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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
Ex 10.5, 5 (Supplementary NCERT) - Chapter 10 Class 12 Vector Algebra - Part 3

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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’6@βˆ’1&βˆ’3&βˆ’8@1&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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.