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Ex 10.5, 5 (Supplementary NCERT) - Show that 4 points with position

Ex 10.5, 5 (Supplementary NCERT) - Chapter 10 Class 12 Vector Algebra - Part 2
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Ex 10.5, 5 (Supplementary NCERT) Show that the four points with position vectors 4𝑖 Μ‚ + 8𝑗 Μ‚ + 12π‘˜ Μ‚, 2𝑖 Μ‚ + 4𝑗 Μ‚ + 6π‘˜ Μ‚, 3𝑖 Μ‚ + 5𝑗 Μ‚ + 4π‘˜ Μ‚ & 5𝑖 Μ‚ + 8𝑗 Μ‚ + 5π‘˜ Μ‚ are coplanarLet points be A = 4𝑖 Μ‚ + 8𝑗 Μ‚ + 12π‘˜ Μ‚ B = 2𝑖 Μ‚ + 4𝑗 Μ‚ + 6π‘˜ Μ‚ C = 3𝑖 Μ‚ + 5𝑗 Μ‚ + 4π‘˜ Μ‚ D = 5𝑖 Μ‚ + 8𝑗 Μ‚ + 5π‘˜ Μ‚ Four points A, B, C, D are coplanar if the three vectors (𝐴𝐡) βƒ— , (𝐴𝐢) βƒ— and (𝐴𝐷) βƒ— are coplanar. i.e. [(𝑨𝑩) βƒ—, (𝑨π‘ͺ) βƒ—, (𝑨𝑫) βƒ— ] = 0 A (4𝑖 Μ‚ + 8𝑗 Μ‚ + 12π‘˜ Μ‚) B (2𝑖 Μ‚ + 4𝑗 Μ‚ + 6π‘˜ Μ‚) (𝑨𝑩) βƒ— = (2𝑖 Μ‚ + 4𝑗 Μ‚ + 6π‘˜ Μ‚) – (4𝑖 Μ‚ + 8𝑗 Μ‚ + 12π‘˜ Μ‚) = (2 βˆ’ 4) 𝑖 Μ‚ + (4 – 8) 𝑗 Μ‚ + (6 – 12)π‘˜ Μ‚ = –2π’Š Μ‚ βˆ’ 4𝒋 Μ‚ βˆ’ 6π’Œ Μ‚ A (4𝑖 Μ‚ + 8𝑗 Μ‚ + 12π‘˜ Μ‚) C (3𝑖 Μ‚ + 5𝑗 Μ‚ + 4π‘˜ Μ‚) (𝑨π‘ͺ) βƒ— = (3𝑖 Μ‚ + 5𝑗 Μ‚ + 4π‘˜ Μ‚) – (4𝑖 Μ‚ + 8𝑗 Μ‚ + 12π‘˜ Μ‚) = (3 βˆ’ 4) 𝑖 Μ‚ + (5 – 8) 𝑗 Μ‚ + (4 – 12) π‘˜ Μ‚ = β€“π’Š Μ‚ – 3𝒋 Μ‚ – 8π’Œ Μ‚ A (4𝑖 Μ‚ + 8𝑗 Μ‚ + 12π‘˜ Μ‚) D (5𝑖 Μ‚ + 8𝑗 Μ‚ + 5π‘˜ Μ‚) (𝑨𝑫) βƒ— = (5𝑖 Μ‚ + 8𝑗 Μ‚ + 5π‘˜ Μ‚) – (4𝑖 Μ‚ + 8𝑗 Μ‚ + 12π‘˜ Μ‚) = (5 βˆ’ 4) 𝑖 Μ‚ + (8 – 8) 𝑗 Μ‚ + (5 – 12) π‘˜ Μ‚ = π’Š Μ‚ + 0𝒋 Μ‚ – 7π’Œ Μ‚ Now, [(𝐴𝐡) βƒ—, (𝐴𝐢) βƒ—, (𝐴𝐷) βƒ— ] = |β– 8(βˆ’2&βˆ’4&βˆ’[email protected]βˆ’1&βˆ’3&βˆ’[email protected]&0&βˆ’7)| = –2[(βˆ’3Γ—βˆ’7)βˆ’(0Γ—βˆ’8) ] βˆ’ (βˆ’4) [(βˆ’1Γ—βˆ’7)βˆ’(1Γ—βˆ’8)] + (βˆ’6)[(βˆ’1Γ—0)βˆ’(1Γ—βˆ’3) ] = βˆ’2[21βˆ’0]+4[7+8]βˆ’6[0+3] = βˆ’2[21]+4[15]βˆ’6[3] = – 42 + 60 – 18 = 60 – 60 = 0 ∴ [(𝑨𝑩) βƒ—, (𝑨π‘ͺ) βƒ—, (𝑨𝑫) βƒ— ] = 0 Therefore, points A, B, C and D are coplanar.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.