Example 27 - If i+j+k, 2i+5j, 3i+2j-3k and i-6j-k are position vectors

Example 27 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 27 - Chapter 10 Class 12 Vector Algebra - Part 3

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Transcript

Example 27 If 𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ Μ‚, 2𝑖 Μ‚ + 5𝑗 Μ‚, 3𝑖 Μ‚ + 2𝑗 Μ‚ – 3π‘˜ Μ‚ and 𝑖 Μ‚ – 6𝑗 Μ‚ – π‘˜ Μ‚ are the position vectors of points A, B, C and D respectively, then find the angle between (𝐴𝐡) βƒ— and (𝐢𝐷) βƒ— . Deduce that (𝐴𝐡) βƒ— and (𝐢𝐷) βƒ— are collinear.Angle between (𝐴𝐡) βƒ— & (𝐢𝐷) βƒ— is given by cos ΞΈ = ((𝑨𝑩) βƒ— . (π‘ͺ𝑫) βƒ—)/|(𝑨𝑩) βƒ— ||(π‘ͺ𝑫) βƒ— | A(π’Š Μ‚ + 𝒋 Μ‚ + π’Œ Μ‚), B(2π’Š Μ‚ + 5𝒋 Μ‚) (𝑨𝑩) βƒ— = (2 βˆ’ 1) 𝑖 Μ‚ + (5 βˆ’ 1) 𝑗 Μ‚ + (0 βˆ’ 1) π‘˜ Μ‚ = 1𝑖 Μ‚ + 4𝑗 Μ‚ βˆ’ π‘˜ Μ‚ |(𝑨𝑩) βƒ— | = √(1^2+4^2+(βˆ’1)^2 ) = √18 = √(9 Γ— 2) = 3√𝟐 C(3π’Š Μ‚ + 2𝒋 Μ‚ – 3π’Œ Μ‚), D(π’Š Μ‚ – 6𝒋 Μ‚ – π’Œ Μ‚) (𝐢𝐷) βƒ— = (1 βˆ’ 3) 𝑖 Μ‚ + (–6 βˆ’ 2) 𝑗 Μ‚ + (βˆ’1 βˆ’ (-3)) π‘˜ Μ‚ = –2𝑖 Μ‚ – 8𝑗 Μ‚ + 2π‘˜ Μ‚ |(π‘ͺ𝑫) βƒ— | = √((βˆ’2)^2+(βˆ’8)^2+2^2 ) = √72 = √(36 Γ— 2) = 6√𝟐 Now, cos ΞΈ = ((𝑨𝑩) βƒ— . (π‘ͺ𝑫) βƒ—)/|(𝑨𝑩) βƒ— ||(π‘ͺ𝑫) βƒ— | = ((𝑖 Μ‚ + 4𝑗 Μ‚ βˆ’ π‘˜ Μ‚ ).(βˆ’2𝑖 Μ‚ βˆ’ 8𝑗 Μ‚ + 2π‘˜ Μ‚ ))/(3√2 Γ— 6√2) = (1(βˆ’2) + 4(βˆ’8) βˆ’ 1(2))/(3√2 Γ— 6√2) = (βˆ’2 βˆ’ 32 βˆ’ 2)/(3 Γ— 6 Γ— √2 Γ— √2) = (βˆ’36)/36 = –1 Since cos ΞΈ = –1, ΞΈ = 180Β° So, ΞΈ = 180Β° = 180Β° Γ— πœ‹/180 = Ο€ So, angle between (𝐴𝐡) βƒ— & (𝐢𝐷) βƒ— is Ο€ Also, Since angle between (𝐴𝐡) βƒ— & (𝐢𝐷) βƒ— is 180Β° , they are in opposite directions Since (𝐴𝐡) βƒ— & (𝐢𝐷) βƒ— are parallel to the same line π‘š βƒ—, they are collinear.

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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.