1. Class 12
2. Important Question for exams Class 12

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

Class 12
Important Question for exams Class 12