JEE Mains · Maths · STD 12 - 10. vector algebra
The vector \(\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}\) is rotated through a right angle, passing through the \(y\)-axis in its way and the resulting vector is \(\vec{b}\). Then the projection of \(3 \vec{a}+\sqrt{2} \vec{b}\) on \(\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}\) is
- A \(3 \sqrt{2}\)
- B \(1\)
- C \(\sqrt{6}\)
- D \(2 \sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(3 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ b }=\lambda \overrightarrow{ a } \times(\overrightarrow{ a } \times \hat{ j })\) \(\Rightarrow \overrightarrow{ b }=\lambda(-2 \hat{ i }-2 \hat{ j }+2 \hat{ k })\)…
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