1. Class 12
2. Important Question for exams Class 12
3. Chapter 11 Class 12 Three Dimensional Geometry

Transcript

Example 20 Find the vector equation of the plane passing through the intersection of the planes 𝑟﷯ . ( 𝑖﷯ + 𝑗﷯ + 𝑘﷯) = 6 and 𝑟﷯ . (2 𝑖﷯ + 3 𝑗﷯ + 4 𝑘﷯)= − 5, and the point (1, 1, 1). The vector equation of a plane passing through the intersection of planes 𝑟﷯. 𝑛1﷯ = d1 and 𝑟﷯. 𝑛2﷯ = d2 and also through the point (x1, y1, z1) is 𝒓﷯.( 𝒏𝟏﷯ + 𝜆 𝒏𝟐﷯) = d1 + 𝜆d2 Given, the plane passes through Equation of plane is 𝑟﷯. 𝑖﷯+ 𝑗﷯+ 𝑘﷯﷯+𝜆(−2 𝑖﷯−3 𝑗﷯−4 𝑘﷯)﷯ = 6 + 𝜆5 𝒓﷯. 𝒊﷯ + 𝒋﷯ + 𝒌﷯﷯−𝜆(𝟐 𝒊﷯+𝟑 𝒋﷯+𝟒 𝒌﷯)﷯ = 6 + 5𝜆 Now to find 𝜆 , put 𝒓﷯ = x 𝒊﷯ + y 𝒋﷯ + z 𝒌﷯ (x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯). 𝑖﷯+ 𝑗﷯+ 𝑘﷯﷯−𝜆(2 𝑖﷯+3 𝑗﷯+4 𝑘﷯)﷯ = 5𝜆 + 6 (x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯). 𝑖﷯+ 𝑗﷯+ 𝑘﷯﷯ − 𝜆 (x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯).(2 𝑖﷯+3 𝑗﷯+4 𝑘﷯) = 5𝜆 + 6 (x × 1) + (y × 1) + (z × 1) − 𝜆 𝑥×2﷯+ 𝑦×3﷯+(𝑧×4)﷯ = 5𝜆 + 6 x + y + z − 𝜆 2𝑥+3𝑦+4𝑧﷯ = 5𝜆 + 6 x + y + z − 2𝜆𝑥 − 3𝜆y − 4𝜆z = 5𝜆 + 6 (1 − 2𝜆)x + (1 − 3𝜆)y + (1 − 4𝜆) z = 5𝜆 + 6 Since the plane passes through (1, 1, 1), Putting (1, 1, 1) in (2) (1 − 2𝜆)x + (1 − 3𝜆)y + (1 − 4𝜆) z = 5𝜆 + 6 (1 −2𝜆) × 1 + (1 − 3𝜆) × 1 + (1 − 4𝜆) × 1 = 5𝜆 + 6 1 −2𝜆 + 1 − 3𝜆 + 1 − 4𝜆= 5𝜆 + 6 3 − 9𝜆 = 5𝜆 + 6 −14𝜆 = 3 ∴ 𝜆 = − 𝟑﷮𝟏𝟒﷯ Putting value of 𝜆 in (1), 𝑟﷯. 𝑖﷯ + 𝑗﷯ + 𝑘﷯﷯− − 3﷮14﷯﷯(2 𝑖﷯+3 𝑗﷯+4 𝑘﷯)﷯= 6 + 5 × − 3﷮14﷯ 𝑟﷯. 𝑖﷯+ 𝑗﷯+ 𝑘﷯﷯+ 3﷮14﷯(2 𝑖﷯+3 𝑗﷯+4 𝑘﷯)﷯= 6 − 15﷮14﷯ 𝑟﷯. 𝑖﷯+ 𝑗﷯ + 𝑘﷯+ 6﷮14﷯ 𝑖﷯+ 9﷮14﷯ 𝑗﷯+ 12﷮14﷯ 𝑘﷯﷯= 69﷮14﷯ 𝑟﷯. 1+ 6﷮14﷯﷯ 𝑖﷯ + 1+ 9﷮14﷯﷯ 𝑗﷯+ 1+ 12﷮14﷯﷯ 𝑘﷯﷯= 69﷮14﷯ 𝑟﷯. 20﷮14﷯ 𝑖﷯ + 23﷮14﷯ 𝑗﷯ + 26﷮14﷯ 𝑘﷯﷯= 69﷮14﷯ 𝑟﷯. 1﷮14﷯(20 𝑖﷯+23 𝑗﷯+26 𝑘﷯)﷯= 69﷮14﷯ 1﷮14﷯ 𝑟﷯. (20 𝑖﷯ + 23 𝑗﷯ + 26 𝑘﷯) = 69﷮14﷯ 𝑟﷯. (20 𝑖﷯ + 23 𝑗﷯ + 26 𝑘﷯) = 69 Therefore, the vector equation of the required plane is 𝒓﷯.(𝟐𝟎 𝒊﷯ + 𝟐𝟑 𝒋﷯ + 𝟐𝟔 𝒌﷯) = 𝟔𝟗

Chapter 11 Class 12 Three Dimensional Geometry

Class 12
Important Question for exams Class 12