Miscellaneous

Chapter 11 Class 12 Three Dimensional Geometry
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Question 11 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 + πΆ_1z = d1 and π΄_2x + B2y + πΆ_2z = d2 is (π¨_π "x " +" B1y" + πͺ_π "z β d1 " ) + π (π¨_π "x" +"B2y" +πͺ_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 "+" " πΆ_1 "z = d1" π΄_1 = 1 , π΅_1 = 1 , πΆ_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) = β4 βπ β. (2π Μ + 3π Μ β π Μ) = 4 π β. ( β2π Μ β3π 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 β2) + (y Γ β 3) + (z Γ 1) = 4 β 2x β 3y + 1z = 4 Comparing with π΄_2 "x "+"B2y "+" " πΆ_2 "z = d2" π΄_2 = β2 , π΅_2 = β3 , πΆ_2 = 1 , π_2 = 4 Equation of plane is (π΄_1 "x " +" B1y" + πΆ_1 "z β d1 " ) + π (π΄_2 "x" +"B2y" +πΆ_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 π1, b1, c1 and π2, b2, c2 are perpendicular if π1 π2 + b1b2 + c1 c2 = 0 π΅ β = (1 β 2π) π Μ + (1 β 3π) π Μ + (1 + π) π Μ Direction ratios = (1 β2π), (1 β 3π), (1 + π)π Μ β΄ π1 = 1 β 2π, b1 = 1 β 3π, c1 = 1 + π (πΆπΏ) β = 1π Μ + 0π Μ + 0π Μ Direction ratios = 1, 0, 0 β΄ π2 = 1, b2 = 0, c2 = 0, So, π1 π2 + 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) + (β1 β 4.1/2) = 0 (1 β 1)x + (1 β 3/2) y + (1+1/2) z + (β1 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