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JEE Mains · Maths · STD 12 - 10. vector algebra

Let \(\vec{u}=\hat{i}-\hat{j}-2 \hat{k}, \vec{v}=2 \hat{i}+\hat{j}-\hat{k}, \vec{v} \cdot \vec{w}=2\) and \(\vec{v} \times \vec{w}=\vec{u}+\lambda \vec{v}\), then \(\vec{u} \cdot \vec{w}\) is equal to

  1. A 1
  2. B \(\frac{2}{3}\)
  3. C 2
  4. D \(-\frac{2}{3}\)
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Answer & Solution

Correct Answer

(A) 1

Step-by-step Solution

Detailed explanation

Given \(\vec{v} \times \vec{w}=\lambda \vec{v}+\vec{u}\) Taking dot with \(\vec{v}\) we get \([\vec{v} \vec{v} \vec{w}]=\lambda|\vec{v}|^2+\vec{u} \cdot \vec{v}\) Substituting values we have \[ 6 \lambda+3=0 \Rightarrow \lambda=-\frac{1}{2} \] So, equation (i) becomes…
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