Dear Student, Let θ be the angle between a→ and b→ Now, a→ + b→ + c→ = 0⇒a→ + b→ = -c→⇒a→ + b→ a→ + b→ = -c→ -c→⇒a→ a→ + a→ b→ + b→ a→ + b→ b→ = c→ c→⇒a→2 + a→ b→ + a→ b→ + b→2 = c→2⇒a→2 + 2a→ b→ + b→2 = c→2⇒32 + 2a→ b→ + 42 = 52⇒9 + 2a→ b→ + 16 = 25⇒2a→ b→ = 0⇒a→ b→ = 0⇒a→ b→ cos θ = 0⇒3 × 4 × cos θ = 0⇒cos θ = 0⇒θ = π2 Regards

Q 24

Dear Student,

Let θ be the angle between \vec{a} and \vec{b} . Now, \vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow \vec{a} + \vec{b} = - \vec{c} \Rightarrow (\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) = (- \vec{c}) . (- \vec{c}) \Rightarrow \vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{b} . \vec{a} + \vec{b} . \vec{b} = \vec{c} . \vec{c} \Rightarrow {|\vec{a}|}^{2} + \vec{a} . \vec{b} + \vec{a} . \vec{b} + {|\vec{b}|}^{2} = {|\vec{c}|}^{2} \Rightarrow {|\vec{a}|}^{2} + 2 \vec{a} . \vec{b} + {|\vec{b}|}^{2} = {|\vec{c}|}^{2} \Rightarrow {(3)}^{2} + 2 \vec{a} . \vec{b} + {(4)}^{2} = {(5)}^{2} \Rightarrow 9 + 2 \vec{a} . \vec{b} + 16 = 25 \Rightarrow 2 \vec{a} . \vec{b} = 0 \Rightarrow \vec{a} . \vec{b} = 0 \Rightarrow |\vec{a}| |\vec{b}| \cos θ = 0 \Rightarrow 3 \times 4 \times \cos θ = 0 \Rightarrow \cos θ = 0 \Rightarrow θ = \frac{π}{2}

Regards

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