Example 12 - Chapter 10 Class 12 Vector Algebra (Term 2)

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Example 12 (Method 1) Show that the points A(2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂), B(𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) , C(3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) are the vertices of a right angled triangle. Given A (2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂), B (𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) C (3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) We know that two vectors are perpendicular to each other, if their scalar product is zero. Finding (𝑨𝑩) ⃗ , (𝑩𝑪) ⃗ , (𝑨𝑪) ⃗ (𝑨𝑩) ⃗ = (𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) − (2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) = (1 − 2) 𝑖 ̂ + (−3 + 1) 𝑗 ̂ + (−5 − 1) 𝑘 ̂ = −1𝒊 ̂ − 2𝒋 ̂ − 6𝒌 ̂ (𝑩𝑪) ⃗ = (3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) − (𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) = (3 − 1) 𝑖 ̂ + (−4 + 3) 𝑗 ̂ + (−4 + 5) 𝑘 ̂ = 2𝒊 ̂ − 1𝒋 ̂ + 1𝒌 ̂ (𝑪𝑨) ⃗ = (2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) − (3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) = (2 − 3) 𝑖 ̂ + (−1 + 4) 𝑗 ̂ + (1 + 4) 𝑘 ̂ = −1𝒊 ̂ + 3𝒋 ̂ + 5𝒌 ̂ Finding (𝑩𝑪) ⃗. (𝑪𝑨) ⃗ (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 Example 12 (Method 2) Show that the points A(2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂), B(𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) , C(3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) are the vertices of a right angled triangle. Given 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" Finding (𝑨𝑩) ⃗ , (𝑩𝑪) ⃗ , (𝑨𝑪) ⃗ (𝑨𝑩) ⃗ = (𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) − (2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) = (1 − 2) 𝑖 ̂ + (−3 + 1) 𝑗 ̂ + (−5 −1) 𝑘 ̂ = −1𝒊 ̂ − 2𝒋 ̂ − 6𝒌 ̂ (𝑩𝑪) ⃗ = (3𝑖 ̂ − 4𝑗 ̂ − 4𝑘 ̂) − (𝑖 ̂ − 3𝑗 ̂ − 5𝑘 ̂) = (3 − 1) 𝑖 ̂ + (−4 + 3) 𝑗 ̂ + (−4 + 5) 𝑘 ̂ = 2𝒊 ̂ − 1𝒋 ̂ + 1𝒌 ̂ (𝑪𝑨) ⃗ = (2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) − (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 Hence, by Pythagoras Theorem, Δ ABC is a right angled triangle.