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

Solution

✅ Correct Option: 4

Given three vectors that add up to zero:

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

∣a⃗∣=5|\vec{a}| = 5

∣b⃗∣=3|\vec{b}| = 3

∣c⃗∣=7|\vec{c}| = 7


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

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


Taking magnitude on both sides:

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

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

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


Substituting the given values:

72=52+2a⃗⋅b⃗+327^2 = 5^2 + 2\vec{a} \cdot \vec{b} + 3^2

49=25+2a⃗⋅b⃗+949 = 25 + 2\vec{a} \cdot \vec{b} + 9

49=34+2a⃗⋅b⃗49 = 34 + 2\vec{a} \cdot \vec{b}

2a⃗⋅b⃗=152\vec{a} \cdot \vec{b} = 15

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}||\vec{b}|\cos\theta:

152=5×3×cos⁡θ\dfrac{15}{2} = 5 \times 3 \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 acute angle between a⃗\vec{a} and b⃗\vec{b} is π3\dfrac{\pi}{3}.

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