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for unit vectors a,b and c, if |a-b|^2+|b-c|^2+|c-a|^2 = 9 then what is |2a+5b+5c|?

Sudhanshu Bansal , 11 Years ago

Grade

2 Answers

Y RAJYALAKSHMI

|a – b|² + |b – c|² + |c – a|² = 9

=> 2(|a|² + |b|² + |c|²) – 2(a.b + b.c + c.a) = 9

=> 2 * 3 – 2(a.b + b.c + c.a) = 9 (Since a, b, c are unit vectors)

=> (cos (a, b) + cos (b, c) + cos(c, a) = – 3/2 [Since a, b, c are unit vectors & a.b = |a||b| cos(a, b) where cos (a, b) , cos (b, c) , cos(c, a) are cosine of angles between these vecotrs ]

This gives us cos (a, b) = cos (b, c) = cos(c, a) = –1/2

=> a.b = b.c = c.a = –1/2

|2a + 5b + 5c|² = 4|a|² + 25|b|² + 25|c|² + 20 a. b + 50 b. c + 20 c.a

= 4 + 25 +25 + 20 * (–1/2) + 50 * (–1/2) + 20 * (–1/2) = 9

Therefore, |2a + 5b + 5c| = 3

Last Activity: 11 Years ago

Kushagra Madhukar

Dear student,

Please find the solution to your problem.

|a – b|² + |b – c|² + |c – a|² = 9

=> 2(|a|² + |b|² + |c|²) – 2(a.b + b.c + c.a) = 9

=> 2 . 3 – 2(a.b + b.c + c.a) = 9 (Since a, b, c are unit vectors)

=> (cos (a, b) + cos (b, c) + cos(c, a) = – 3/2 [Since a, b, c are unit vectors & a.b = |a||b| cos(a, b) where cos (a, b) , cos (b, c) , cos(c, a) are cosine of angles between these vecotrs ]

This gives us cos (a, b) = cos (b, c) = cos(c, a) = –1/2

=> a.b = b.c = c.a = –1/2

|2a + 5b + 5c|² = 4|a|² + 25|b|² + 25|c|² + 20 a. b + 50 b. c + 20 c.a

= 4 + 25 +25 + 20 * (–1/2) + 50 * (–1/2) + 20 * (–1/2) = 9

Therefore, |2a + 5b + 5c| = 3

Thanks and regards,

Kushagra

Last Activity: 5 Years ago