CUET UG 2025 — Mathematics Algebra

Given $|\vec{a}| = 10$∣a∣=10, $|\vec{b}| = 2$∣b∣=2, and $\vec{a} \cdot \vec{b} = 12$a⋅b=12.

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

$$12 = 10 \times 2 \times \cos\theta$$12=10×2×cosθ

$$12 = 20\cos\theta$$12=20cosθ

$$\cos\theta = \frac{12}{20}$$cosθ=2012​

$$\cos\theta = \frac{3}{5}$$cosθ=53​

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Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$sin2θ+cos2θ=1:

$$\sin^2\theta = 1 - \cos^2\theta$$sin2θ=1−cos2θ

$$\sin^2\theta = 1 - \left(\frac{3}{5}\right)^2$$sin2θ=1−(53​)2

$$\sin^2\theta = 1 - \frac{9}{25}$$sin2θ=1−259​

$$\sin^2\theta = \frac{16}{25}$$sin2θ=2516​

$$\sin\theta = \frac{4}{5}$$sinθ=54​

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Using the cross product magnitude formula $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$∣a×b∣=∣a∣∣b∣sinθ:

$$|\vec{a} \times \vec{b}| = 10 \times 2 \times \frac{4}{5}$$∣a×b∣=10×2×54​

$$|\vec{a} \times \vec{b}| = 20 \times \frac{4}{5}$$∣a×b∣=20×54​

$$|\vec{a} \times \vec{b}| = \frac{80}{5}$$∣a×b∣=580​

$$|\vec{a} \times \vec{b}| = 16$$∣a×b∣=16

Therefore, $|\vec{a} \times \vec{b}| = 16$∣a×b∣=16.