Skip to main contentSkip to solution

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

Solution

✅ Correct Option: 1

When a⃗×b⃗=0⃗\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:

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

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

Setting up the ratios:

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


−39=−13\dfrac{-3}{9} = -\dfrac{1}{3}

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

Both ratios are consistent.


Using the middle ratio:

−6−λ=−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

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question