Question

Let a vector $\begin{array}{l}\alpha \hat{i}+\beta \end{array}\hat{j}$ be obtained by rotating the vector $\begin{array}{l}\sqrt{3}\hat{i}+\hat{j}\end{array}$by an angle $45°$ about the origin in counter clockwise direction in the first quadrant. Then the area of triangle having vertices $(ɑ,\beta ),(0,\beta )$ and $(0,0)$ is equal to:

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

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation for the correct option:

Step-1 : find the coordinates of $\left(\alpha ,\beta \right)$

The given vector is $\begin{array}{l}\sqrt{3}\hat{i}+\hat{j}\end{array}$

We know that the angle subtended by a vector $x\hat{i}+y\hat{j}$ with $x$- axes

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{y}{x}\right)$

Therefore, the angle subtended by a vector $\begin{array}{l}\sqrt{3}\hat{i}+\hat{j}\end{array}$ with $x$- axes

$$\begin{gathered} \Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \Rightarrow \theta ={\tan }^{-1}\left(\tan 30°\right) \\ \Rightarrow \theta =30° \end{gathered}$$

Using the formula for magnitude of the vector, i.e., $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k},\left|a\right|=\sqrt{x^2+y^2+z^2}$

Now the magnitude of this vector be

$$\begin{gathered} \Rightarrow r=\sqrt{{\left(\sqrt{3}\right)}^2+1^2} \\ \Rightarrow r=\sqrt{3+1} \\ \Rightarrow r=2 \end{gathered}$$

And given that the vector $\begin{array}{l}\alpha \hat{i}+\beta \end{array}\hat{j}$is obtained by rotating the vector $\begin{array}{l}\sqrt{3}\hat{i}+\hat{j}\end{array}$by an angle $45°$counterclockwise.

Considering $\alpha =r\cos \varphi ,\beta =r\sin \varphi$ and we have $r=2$

$$\begin{gathered} \because \alpha =2\cos (30°+45°),\beta =2\sin (30°+45°) \\ \Rightarrow \alpha =2\cos (75°),\beta =2\sin (75°) \end{gathered}$$

Since the vector is rotated in the first quadrant

Therefore, the angle of the triangle having vertices $A(2\cos (75°),2\sin (75°\left)\right),B(0,2\sin (75°\left)\right)$and $O(0,0)$ subtended at the origin

$\Rightarrow \angle AOC=90°-75°=15°$

Step-2 Area of trainagle formed by the given points:

We know that area of the triangle $∆\left(OAB\right)=\frac{1}{2}\times OA\cos \left(\angle AOC\right)\times \sin \left(\angle AOC\right)\left[\because \mathrm{Area}\mathrm{of}∆=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}\right]$

Therefore,

$$\begin{gathered} ∆\left(AOB\right)=\frac{1}{2}\times (2\cos 75°)\times (2\sin 75°) \\ =\sin (2\times 75°)\mathbf{}\left[\because sin\left(2\theta \right)=2sin\theta cos\theta \right] \\ =\sin (150°) \\ =\frac{1}{2}\text{square unit} \end{gathered}$$

Therefore, option (B) is the correct answer.