JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\vec{b}=\hat{i}+\hat{j}-\hat{k}\). If \(\vec{c}\) is a vector such that \(\vec{a} \cdot \vec{c}=11, \vec{b} \cdot(\vec{a} \times \vec{c})=27\) and \(\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|\), then \(|\vec{a} \times \vec{c}|^2\) is equal to
- A \(285\)
- B \(284\)
- C \(283\)
- D \(282\)
Answer & Solution
Correct Answer
(A) \(285\)
Step-by-step Solution
Detailed explanation
Sol. \(\overrightarrow{ a }=\hat{ i }+2 \hat{ j }+3 \hat{ k }, \overrightarrow{ b }=\hat{ i }+\hat{ j }-\hat{ k }\) \(\vec{b} \cdot(\vec{a} \times \vec{c})=27, \vec{a} \cdot \vec{b}=0\) \(\vec{b} \times(\vec{a} \times \vec{c})=-3 \vec{a}\) Let \(\theta\) be angle between…
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