If a +b+c =0. The angle between a and b and c are 150 and 120 respectively then the magnitude of vector a b c are in ratio of
Answers
Given:
a + b + c =0.
The angle between a and b and c are 150 and 120 respectively.
To find:
The magnitude of vector a b c are in ratio of ?
Solution:
From given, we have,
a + b + c = 0
a + b = - c
taking mod on both sides, we get,
⇒ |a + b | = |-c |
⇒ √(a² + b² + 2abcosα) = c²
given the angle between a and b = 150°
so, α = 150°
⇒ √{a² + b² + 2abcos150°} = c²
squaring both sides, we get
⇒ a² + b² + 2abcos150° = c²
∴ a² + b² - √3ab = c² ........(1)
again, a + b + c = 0
b + c = -a
taking mod on both sides, we get,
⇒ |b + c| = |-a|
⇒ √(b² + c² + 2bccosβ) = a²
given the angle between b and c is 120° .
so, β = 120°
⇒ √(b² + c² + 2bccos120°} = a²
squaring both sides, we get,
⇒ b² + c² + 2bccos120° = a²
∴ b² + c² - ab = a² .......(2)
Using equations (1) and (2), we get,
2b² - (√3 + 1)ab = 0
⇒ 2b² = (√3 + 1)ab
⇒ b = {(√3 + 1)/2}a .......(3)
again using equation (1), we get,
a² + (√3 + 1)²/4 a² - √3a²(√3 + 1)/2 = c²
⇒ a² [ 1 + (4 + 2√3)/4 - (√3 + 3)/2] = c²
⇒ a² {4 + 4 + 2√3 - 2√3 - 6}/4 = c²
⇒ a²/2 = c²
⇒ c = a/√2 ......(4)
using equations (3) and (4), we get,
a : b : c = a : (√3 + 1)/2 a : a/√2
taking LCM and "a" as common, we have,
a : b : c = 2 : (√3 + 1) : √2
Hence the ratio of magnitude of vectors a b c