Question

If vector a and b are such that |a+b|=|a|=|b| then |a-b| may be equated to

Answer 1

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Answer 2

|a - b| = √3|a| = √3|b|

it is given that,

|a + b| = |a| = |b|

⇒√(a² + b² + abcosθ) = a = b

squaring then we get,

⇒a² + b² + abcosθ = a² = b²

case 1 : a² + b² + abcosθ = a²

⇒b² + abcosθ = 0

⇒cosθ = -|b|/|a| ......(1)

case 2 : a² + b² + abcosθ = b²

⇒a² + abcosθ = 0

⇒cosθ = -|a|/|b| ......(2)

from equation (1) and (2),

cosθ = -1 = 180°

so, |a - b| = √(a² + b² - abcos180° )

= √(a² + a² - a²(-1))

= √3|a| = √3|b|

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