CUET UG 2025 — Mathematics Algebra

When $\vec{a} \times \vec{b} = \vec{0}$a×b=0, the vectors are parallel, so the ratios of their corresponding components must be equal:

$$\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2} = \dfrac{a_3}{b_3}$$b1​a1​​=b2​a2​​=b3​a3​​

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Given:

$$\vec{a} = -3\hat{i} - 6\hat{j} + 4\hat{k}$$a=−3i^−6j^​+4k^

$$\vec{b} = 9\hat{i} - \lambda\hat{j} - 12\hat{k}$$b=9i^−λj^​−12k^

Setting up the ratios:

$$\dfrac{-3}{9} = \dfrac{-6}{-\lambda} = \dfrac{4}{-12}$$9−3​=−λ−6​=−124​

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$$\dfrac{-3}{9} = -\dfrac{1}{3}$$9−3​=−31​

$$\dfrac{4}{-12} = -\dfrac{1}{3}$$−124​=−31​

Both ratios are consistent.

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Using the middle ratio:

$$\dfrac{-6}{-\lambda} = -\dfrac{1}{3}$$−λ−6​=−31​

$$\dfrac{6}{\lambda} = -\dfrac{1}{3}$$λ6​=−31​

$$6 \times 3 = -1 \times \lambda$$6×3=−1×λ

$$18 = -\lambda$$18=−λ

$$\lambda = -18$$λ=−18

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The value of $\lambda = -18$λ=−18