Misc 18 - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

Misc 18 (Method 1) Find the distance of the point (–1, –5, –10) from the point of intersection of the line 𝑟﷯ = 2 𝑖﷯ – 𝑗﷯ + 2 𝑘﷯ + 𝜆 (3 𝑖﷯ + 4 𝑗﷯ + 2 𝑘﷯) and the plane 𝑟﷯ . ( 𝑖﷯ – 𝑗﷯ + 𝑘﷯) = 5 . Given, the equation of line is 𝑟﷯ = (2 𝑖﷯ − 𝑗﷯ + 2 𝑘﷯) + 𝜆 (3 𝑖﷯ + 4 𝑗﷯ + 2 𝑘﷯) 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﷯+3𝜆 𝑖﷯ + 4𝜆 𝑗﷯+2𝜆 𝑘﷯)﷯ . (1 𝑖﷯ − 1 𝑗﷯ + 1 𝑘﷯) = 5 (2 + 3𝜆) 𝑖﷯ + (−1 + 4𝜆) 𝑗﷯+(2+2𝜆) 𝑘﷯ ﷯ . (1 𝑖﷯ − 1 𝑗﷯ + 1 𝑘﷯) = 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﷯ = (2 𝑖﷯ − 𝑗﷯ + 2 𝑘﷯) + 𝜆 (3 𝑖﷯ + 4 𝑗﷯ + 2 𝑘﷯) 𝒓﷯ = 2 𝒊﷯ − 𝒋﷯ + 2 𝒌﷯ Let the point of intersection be (x, y, z) So, 𝑟﷯ = x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯ x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯ = 2 𝑖﷯ − 𝑗﷯ + 2 𝑘﷯ 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. Misc 18 (Method 2) Find the distance of the point (–1, –5, –10) from the point of intersection of the line 𝑟﷯ = 2 𝑖﷯ – 𝑗﷯ + 2 𝑘﷯ + 𝜆 (3 𝑖﷯ + 4 𝑗﷯ + 2 𝑘﷯) and the plane 𝑟﷯ . ( 𝑖﷯ – 𝑗﷯ + 𝑘﷯) = 5 . Given, the equation of line is 𝑟﷯ = (2 𝑖﷯ − 𝑗﷯ + 2 𝑘﷯) + 𝜆 (3 𝑖﷯ + 4 𝑗﷯ + 2 𝑘﷯) Comparing with 𝒓﷯ = 𝒂﷯ + 𝜆 𝒃﷯ , Equation of line in Cartesian form is 𝑥 − 𝑥﷮1﷯﷮𝑎﷯ = 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﷯ = 1 𝑖﷯ − 1 𝑗﷯ + 1 𝑘﷯ & d = 5 Comparing 𝑛﷯ with A 𝑖﷯ + B 𝑗﷯ + C 𝑘﷯, 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