© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the initial vector
The given vector before rotation is
$ \sqrt{3} \hat{i} + \hat{j} $.
Its components are $x = \sqrt{3}$ and $y = 1$.
Step 2: Apply the rotation by 45° in the first quadrant
A rotation by an angle $ \theta $ in the counterclockwise direction in 2D is given by multiplying with the rotation matrix:
$
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}.
$
For $ \theta = 45^\circ $, $ \cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2} $.
If the rotated vector is $ \alpha \hat{i} + \beta \hat{j} $, we have:
$
\alpha = x \cos 45^\circ \;-\; y \sin 45^\circ, \quad
\beta = x \sin 45^\circ \;+\; y \cos 45^\circ.
$
Substituting $ x = \sqrt{3} $ and $ y = 1 $:
$
\alpha = \sqrt{3} \cdot \frac{\sqrt{2}}{2} - 1 \cdot \frac{\sqrt{2}}{2}
= \frac{\sqrt{6} - \sqrt{2}}{2},
$
$
\beta = \sqrt{3} \cdot \frac{\sqrt{2}}{2} + 1 \cdot \frac{\sqrt{2}}{2}
= \frac{\sqrt{6} + \sqrt{2}}{2}.
$
Step 3: Set up the coordinates of the triangle’s vertices
The problem gives the vertices of the triangle as:
$ (\alpha, \beta) $
$ (0, \beta) $
$ (0, 0) $
Step 4: Find the area of the triangle
Observe that the vertical side of the triangle goes from $(0, 0)$ to $(0, \beta)$, which has length $|\beta|$. The horizontal side from $(0, \beta)$ to $(\alpha, \beta)$ has length $|\alpha|$. Because these sides are perpendicular, the area of the right-angled triangle is:
$
\text{Area} = \frac{1}{2} \times |\text{base}| \times |\text{height}|
= \frac{1}{2} \times \alpha \times \beta.
$
Substituting $ \alpha = \frac{\sqrt{6} - \sqrt{2}}{2} $ and $ \beta = \frac{\sqrt{6} + \sqrt{2}}{2} $, we get:
$
\alpha \beta
= \Bigl(\frac{\sqrt{6} - \sqrt{2}}{2}\Bigr)
\Bigl(\frac{\sqrt{6} + \sqrt{2}}{2}\Bigr)
= \frac{(\sqrt{6})^2 - (\sqrt{2})^2}{4}
= \frac{6 - 2}{4}
= \frac{4}{4}
= 1.
$
Therefore,
$
\text{Area}
= \frac{1}{2} \times 1
= \frac{1}{2}.
$
Step 5: Final Answer
The area of the triangle is $ \frac{1}{2} $.
