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Ex 10.3, 17 - Show vectors form vertices of right angled triangle - Right Angled triangle

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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Ex 10.3, 17 (Method 1) Show that the vectors 2 𝑖﷯ − 𝑗﷯ + 𝑘﷯, 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯ , 3 𝑖﷯ − 4 𝑗﷯ − 4 𝑘﷯ form the vertices of a right angled triangle. Let A(2 𝑖﷯ − 𝑗﷯ + 𝑘﷯), B( 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯) C(3 𝑖﷯ − 4 𝑗﷯ − 4 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. 𝐴𝐵﷯ = ( 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯) − (2 𝑖﷯ − 𝑗﷯ + 𝑘﷯) = 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﷯ 𝐵𝐶﷯ = (3 𝑖﷯ − 4 𝑗﷯ − 4 𝑘﷯) − ( 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯) = 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﷯ 𝐶𝐴﷯ = (2 𝑖﷯ − 𝑗﷯ + 𝑘﷯) − (3 𝑖﷯ − 4 𝑗﷯ − 4 𝑘﷯) = 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﷯ Now, 𝐵𝐶﷯. 𝐶𝐴﷯ = (2 𝑖﷯ − 1 𝑗﷯ + 1 𝑘﷯) . (-1 𝑖﷯ + 3 𝑗﷯ + 5 𝑘﷯) = (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 Ex 10.3, 17 (Method 2) Show that the vectors 2 𝑖﷯ − 𝑗﷯ + 𝑘﷯, 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯ , 3 𝑖﷯ − 4 𝑗﷯ − 4 𝑘﷯ form the vertices of a right angled triangle. Let A(2 𝑖﷯ − 𝑗﷯ + 𝑘﷯), B( 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯) C(3 𝑖﷯ − 4 𝑗﷯ − 4 𝑘﷯) Considering ∆ABC as a right angled triangle, By Pythagoras theorem, AB2 = BC2 + CA2 or AB﷯﷯2 = BC﷯﷯2 + CA﷯﷯2 𝐴𝐵﷯ = ( 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯) − (2 𝑖﷯ − 𝑗﷯ + 𝑘﷯) = 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﷯ 𝐵𝐶﷯ = (3 𝑖﷯ − 4 𝑗﷯ − 4 𝑘﷯) − ( 𝑖﷯ − 3 𝑗﷯ − 5 𝑘﷯) = 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﷯ 𝐶𝐴﷯ = (2 𝑖﷯ − 𝑗﷯ + 𝑘﷯) − (3 𝑖﷯ − 4 𝑗﷯ − 4 𝑘﷯) = 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﷯ 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﷯ So, ABC is a right angled triangle.

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