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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams
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Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Last updated at March 22, 2023 by Teachoo
Example 12 (Method 1) Show that the points A(2π Μ β π Μ + π Μ), B(π Μ β 3π Μ β 5π Μ) , C(3π Μ β 4π Μ β 4π Μ) are the vertices of a right angled triangle. Given A (2π Μ β π Μ + π Μ), B (π Μ β 3π Μ β 5π Μ) C (3π Μ β 4π Μ β 4π Μ) We know that two vectors are perpendicular to each other, if their scalar product is zero. Finding (π¨π©) β , (π©πͺ) β , (π¨πͺ) β (π¨π©) β = (π Μ β 3π Μ β 5π Μ) β (2π Μ β π Μ + π Μ) = (1 β 2) π Μ + (β3 + 1) π Μ + (β5 β 1) π Μ = β1π Μ β 2π Μ β 6π Μ (π©πͺ) β = (3π Μ β 4π Μ β 4π Μ) β (π Μ β 3π Μ β 5π Μ) = (3 β 1) π Μ + (β4 + 3) π Μ + (β4 + 5) π Μ = 2π Μ β 1π Μ + 1π Μ (πͺπ¨) β = (2π Μ β π Μ + π Μ) β (3π Μ β 4π Μ β 4π Μ) = (2 β 3) π Μ + (β1 + 4) π Μ + (1 + 4) π Μ = β1π Μ + 3π Μ + 5π Μ Finding (π©πͺ) β. (πͺπ¨) β (π©πͺ) β. (πͺπ¨) β = (2π Μ β 1π Μ + 1π Μ) . (-1π Μ + 3π Μ + 5π Μ) = (2 Γ β1) + (β1 Γ 3) + (1 Γ 5) = (β2) + (β3) + 5 = β5 + 5 = 0 Since, (π©πͺ) β. (πͺπ¨) β = 0 Therefore, (π΅πΆ) β is perpendicular to (πΆπ΄) β . Hence, Ξ ABC is a right angled triangle Example 12 (Method 2) Show that the points A(2π Μ β π Μ + π Μ), B(π Μ β 3π Μ β 5π Μ) , C(3π Μ β 4π Μ β 4π Μ) are the vertices of a right angled triangle. Given A (2π Μ β π Μ + π Μ), B (π Μ β 3π Μ β 5π Μ) C (3π Μ β 4π Μ β 4π Μ) Considering βABC as a right angled triangle, By Pythagoras theorem, AB2 = BC2 + CA2 or |("AB" ) β |"2" = |("BC" ) β |"2" + |("CA" ) β |"2" Finding (π¨π©) β , (π©πͺ) β , (π¨πͺ) β (π¨π©) β = (π Μ β 3π Μ β 5π Μ) β (2π Μ β π Μ + π Μ) = (1 β 2) π Μ + (β3 + 1) π Μ + (β5 β1) π Μ = β1π Μ β 2π Μ β 6π Μ (π©πͺ) β = (3π Μ β 4π Μ β 4π Μ) β (π Μ β 3π Μ β 5π Μ) = (3 β 1) π Μ + (β4 + 3) π Μ + (β4 + 5) π Μ = 2π Μ β 1π Μ + 1π Μ (πͺπ¨) β = (2π Μ β π Μ + π Μ) β (3π Μ β 4π Μ β 4π Μ) = (2 β 3) π Μ + (β1 + 4) π Μ + (1 + 4) π Μ = β1π Μ + 3π Μ + 5π Μ Now, "Magnitude of " (π¨π©) β" = " β((β1)2+(β2)2+(β6)2) " " |(π΄π΅) β |" = " β(1+4+36) " = " βππ Magnitude of (π©πͺ) β = β(22+(β1)2+1) |(π΅πΆ) β | = β(4+1+1) = βπ Magnitude of (πͺπ¨) β = β((β1)2+32+52) |(πΆπ΄) β | = β(1+9+25) = βππ Now, |(π©πͺ) β |^π + |(πͺπ¨) β |^π = (β6)2 + (β35)2 = 6 + 35 = 41 = (β41)2 = |(π¨π©) β |^π Thus, |(π¨π©) β |^π = |(π©πͺ) β |^π + |(πͺπ¨) β |^π Hence, by Pythagoras Theorem, Ξ ABC is a right angled triangle.