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Question: If $\vec{a}+\vec{b}+\vec{c}=0$, $|\vec{a}|=3$, $|\vec{b}|=5$, $|\vec{c}|=7$, then the angle between ...

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

A

π/6\pi/6

B

2π/32\pi/3

C

5π/35\pi/3

D

π/3\pi/3

Answer

π/3\pi/3

Explanation

Solution

Given a⃗+b⃗+c⃗=0\vec{a}+\vec{b}+\vec{c}=0, which implies a⃗+b⃗=−c⃗\vec{a}+\vec{b} = -\vec{c}.

Squaring both sides (taking dot product with itself): ∣a⃗+b⃗∣2=∣−c⃗∣2|\vec{a}+\vec{b}|^2 = |-\vec{c}|^2 ∣a⃗∣2+∣b⃗∣2+2a⃗⋅b⃗=∣c⃗∣2|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{c}|^2

Substitute given magnitudes ∣a⃗∣=3,∣b⃗∣=5,∣c⃗∣=7|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7: 32+52+2a⃗⋅b⃗=723^2 + 5^2 + 2\vec{a} \cdot \vec{b} = 7^2 9+25+2a⃗⋅b⃗=499 + 25 + 2\vec{a} \cdot \vec{b} = 49 34+2a⃗⋅b⃗=4934 + 2\vec{a} \cdot \vec{b} = 49 2a⃗⋅b⃗=152\vec{a} \cdot \vec{b} = 15 a⃗⋅b⃗=152\vec{a} \cdot \vec{b} = \frac{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=(3)(5)cos⁡θ\frac{15}{2} = (3)(5) \cos\theta 152=15cos⁡θ\frac{15}{2} = 15 \cos\theta cos⁡θ=12\cos\theta = \frac{1}{2} θ=π3\theta = \frac{\pi}{3}