^2 〖+ (𝑦_2−𝑦_1 )〗^2+ (𝑧_2−𝑧_1 )^2 ) Distance between (2, −1, 2) and (−1, −5, −10) = √((−1−2)^2 〖+ (−5+1)〗^2+ (−10−2)^2 ) = √((−3)^2 〖+ (−4)〗^2+ (−12)^2 ) = √(9+16+144) = √169 = 13. Question 14 (Method 2) Find the distance of the point (–1, –5, –10) from the point of intersection of the line 𝑟 ⃗ = 2𝑖 ̂ – 𝑗 ̂ + 2𝑘 ̂ + 𝜆 (3𝑖 ̂ + 4𝑗 ̂ + 2𝑘 ̂) and the plane 𝑟 ⃗ . (𝑖 ̂ – 𝑗 ̂ + 𝑘 ̂) = 5 .Given, the equation of line is 𝑟 ⃗ = (2𝑖 ̂ − 𝑗 ̂ + 2𝑘 ̂) + 𝜆 (3𝑖 ̂ + 4𝑗 ̂ + 2𝑘 ̂) Comparing with 𝒓 ⃗ = 𝒂 ⃗ + 𝜆𝒃 ⃗ , 𝒂 ⃗ = 2𝒊 ̂ − 𝒋 ̂ + 2𝒌 ̂ Comparing with ğ‘Ž ⃗ = 𝑥_1 𝑖 ̂ + 𝑦_1 𝑗 ̂ + 𝑧_1 𝑘 ̂, ∴ 𝑥_1= 2, 𝑦_1= −1, 𝑧_1= 2, 𝒃 ⃗ = 3𝒊 ̂ + 4𝒋 ̂ + 2𝒌 ̂ Comparing with 𝑏 ⃗ = ğ‘Žğ‘– ̂ + 𝑏𝑗 ̂ + 𝑐𝑘 ̂, ∴ ğ‘Ž = 3, 𝑏 = 4, 𝑐 = 2, Equation of line in Cartesian form is (𝑥 − 𝑥_1)/ğ‘Ž = (𝑦 − 𝑦_1)/𝑏 = (𝑧 − 𝑧_1)/𝑐 (𝑥 − 2)/3 = (𝑦 − (−1))/4 = (𝑧 − 2)/2 (𝒙 − 𝟐)/𝟑 = (𝒚 + 𝟏)/𝟒 = (𝒛 − 𝟐)/𝟐 = k So, Also, the equation of plane is 𝑟 ⃗.(𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) = 5 Comparing with 𝑟 ⃗.𝑛 ⃗ = d, 𝑛 ⃗ = 1𝑖 ̂ − 1𝑗 ̂ + 1𝑘 ̂ & d = 5 Comparing 𝑛 ⃗ with A𝑖 ̂ + B𝑗 ̂ + C𝑘 ̂, A = 1, B = −1, C = 1 Equation of plane in Cartesian form is Ax + By + Cz = d 1x − 1y + 1z = 5 x − y + z = 5 Let the point of intersection of line and plane be (x, y, z) Putting values of x, y, z in equation of plane, (3k + 2) − (4k − 1) + (2k + 2) = 5 3k + 2 − 4k + 1 + 2k + 2 = 5 k + 5 = 5 ∴ k = 0 So, x = 3k + 2 = 3 × 0 + 2 = 2 y = 4k − 1 = 4 × 0 − 1 = −1 z = 2k + 2 = 2 × 0 + 2 = 2 Therefore, the point of intersection is (2, −1, 2). Distance between two points (𝑥_1, 𝑦_1, 𝑧_1) & (𝑥_2, 𝑦_2, 𝑧_2) = √((𝑥_2−𝑥_1 )^2 (𝑦_2−𝑦_1 )^2+(𝑧_2−𝑧_1 )^2 ) ∴ Distance between (2, −1, 2) and (−1, −5, −10) = √((−1−2)^2+(−5+1)^2+(−10−2)^2 ) = √((−3)^2+(−4)^2+(−12)^2 ) = √(9+16+144) = √169 = 13

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.