NIMCET 2012 Question Paper with solutions

Given expression
a + b + c = 0
⇒ a + b = -c
On squaring both sides,
⇒ ${\left(\mathit{a}+\mathit{b}\right)}^2$ = ${\left(-\mathit{c}\right)}^2$
⇒ (a + b) . (a + b) = (-c) . (-c)
⇒ (a . a) + (b . a) + (a . b) + (b . b) = (c . c)
⇒ ${\mathit{a}}^2$ + 2a . b + ${\mathit{b}}^2$ = ${\mathit{x}}^2$ Since (a.b = b.a)
⇒ ${\left|\mathit{a}\right|}^2$ + 2a . b + ${\left|\mathit{b}\right|}^2$ = ${\left|\mathit{c}\right|}^2$ Since (${\mathit{a}}^2$ = ${\left|\mathit{a}\right|}^2$)
⇒ ${\left(3\right)}^2$ + 2a . b + ${\left(5\right)}^2$ = ${\left(7\right)}^2$
Since |a| = 3 , |b| = 5 and |c| = 7
⇒ 2a . b = 49 - 25 - 9
⇒ 2a . b = 15
⇒ 2 . |a| |b| cos θ = 15
r cos θ = $\frac{1}{2}$ = cos 60
∴ θ = $\frac{\mathit{\pi }}{3}$