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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise

Transcript

Ex 10.2, 17 Show that the points A, B and C with position vectors, ๐‘Ž โƒ— = 3๐‘– ฬ‚ โˆ’ 4 ๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚, ๐‘ โƒ— = 2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + ๐‘˜ ฬ‚ and ๐‘ โƒ— = ๐‘– ฬ‚ โˆ’ 3 ๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ , respectively form the vertices of a right angled triangle. Position vectors of vertices A, B, C of triangle ABC are ๐‘Ž โƒ— = 3๐‘– ฬ‚ โˆ’ 4๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚, ๐‘ โƒ— = 2๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ + 1๐‘˜ ฬ‚ ๐‘ โƒ— = 1๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ We know that two vectors are perpendicular to each other, i.e. have an angle of 90ยฐ between them, if their scalar product is zero. So, if (CA) โƒ—. (AB) โƒ— = 0, then (CA) โƒ— โŠฅ (AB) โƒ— & โˆ  CAB = 90ยฐ Now, (AB) โƒ— = ๐‘ โƒ— โˆ’ ๐‘Ž โƒ— = (2i ฬ‚ โˆ’ 1j ฬ‚ + 1k ฬ‚) โˆ’ (3i ฬ‚ โˆ’ 4j ฬ‚ โˆ’ 4k ฬ‚) = (2 โˆ’ 3) i ฬ‚ + (โˆ’1 + 4) j ฬ‚ + (1 + 4) k ฬ‚ = โ€“1i ฬ‚ + 3j ฬ‚ + 5k ฬ‚ (BC) โƒ— = ๐‘ โƒ— โˆ’ ๐‘ โƒ— = (1i ฬ‚ โˆ’ 3j ฬ‚ โˆ’ 5k ฬ‚) โˆ’ (2i ฬ‚ โˆ’ 1j ฬ‚ + 1k ฬ‚) = (1 โˆ’ 2) i ฬ‚ + (โˆ’3 + 1) j ฬ‚ + (โˆ’5 โˆ’ 1) k ฬ‚ = โˆ’1i ฬ‚ โˆ’ 2j ฬ‚ โˆ’ 6k ฬ‚ (CA) โƒ— = ๐‘Ž โƒ— โˆ’ ๐‘ โƒ— = (3i ฬ‚ โˆ’ 4j ฬ‚ โˆ’ 4k ฬ‚) โˆ’ (1i ฬ‚ โˆ’ 3j ฬ‚ โˆ’ 5k ฬ‚) = (3 โˆ’ 1) i ฬ‚ + (โˆ’4 + 3) j ฬ‚ + (โˆ’4 + 5) k ฬ‚ = 2i ฬ‚ โˆ’ 1j ฬ‚ + 1k ฬ‚ Now, (๐€๐) โƒ— . (๐‚๐€) โƒ— = (โ€“1i ฬ‚ + 3j ฬ‚ + 5k ฬ‚) . (2i ฬ‚ โˆ’ 1j ฬ‚ + 1k ฬ‚) = (โˆ’1 ร— 2) + (3 ร— โˆ’1) + (5 ร— 1) = (โˆ’2) + (โˆ’3) + 5 = โˆ’5 + 5 = 0 So, (AB) โƒ—.(CA) โƒ— = 0 Thus, (AB) โƒ— and (CA) โƒ— are perpendicular to each other. Hence, ABC is a right angled triangle.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.