Right Angled triangle

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.2, 17 Show that the points A, B and C with position vectors, π β = 3π Μ β 4 π Μ β 4π Μ, π β = 2π Μ β π Μ + π Μ and π β = π Μ β 3 π Μ β 5π Μ , respectively form the vertices of a right angled triangle. Position vectors of vertices A, B, C of triangle ABC are π β = 3π Μ β 4π Μ β 4π Μ, π β = 2π Μ β 1π Μ + 1π Μ π β = 1π Μ β 3π Μ β 5π Μ We know that two vectors are perpendicular to each other, i.e. have an angle of 90Β° between them, if their scalar product is zero. So, if (CA) β. (AB) β = 0, then (CA) β β₯ (AB) β & β  CAB = 90Β° Now, (AB) β = π β β π β = (2i Μ β 1j Μ + 1k Μ) β (3i Μ β 4j Μ β 4k Μ) = (2 β 3) i Μ + (β1 + 4) j Μ + (1 + 4) k Μ = β1i Μ + 3j Μ + 5k Μ (BC) β = π β β π β = (1i Μ β 3j Μ β 5k Μ) β (2i Μ β 1j Μ + 1k Μ) = (1 β 2) i Μ + (β3 + 1) j Μ + (β5 β 1) k Μ = β1i Μ β 2j Μ β 6k Μ (CA) β = π β β π β = (3i Μ β 4j Μ β 4k Μ) β (1i Μ β 3j Μ β 5k Μ) = (3 β 1) i Μ + (β4 + 3) j Μ + (β4 + 5) k Μ = 2i Μ β 1j Μ + 1k Μ Now, (ππ) β . (ππ) β = (β1i Μ + 3j Μ + 5k Μ) . (2i Μ β 1j Μ + 1k Μ) = (β1 Γ 2) + (3 Γ β1) + (5 Γ 1) = (β2) + (β3) + 5 = β5 + 5 = 0 So, (AB) β.(CA) β = 0 Thus, (AB) β and (CA) β are perpendicular to each other. Hence, ABC is a right angled triangle.