Check sibling questions

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


Transcript

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.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.