JEE Mains · Physics · STD 11 - 3.1 vectors
A vector in \(x-y\) plane makes an angle of \(30^{\circ}\) with \(y\)-axis The magnitude of \(y\)-component of vector is \(2 \sqrt{3}\). The magnitude of \(x\)-component of the vector will be
- A \(\frac{1}{\sqrt{3}}\)
- B \(6\)
- C \(\sqrt{3}\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(A _y= A \cos 30^{\circ}=2 \sqrt{3}\) \(\Rightarrow A \frac{\sqrt{3}}{2}=2 \sqrt{3}\) \(\Rightarrow A =4\) Now \(A_x=A \sin 30^{\circ}=4 \times \frac{1}{2}=2\)
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