



Subscribe to our Youtube Channel - https://you.tube/teachoo
Last updated at Feb. 1, 2020 by Teachoo
Transcript
Misc 17 Find the equation of the plane which contains the line of intersection of the planes ๐ โ . (๐ ฬ + 2๐ ฬ + 3๐ ฬ) โ 4 = 0 , ๐ โ . (2๐ ฬ + ๐ ฬ โ ๐ ฬ) + 5 = 0 and which is perpendicular to the plane ๐ โ . (5๐ ฬ + 3๐ ฬ โ 6๐ ฬ) + 8 = 0 .Equation of a plane passing through the intersection of the places A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is (A1x + B1y + C1z โ d1) + ๐ (A2x + B2y + C2z โ d2) = 0 Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 ๐ โ. (๐ ฬ + 2๐ ฬ + 3๐ ฬ) โ 4 = 0 ๐ โ. (๐ ฬ + 2๐ ฬ + 3๐ ฬ) = 4 Putting ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ (x๐ ฬ + y๐ ฬ + z๐ ฬ).(๐ ฬ + 2๐ ฬ + 3๐ ฬ) = 4 (x ร 1) + (y ร 2) + (z ร 3) = 4 1x + 2y + 3z = 4 Comparing with ๐ด_1 "x"+"B1y"+๐ถ_1 "z = d1" ๐ด_1 = 1, ๐ต_1= 2 , ๐ถ_1 = 3 , ๐_1 = 4 ๐ โ. (2๐ ฬ + ๐ ฬ โ ๐ ฬ) + 5 = 0 ๐ โ. (2๐ ฬ + ๐ ฬ โ ๐ ฬ) = โ 5 โ๐ โ. (2๐ ฬ + ๐ ฬ โ ๐ ฬ) = 5 ๐ โ. ( โ2๐ ฬ โ ๐ ฬ + ๐ ฬ) = 5 Putting ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ, (x๐ ฬ + y๐ ฬ + z๐ ฬ).(-2๐ ฬ โ ๐ ฬ + ๐ ฬ) = 5 (x รโ 2) + (Y ร โ 1) + (z ร 1) = 5 โ2x โ 1y + 1z = 5 Comparing with ๐ด_2 "x"+ "B2y"+ ๐ถ_2 "z = d2" ๐ด_2 = โ2, ๐ต_2= โ1 , ๐ถ_2 = 1 , ๐_2 = 5 Equation of plane is (A1x + B1y + C1z โ d1) + ๐ (A2x + B2y + C2z = d2) = 0 Putting values (1x + 2y + 3z โ 4) + ๐ ( โ 2x โ 1y + 1z โ 5) = 0 (1 โ 2๐) x + (2 โ ๐)y + (3 + ๐) z + ( โ4 โ 5๐) = 0 Now, the plane is perpendicular to the plane ๐ โ.(5๐ ฬ + 3๐ ฬ โ 6๐ ฬ) + 8 = 0 So, normal to plane ๐ โ will be perpendicular to normal ๐ โ of ๐ โ.(5๐ ฬ + 3๐ ฬ โ 6๐ ฬ) + 8 = 0 Now, ๐ โ.(5๐ ฬ + 3๐ ฬ โ 6๐ ฬ) + 8 = 0 ๐ โ .(5๐ ฬ + 3๐ ฬ โ 6๐ ฬ) = โ8 โ ๐ โ .(5๐ ฬ + 3๐ ฬ โ 6๐ ฬ) = 8 ๐ โ .( โ5๐ ฬ โ 3๐ ฬ + 6๐ ฬ) = 8 Finding direction cosines of ๐ โ & ๐ โ Since, ๐ โ is perpendicular to ๐ โ ๐1 ๐2 + b1 b2 + c1 c2 = 0 (1 โ 2๐) ร โ5 + (2 โ ๐) ร โ3 + (3 + ๐) ร 6 = 0 Theory : Two lines with direction ratios ๐1, b1, c1 and ๐2, b2, c2 are perpendicular if ๐1 ๐2 + b1b2 + c1 c2 = 0 ๐ต โ = (1 โ 2๐) ๐ ฬ + (2 โ ๐) ๐ ฬ + (3 + ๐) ๐ ฬ Direction ratios = 1 โ 2๐, 2 โ ๐, 3 + ๐ โด ๐1 = 1 โ 2๐, b1 = 2 โ ๐, c1 = 3 + ๐ ๐ โ = โ 5๐ ฬ โ 3๐ ฬ + 6๐ ฬ Direction ratios = โ5, โ3, 6 โด ๐2 = โ 5, b2 = โ3, c2 = 6, โ 5 + 10๐ โ 6 + 3๐ + 18 + 6๐ = 0 19๐ + 7 = 0 โด ๐ = (โ๐)/๐๐ Putting value of ๐ in (1), (1 โ 2๐) x + (2 โ ๐)y + (3 + ๐) z + ( โ4 โ 5๐) = 0 (1โ2 ร(โ7)/19) x + (2โ(( โ7)/19)) y + (3+(( โ 7)/19)) z + ( โ4โ5ร(โ7)/19) = 0 (1 + 14/19) x + (2 + 7/19)y + (3 โ 7/19)z + ( โ 4 + 35/19) = 0 33/19 x + 45/19 y + 50/19 z โ 41/19 = 0 1/19 (33x + 45y + 50z โ 41) = 0 33x + 45y + 50z โ 41 = 0
Miscellaneous
Misc 2
Misc 3 Deleted for CBSE Board 2021 Exams only
Misc 4 Important
Misc 5 Important Deleted for CBSE Board 2021 Exams only
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
Misc 16 Important
Misc 17 Important You are here
Misc 18 Important
Misc 19 Important
Misc 20 Important
Misc 21 Important
Misc 22 Important
Misc 23 Important
About the Author