JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k} \quad a_{i}>0, i=1,2,3\) be \(a\) vector which makes equal angles with the coordinates axes OX, OY and OZ. Also, let the projection of \(\vec{a}\) on the vector \(3 \hat{i}+4 \hat{j}\) be \(7\) . Let \(\vec{b}\) be a vector obtained by rotating \(\vec{a}\) with \(90^{\circ}\). If \(\vec{a}, \vec{b}\) and \(x\)-axis are coplanar, then projection of a vector \(\vec{b}\) on \(3 \hat{i}+4 \hat{j}\) is equal to
- A \(\sqrt{7}\)
- B \(\sqrt{2}\)
- C \(2\)
- D \(7\)
Answer & Solution
Correct Answer
(B) \(\sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ a }=a_{1} \hat{ i }+ a _{2} \hat{ j }+ a _{3} \hat{ k }\) \(\overrightarrow{ a }=\lambda\left(\frac{1}{\sqrt{3}} \hat{ i }+\frac{1}{\sqrt{3}} \hat{ j }+\frac{1}{\sqrt{3}} \hat{ k }\right)=\frac{\lambda}{\sqrt{3}}(\hat{ i }+\hat{ j }+\hat{ k }\) Now projection…
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