Chapter 10 Class 12 Vector Algebra
Ex 10.2, 9
Ex 10.2, 10 Important
Ex 10.2, 13 Important
Ex 10.2, 17 Important You are here
Example 14 Important
Example 16 Important
Example 21 Important
Ex 10.3, 2
Ex 10.3, 3 Important
Ex 10.3, 10 Important
Ex 10.3, 13 Important
Ex 10.3, 16 Important
Example 23 Important
Example 24
Example 25 Important
Ex 10.4, 2 Important
Ex 10.4, 5 Important
Ex 10.4, 9 Important
Ex 10.4, 10
Ex 10.4, 11 (MCQ) Important
Example 28 Important
Example 29 Important
Example 30 Important
Misc 6
Misc 12 Important
Misc 13
Misc 15 Important
Misc 19 (MCQ) Important
Chapter 10 Class 12 Vector Algebra
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.