CUET UG 2025 — Mathematics Algebra

Given three vectors: $\vec{a}$a, $\vec{b}$b, and $\vec{c}$c where:

• $\vec{a} + \vec{b} + \vec{c} = \vec{0}$a+b+c=0
• $|\vec{a}| = 3$∣a∣=3
• $|\vec{b}| = 5$∣b∣=5
• $|\vec{c}| = 7$∣c∣=7

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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:

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

Expanding using $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$∣a+b∣2=∣a∣2+∣b∣2+2a⋅b:

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

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Substituting the known values:

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

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

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

$$\vec{a} \cdot \vec{b} = \frac{15}{2}$$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:

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

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

$$\cos\theta = \frac{1}{2}$$cosθ=21​

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When $\cos\theta = \frac{1}{2}$cosθ=21​:

$\theta = \frac{\pi}{3}$θ=3π​ (or 60°)

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