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Ex 10.2, 17 -Show that a = 3i - 4j - 4k, b = 2i + k, c = i - 3j - 5k

Ex 10.2, 17 - Chapter 10 Class 12 Vector Algebra - Part 2
Ex 10.2, 17 - Chapter 10 Class 12 Vector Algebra - Part 3

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Ex 10.2, 17 Show that the points A, B and C with position vectors, π‘Ž βƒ— = 3𝑖 Μ‚ βˆ’ 4 𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚, 𝑏 βƒ— = 2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚ and 𝑐 βƒ— = 𝑖 Μ‚ βˆ’ 3 𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚ , respectively form the vertices of a right angled triangle. Position vectors of vertices A, B, C of triangle ABC are π‘Ž βƒ— = 3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚, 𝑏 βƒ— = 2𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 1π‘˜ Μ‚ 𝑐 βƒ— = 1𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚ We know that two vectors are perpendicular to each other, i.e. have an angle of 90Β° between them, if their scalar product is zero. So, if (CA) βƒ—. (AB) βƒ— = 0, then (CA) βƒ— βŠ₯ (AB) βƒ— & ∠ CAB = 90Β° Now, (AB) βƒ— = 𝑏 βƒ— βˆ’ π‘Ž βƒ— = (2i Μ‚ βˆ’ 1j Μ‚ + 1k Μ‚) βˆ’ (3i Μ‚ βˆ’ 4j Μ‚ βˆ’ 4k Μ‚) = (2 βˆ’ 3) i Μ‚ + (βˆ’1 + 4) j Μ‚ + (1 + 4) k Μ‚ = –1i Μ‚ + 3j Μ‚ + 5k Μ‚ (BC) βƒ— = 𝑐 βƒ— βˆ’ 𝑏 βƒ— = (1i Μ‚ βˆ’ 3j Μ‚ βˆ’ 5k Μ‚) βˆ’ (2i Μ‚ βˆ’ 1j Μ‚ + 1k Μ‚) = (1 βˆ’ 2) i Μ‚ + (βˆ’3 + 1) j Μ‚ + (βˆ’5 βˆ’ 1) k Μ‚ = βˆ’1i Μ‚ βˆ’ 2j Μ‚ βˆ’ 6k Μ‚ (CA) βƒ— = π‘Ž βƒ— βˆ’ 𝑐 βƒ— = (3i Μ‚ βˆ’ 4j Μ‚ βˆ’ 4k Μ‚) βˆ’ (1i Μ‚ βˆ’ 3j Μ‚ βˆ’ 5k Μ‚) = (3 βˆ’ 1) i Μ‚ + (βˆ’4 + 3) j Μ‚ + (βˆ’4 + 5) k Μ‚ = 2i Μ‚ βˆ’ 1j Μ‚ + 1k Μ‚ Now, (𝐀𝐁) βƒ— . (𝐂𝐀) βƒ— = (–1i Μ‚ + 3j Μ‚ + 5k Μ‚) . (2i Μ‚ βˆ’ 1j Μ‚ + 1k Μ‚) = (βˆ’1 Γ— 2) + (3 Γ— βˆ’1) + (5 Γ— 1) = (βˆ’2) + (βˆ’3) + 5 = βˆ’5 + 5 = 0 So, (AB) βƒ—.(CA) βƒ— = 0 Thus, (AB) βƒ— and (CA) βƒ— are perpendicular to each other. Hence, ABC is a right angled triangle.

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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.