Question

Let a vector \(\alpha\hat{i}+\beta\hat{j}\) be obtained by rotating the vector \(\sqrt{3}\,\hat{i}+\hat{j}\) by an angle \(45^\circ\) about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices \((\alpha, \beta)\), \((0, \beta)\) and \((0, 0)\) is equal to :

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\frac{1}{2}\)
• C. 1
• D. \(2\sqrt{2}\)

Hint:

The vector \(\sqrt3\hat i+\hat j\) has length 2 and makes \(30^\circ\) with the x-axis, so after rotation it makes \(75^\circ\). The triangle is right-angled with legs \(\alpha\) and \(\beta\).

Answer 1

The Correct Option is B

Length of \(\sqrt3\hat i+\hat j\) is \(\sqrt{3+1}=2\) and its angle with the x-axis is \(\tan^{-1}(1/\sqrt3)=30^\circ\).
After counterclockwise rotation by \(45^\circ\) the angle is \(75^\circ\), length unchanged, so \(\alpha=2\cos 75^\circ,\ \beta=2\sin 75^\circ\).
The triangle with vertices \((\alpha,\beta),(0,\beta),(0,0)\) has a right angle at \((0,\beta)\) with perpendicular sides \(\alpha\) and \(\beta\).
Area \(=\frac12\alpha\beta=\frac12\cdot 4\sin75^\circ\cos75^\circ=\sin150^\circ=\frac12\).
Answer: (2) \(\frac{1}{2}\)