1. Chapter 10 Class 12 Vector Algebra
2. Serial order wise

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Ex 10.2, 17 Show that the points A, B and C with position vectors, 𝑎﷯ = 3 𝑖﷯ − 4 𝑗﷯ − 4 𝑘﷯, 𝑏﷯ = 2 𝑖﷯ − 𝑗﷯ + 𝑘﷯ and 𝑐﷯ = 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯ , respectively form the vertices of a right angled triangle. Position vectors of vertices A, B, C of triangle ABC are 𝑎﷯ = 3 𝑖﷯ − 4 𝑗﷯ − 4 𝑘﷯, 𝑏﷯ = 2 𝑖﷯ − 1 𝑗﷯ + 1 𝑘﷯ 𝑐﷯ = 1 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯ 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. So, if CA﷯. AB﷯ = 0, then CA﷯ ⊥ AB﷯ & ∠ CAB = 90° Now, AB﷯ = 𝑏﷯ − 𝑎﷯ = (2 i﷯ − 1 j﷯ + 1 k﷯) − (3 i﷯ − 4 j﷯ − 4 k﷯) = (2 − 3) i﷯ + (−1 + 4) j﷯ + (1 + 4) k﷯ = –1 i﷯ + 3 j﷯ + 5 k﷯ BC﷯ = 𝑐﷯ − 𝑏﷯ = (1 i﷯ − 3 j﷯ − 5 k﷯) − (2 i﷯ − 1 j﷯ + 1 k﷯) = (1 − 2) i﷯ + (−3 + 1) j﷯ + (−5 − 1) k﷯ = -1 i﷯ − 2 j﷯ − 6 k﷯ CA﷯ = 𝑎﷯ − 𝑐﷯ = (3 i﷯ − 4 j﷯ − 4 k﷯) − (1 i﷯ − 3 j﷯ − 5 k﷯) = (3 − 1) i﷯ + (−4 + 3) j﷯ + (−4 + 5) k﷯ = 2 i﷯ − 1 j﷯ + 1 k﷯ Now, 𝐀𝐁﷯ . 𝐂𝐀﷯ = (–1 i﷯ + 3 j﷯ + 5 k﷯) . (2 i﷯ − 1 j﷯ + 1 k﷯) = (−1 × 2) + (3 × −1) + (5 × 1) = (−2) + (−3) + 5 = −5 + 5 = 0 So, AB﷯. CA﷯ = 0 Thus, AB﷯ and CA﷯ are perpendicular to each other. Hence, ABC is a right angled triangle.