CUET UG 2025 — Mathematics Calculus

Given three vectors that add up to zero:

$$\vec{a} + \vec{b} + \vec{c} = \vec{0}$$a+b+c=0

$$|\vec{a}| = 5$$∣a∣=5

$$|\vec{b}| = 3$$∣b∣=3

$$|\vec{c}| = 7$$∣c∣=7

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From $\vec{a} + \vec{b} + \vec{c} = \vec{0}$a+b+c=0:

$$\vec{c} = -(\vec{a} + \vec{b})$$c=−(a+b)

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Taking magnitude on both sides:

$$|\vec{c}| = |\vec{a} + \vec{b}|$$∣c∣=∣a+b∣

$$|\vec{c}|^2 = |\vec{a} + \vec{b}|^2$$∣c∣2=∣a+b∣2

$$|\vec{c}|^2 = |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2$$∣c∣2=∣a∣2+2a⋅b+∣b∣2

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Substituting the given values:

$$7^2 = 5^2 + 2\vec{a} \cdot \vec{b} + 3^2$$72=52+2a⋅b+32

$$49 = 25 + 2\vec{a} \cdot \vec{b} + 9$$49=25+2a⋅b+9

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

$$2\vec{a} \cdot \vec{b} = 15$$2a⋅b=15

$$\vec{a} \cdot \vec{b} = \dfrac{15}{2}$$a⋅b=215​

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Using the dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$a⋅b=∣a∣∣b∣cosθ:

$$\dfrac{15}{2} = 5 \times 3 \times \cos\theta$$215​=5×3×cosθ

$$\dfrac{15}{2} = 15\cos\theta$$215​=15cosθ

$$\cos\theta = \dfrac{1}{2}$$cosθ=21​

$$\theta = \dfrac{\pi}{3}$$θ=3π​

Therefore, the acute angle between $\vec{a}$a and $\vec{b}$b is $\dfrac{\pi}{3}$3π​.