JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{\alpha}=4 \hat{ i }+3 \hat{ j }+5 \hat{ k }\) and \(\vec{\beta}=\hat{ i }+2 \hat{ j }-4 \hat{ k }\). Let \(\vec{\beta}_1\) be parallel to \(\vec{\alpha}\) and \(\vec{\beta}_2\) be perpendicular to \(\vec{\alpha}\). If \(\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2\), then the value of \(5 \vec{\beta}_2 \cdot(\hat{ i }+\hat{ j }+\hat{ k })\) is
- A \(6\)
- B \(11\)
- C \(7\)
- D \(9\)
Answer & Solution
Correct Answer
(C) \(7\)
Step-by-step Solution
Detailed explanation
Let \(\vec{\beta}_1=\lambda \vec{\alpha}\) Now \(\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1\) \(=(\hat{ i }+2 \hat{ j }-4 \hat{ k })-\lambda(4 \hat{ i }+3 \hat{ j }+5 \hat{ k })\) \(=(1-4 \lambda) \hat{ i }+(2-3 \lambda) \hat{ j }-(5 \lambda+4) \hat{ k }\)…
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