





Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 10.3
Ex 10.3, 2
Ex 10.3, 3 Important
Ex 10.3, 4
Ex 10.3, 5 Important
Ex 10.3, 6
Ex 10.3, 7
Ex 10.3, 8
Ex 10.3, 9 Important
Ex 10.3, 10 Important
Ex 10.3, 11
Ex 10.3, 12 Important
Ex 10.3, 13 Important
Ex 10.3, 14
Ex 10.3, 15 Important
Ex 10.3, 16 Important
Ex 10.3, 17 You are here
Ex 10.3, 18 (MCQ) Important
Last updated at May 29, 2023 by Teachoo
Ex 10.3, 17 (Method 1) Show that the vectors 2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂, 𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂ , 3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂ form the vertices of a right angled triangle. Let A(2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂), B(𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) C(3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) 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. (𝐴𝐵) ⃗ = (𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) − (2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) = 1𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂ − 2𝑖 ̂ + 1𝑗 ̂ − 1𝑘 ̂ = (1 − 2) 𝑖 ̂ + (−3 + 1) 𝑗 ̂ + (−5 −1) 𝑘 ̂ = −1𝑖 ̂ − 2𝑗 ̂ − 6𝑘 ̂ (𝐵𝐶) ⃗ = (3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) − (𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) = 3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂ − 1𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂ = (3 − 1) 𝑖 ̂ + (−4 + 3) 𝑗 ̂ + (−4 + 5) 𝑘 ̂ = 2𝑖 ̂ − 1𝑗 ̂ + 1𝑘 ̂ (𝐶𝐴) ⃗ = (2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) − (3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) = 2𝑖 ̂ − 1𝑗 ̂ + 1𝑘 ̂ − 3𝑖 ̂ + 4𝑗 ̂ + 4𝑘 ̂ = (2 − 3) 𝑖 ̂ + (−1 + 4) 𝑗 ̂ + (1 + 4) 𝑘 ̂ = −1𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂ Now, (𝐵𝐶) ⃗. (𝐶𝐴) ⃗ = (2𝑖 ̂ − 1𝑗 ̂ + 1𝑘 ̂) . (-1𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂) = (2 × –1) + (−1 × 3) + (1 × 5) = (−2) + (−3) + 5 = −5 + 5 = 0 Since, (𝑩𝑪) ⃗. (𝑪𝑨) ⃗ = 0 Therefore, (𝐵𝐶) ⃗ is perpendicular to (𝐶𝐴) ⃗ . Hence Δ ABC is a right angled triangle Ex 10.3, 17 (Method 2) Show that the vectors 2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂, 𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂ , 3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂ form the vertices of a right angled triangle. Let A(2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂), B(𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) C(3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) Considering ∆ABC as a right angled triangle, By Pythagoras theorem, AB2 = BC2 + CA2 or |("AB" ) ⃗ |"2" = |("BC" ) ⃗ |"2" + |("CA" ) ⃗ |"2" (𝐴𝐵) ⃗ = (𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) − (2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) = 1𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂ − 2𝑖 ̂ + 1𝑗 ̂ − 1𝑘 ̂ = (1 − 2) 𝑖 ̂ + (−3 + 1) 𝑗 ̂ + (−5 −1) 𝑘 ̂ = −1𝑖 ̂ − 2𝑗 ̂ − 6𝑘 ̂ (𝐵𝐶) ⃗ = (3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) − (𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) = 3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂ − 1𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂ = (3 − 1) 𝑖 ̂ + (−4 + 3) 𝑗 ̂ + (−4 + 5) 𝑘 ̂ = 2𝑖 ̂ − 1𝑗 ̂ + 1𝑘 ̂ (𝐶𝐴) ⃗ = (2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) − (3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) = 2𝑖 ̂ − 1𝑗 ̂ + 1𝑘 ̂ − 3𝑖 ̂ + 4𝑗 ̂ + 4𝑘 ̂ = (2 − 3) 𝑖 ̂ + (−1 + 4) 𝑗 ̂ + (1 + 4) 𝑘 ̂ = −1𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂ Now, "Magnitude of " (𝐴𝐵) ⃗" = " √((−1)2+(−2)2+(−6)2) " " |(𝐴𝐵) ⃗ |" = " √(1+4+36) " = " √41 Magnitude of (𝐵𝐶) ⃗ = √(22+(−1)2+1) |(𝐵𝐶) ⃗ | = √(4+1+1) = √6 Magnitude of (𝐶𝐴) ⃗ = √((−1)2+32+52) |(𝐶𝐴) ⃗ | = √(1+9+25) = √35 Now, |(𝐵𝐶) ⃗ |2 + |(𝐶𝐴) ⃗ |2 = (√6)2 + (√35)2 = 6 + 35 = 41 = (√41)2 = |(𝐴𝐵) ⃗ |2 Thus, |(𝐴𝐵) ⃗ |^2 = |(𝐵𝐶) ⃗ |^2 + |(𝐶𝐴) ⃗ |^2 So, ABC is a right angled triangle.