CUET UG 2025 — Mathematics Algebra

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

$|\vec{a}| = 3$∣a∣=3, $|\vec{b}| = 5$∣b∣=5, $|\vec{c}| = 7$∣c∣=7

From the constraint:

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

Taking magnitudes:

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

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

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Expanding the right side:

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

Substituting the given values:

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

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

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

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

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

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

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

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

$$\cos\theta = \dfrac{1}{2}$$cosθ=21​

$$\theta = \dfrac{\pi}{3}$$θ=3π​

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