Question

If a+b+c=0,|a|=3 |b|=5 |c|=7,then find the angle between a and b​

Answer 1

Answer:

"The vectors form a triangle with sides 3,5, and 7.

According to cosine law

7^2=3^3+5^2–2*3*5*cosC where C is the angle between a and b.

cosC = 1/2 => C=60 degrees. "

Answer 2

$|a| = 3 \: \: \: \: \: |b| = 5 \: \: \: \: |c| = 7$

$a \: + b + c = 0$

⟹$a + b = - c$

⟹$|a + b| = | - c|$

⟹$|a + b| {}^{2} = | - c| {}^{2}$

⟹ $|a| {}^{2} + |b| {}^{2} + 2a.b = |c | {}^{2}$

⟹$9 + 25 + 2a.b = 49$

Also ab = $3 \times 5$

Hence 9+25+2abcosθ=49

⟹15=30 cos θ

θ= $( \frac{1}{2} )$

θ $60 {}^{o}$

$so \: angle \: is \: 60 {}^{o}$