Subscribe to our Youtube Channel - https://you.tube/teachoo

Last updated at Feb. 1, 2020 by Teachoo

Transcript

Misc 15 Find the equation of the plane passing through the line of intersection of the planes π β . (π Μ + π Μ + π Μ) =1 and π β . (2π Μ + 3π Μ β π Μ) + 4 = 0 and parallel to x-axis. Equation of a plane passing through the intersection of two planes π΄_1x + B1y + πΆ_1z = d1 and π΄_2x + B2y + πΆ_2z = d2 is (π¨_π "x " +" B1y" + πͺ_π "z β d1 " ) + π (π¨_π "x" +"B2y" +πͺ_π "z β d2 " ) = 0. Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 π β. (π Μ + π Μ + π Μ) = 1 Putting π β = xπ Μ + yπ Μ + zπ Μ, (xπ Μ + yπ Μ + zπ Μ).(π Μ + π Μ + π Μ) = 1 (x Γ 1) + (Y Γ 1) + (z Γ 1) = 1 1x + 1y + 1z = 1 Comparing with π΄_1 "x "+"B1y "+" " πΆ_1 "z = d1" π΄_1 = 1 , π΅_1 = 1 , πΆ_1 = 1 , π_1 = 1 π β. (2π Μ + 3π Μ β π Μ) + 4 = 0 π β. (2π Μ + 3π Μ β π Μ) = β4 βπ β. (2π Μ + 3π Μ β π Μ) = 4 π β. ( β2π Μ β3π Μ + π Μ) = 4 Putting π β = xπ Μ + yπ Μ + zπ Μ, (xπ Μ + yπ Μ + zπ Μ).(-2π Μ β 3π Μ + 1π Μ) = 4 (x Γ β2) + (y Γ β 3) + (z Γ 1) = 4 β 2x β 3y + 1z = 4 Comparing with π΄_2 "x "+"B2y "+" " πΆ_2 "z = d2" π΄_2 = β2 , π΅_2 = β3 , πΆ_2 = 1 , π_2 = 4 Equation of plane is (π΄_1 "x " +" B1y" + πΆ_1 "z β d1 " ) + π (π΄_2 "x" +"B2y" +πΆ_2 "z β d2 " ) = 0 (1x + 1y + 1z β 1) + π ( β 2x β 3y + 1z β 4) = 0 (1 β 2π) x + (1 β 3π)y + (1 + π) z + (β 1β 4π) = 0 Also, the plane is parallel to x β axis. So, Normal vector π β to the plane is perpendicular to x β axis. Finding direction ratios of normal and x-axis Theory : Two lines with direction ratios π1, b1, c1 and π2, b2, c2 are perpendicular if π1 π2 + b1b2 + c1 c2 = 0 π΅ β = (1 β 2π) π Μ + (1 β 3π) π Μ + (1 + π) π Μ Direction ratios = (1 β2π), (1 β 3π), (1 + π)π Μ β΄ π1 = 1 β 2π, b1 = 1 β 3π, c1 = 1 + π (πΆπΏ) β = 1π Μ + 0π Μ + 0π Μ Direction ratios = 1, 0, 0 β΄ π2 = 1, b2 = 0, c2 = 0, So, π1 π2 + b1 b2 + c1 c2 = 0 (1 β 2π) Γ 1 + (1 β 3π) Γ 0 + (1 + π) Γ 0 = 0 (1 β 2π) + 0 + 0 = 0 1 = 2π β΄ π = π/π Putting value of π in (1) (1β 2. 1/2)x + (1β 3. 1/2) y + (1+ 1/2) + (β1 β 4.1/2) = 0 (1 β 1)x + (1 β 3/2) y + (1+1/2) z + (β1 β 2) = 0 0x β 1/2 y + 3/2 z β 3 = 0 0x β 1/2 y + 3/2 z = 3 β y + 3z = 6 0 = y β 3z + 6 y β 3z + 6 = 0 Therefore, the equation of the plane is y β 3z + 6 = 0

Miscellaneous

Misc 1
Important

Misc 2

Misc 3

Misc 4 Important

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important You are here

Misc 16 Important

Misc 17 Important

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22 Important

Misc 23 Important

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.