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

Let \(\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7\), \(2 \vec{b} \cdot \vec{c}+43=0, \vec{a} \times \vec{c}=\vec{b} \times \vec{c}\). Then \(|\vec{a} \cdot \vec{b}|\) is equal to

  1. A \(4\)
  2. B \(2\)
  3. C \(6\)
  4. D \(8\)
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Answer & Solution

Correct Answer

(D) \(8\)

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\(\overrightarrow{ a }=\hat{ i }+2 \hat{ j }+\lambda \hat{ k }, \quad \overrightarrow{ b }=3 \hat{ i }-5 \hat{ j }-\lambda \hat{ k }, \overrightarrow{ a } \cdot \overrightarrow{ c }=7\)…
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