Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

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.