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Question 130 of 162

Q.If a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0}, ∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=4|\vec{b}| = 4, ∣c⃗∣=5|\vec{c}| = 5 then, the angle between a⃗\vec{a} and b⃗\vec{b} is :

(a) 5π3\dfrac{5\pi}{3}
(b) π2\dfrac{\pi}{2}
(c) π6\dfrac{\pi}{6}
(d) 2π3\dfrac{2\pi}{3}
Puducherry TnboardTamil Nadu HSC (DGE) Board 2019MCQ· 1mImportance★★★★★
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From a⃗+b⃗+c⃗=0⃗\vec a+\vec b+\vec c=\vec 0 with ∣a⃗∣=3,∣b⃗∣=4,∣c⃗∣=5|\vec a|=3,|\vec b|=4,|\vec c|=5, the angle between a⃗\vec a and b⃗\vec b works out to π2\dfrac{\pi}{2}.

  1. From a⃗+b⃗+c⃗=0⃗\vec a+\vec b+\vec c=\vec0, we get c⃗=−(a⃗+b⃗)\vec c=-(\vec a+\vec b).
  2. Take magnitudes squared: ∣c⃗∣2=∣a⃗+b⃗∣2=∣a⃗∣2+2a⃗⋅b⃗+∣b⃗∣2|\vec c|^2=|\vec a+\vec b|^2=|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2.
  3. Substitute the given magnitudes: 52=32+2a⃗⋅b⃗+425^2=3^2+2\vec a\cdot\vec b+4^2, i.e. 25=9+16+2a⃗⋅b⃗=25+2a⃗⋅b⃗25=9+16+2\vec a\cdot\vec b=25+2\vec a\cdot\vec b.
  4. So 2a⃗⋅b⃗=0⇒a⃗⋅b⃗=02\vec a\cdot\vec b=0\Rightarrow \vec a\cdot\vec b=0. …

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