JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\lambda \in Z, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}\) and \(\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\). Let \(\overrightarrow{ c }\) be a vector such that \((\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\overrightarrow{0}, \vec{a} \cdot \vec{c}=-17\) and \(\vec{b} \cdot \vec{c}=-20\) Then \(|\overrightarrow{ c } \times(\lambda \hat{i}+\hat{j}+\hat{ k })|^2\) is equal to
- A \(62\)
- B \(46\)
- C \(53\)
- D \(49\)
Answer & Solution
Correct Answer
(B) \(46\)
Step-by-step Solution
Detailed explanation
\((\vec{a}+\vec{b}+\overrightarrow{ c }) \times \overrightarrow{ c }=0\) \((\vec{a}+\vec{b}) \times \vec{c}=0\) \(\overrightarrow{ c }=\alpha(\overrightarrow{ a }+\overrightarrow{ b })=\alpha(\lambda+3) \hat{ i }+\alpha \hat{ k }\)…
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