Question

If (a+3b).(7a-5b)=0 and (a-4b).(7a-2b)=0, then angle between a and b is

Answer 1

angle between a and b is π/3

given, (a + 3b).(7a - 5b) = 0 and (a - 4b).(7a - 2b) = 0

then we have to find angle between a and b.

now, (a + 3b).(7a - 5b) = 0

⇒7a² - 5a.b + 21a.b - 15b² = 0

⇒7a² + 16a.b - 15b² = 0.......(1)

again, (a - 4b).(7a - 2b) = 0

⇒7a² - 2a.b - 28a.b + 8b² = 0

⇒7a² - 30a.b + 8b² = 0 .......(2)

from equations (1) and (2) we get,

46a.b - 23b² = 0

⇒2a.b - b² = 0

now, putting a.b = abcosθ [ where θ is angle between a and b]

so, 2abcosθ - b² = 0

⇒ cosθ = b/2a

if a and b are unit vectors

then, a = b = 1

so, cosθ = 1/2 = cos(π/3)

⇒θ = π/3

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