Ex 10.2

Ex 10.2, 1

Ex 10.2, 2

Ex 10.2, 3 Important

Ex 10.2, 4

Ex 10.2, 5 Important

Ex 10.2, 6

Ex 10.2, 7 Important

Ex 10.2, 8

Ex 10.2, 9

Ex 10.2, 10 Important

Ex 10.2, 11 Important

Ex 10.2, 12

Ex 10.2, 13 Important

Ex 10.2, 14

Ex 10.2, 15 Important

Ex 10.2, 16

Ex 10.2, 17 Important You are here

Ex 10.2, 18 (MCQ) Important

Ex 10.2, 19 (MCQ) Important

Last updated at April 16, 2024 by Teachoo

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