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Given that a=3i^6j^+4k^\vec{a} = -3\hat{i} - 6\hat{j} + 4\hat{k}, b=9i^λj^12k^\vec{b} = 9\hat{i} - λ\hat{j} - 12\hat{k}. If a×b=0\vec{a} \times \vec{b} = \vec{0}, then the value of λ is

Solution

Correct Option: 1

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

a1b1=a2b2=a3b3\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2} = \dfrac{a_3}{b_3}


Given:

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

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

Setting up the ratios:

39=6λ=412\dfrac{-3}{9} = \dfrac{-6}{-\lambda} = \dfrac{4}{-12}


39=13\dfrac{-3}{9} = -\dfrac{1}{3}

412=13\dfrac{4}{-12} = -\dfrac{1}{3}

Both ratios are consistent.


Using the middle ratio:

6λ=13\dfrac{-6}{-\lambda} = -\dfrac{1}{3}

6λ=13\dfrac{6}{\lambda} = -\dfrac{1}{3}

6×3=1×λ6 \times 3 = -1 \times \lambda

18=λ18 = -\lambda

λ=18\lambda = -18


The value of λ=18\lambda = -18

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