Example 20 - 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