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If a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} be vectors such that a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0}, ∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=5|\vec{b}| = 5 and ∣c⃗∣=7|\vec{c}| = 7, then the angle between a⃗\vec{a} and b⃗\vec{b} is

Solution

✅ Correct Option: 2

Given:

a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0}

∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=5|\vec{b}| = 5, ∣c⃗∣=7|\vec{c}| = 7

From the constraint:

c⃗=−(a⃗+b⃗)\vec{c} = -(\vec{a} + \vec{b})

Taking magnitudes:

∣c⃗∣=∣a⃗+b⃗∣|\vec{c}| = |\vec{a} + \vec{b}|

∣c⃗∣2=∣a⃗+b⃗∣2|\vec{c}|^2 = |\vec{a} + \vec{b}|^2


Expanding the right side:

∣c⃗∣2=∣a⃗∣2+2(a⃗⋅b⃗)+∣b⃗∣2|\vec{c}|^2 = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2

Substituting the given values:

72=32+2(a⃗⋅b⃗)+527^2 = 3^2 + 2(\vec{a} \cdot \vec{b}) + 5^2

49=9+2(a⃗⋅b⃗)+2549 = 9 + 2(\vec{a} \cdot \vec{b}) + 25

49=34+2(a⃗⋅b⃗)49 = 34 + 2(\vec{a} \cdot \vec{b})

15=2(a⃗⋅b⃗)15 = 2(\vec{a} \cdot \vec{b})

a⃗⋅b⃗=152\vec{a} \cdot \vec{b} = \dfrac{15}{2}


Using the dot product formula a⃗⋅b⃗=∣a⃗∣×∣b⃗∣×cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}| \times |\vec{b}| \times \cos\theta:

152=3×5×cos⁡θ\dfrac{15}{2} = 3 \times 5 \times \cos\theta

152=15cos⁡θ\dfrac{15}{2} = 15\cos\theta

cos⁡θ=12\cos\theta = \dfrac{1}{2}

θ=π3\theta = \dfrac{\pi}{3}

Therefore, the angle between a⃗\vec{a} and b⃗\vec{b} is π3\dfrac{\pi}{3}.

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