Let a vector \alpha \widehat i + \beta \widehat j be obtained by rotating the vector \sqrt 3 \widehat i + \widehat 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 :

Vector Algebra PYQ — JEE Main 2021 · 16 Mar · Shift 1 · Q23 with Solution

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Correct answer: B

The original vector is $\vec v=\sqrt{3}\,\hat i+\hat j,$ so its coordinate form is $(\sqrt{3},1)$ .

We rotate this vector by $45^\circ$ counterclockwise.

The rotation formula for a point $(x,y)$ by angle $\theta$ is: $x'=x\cos\theta-y\sin\theta,$ $y'=x\sin\theta+y\cos\theta.$

Here, $x=\sqrt{3},\quad y=1,\quad \theta=45^\circ,$ and $\cos 45^\circ=\sin 45^\circ=\frac{1}{\sqrt2}.$

So,

=\frac{\sqrt3-1}{\sqrt2},$$ $$\beta=\sqrt3\cdot \frac{1}{\sqrt2}+1\cdot \frac{1}{\sqrt2} =\frac{\sqrt3+1}{\sqrt2}.$$ 2. **Area of the triangle** The triangle has vertices: $$(\alpha,\beta),\quad (0,\beta),\quad (0,0).$$ This is a right triangle with: - base = distance from $(0,\beta)$ to $(0,0)$ = $\beta$ - height = distance from $(0,\beta)$ to $(\alpha,\beta)$ = $\alpha$ Therefore, $$\text{Area}=\frac12\alpha\beta.$$ Now compute $\alpha\beta$: $$\alpha\beta=\frac{\sqrt3-1}{\sqrt2}\cdot \frac{\sqrt3+1}{\sqrt2} =\frac{(\sqrt3-1)(\sqrt3+1)}{2} =\frac{3-1}{2}=1.$$ Hence, $$\text{Area}=\frac12\cdot 1=\frac12.$$ 3. **Correct option** $$\boxed{\frac12}$$ So the correct option is **B**.

View written solutionFree

Correct answer: B

The original vector is $\vec v=\sqrt{3}\,\hat i+\hat j,$ so its coordinate form is $(\sqrt{3},1)$ .

We rotate this vector by $45^\circ$ counterclockwise.

The rotation formula for a point $(x,y)$ by angle $\theta$ is: $x'=x\cos\theta-y\sin\theta,$ $y'=x\sin\theta+y\cos\theta.$

Here, $x=\sqrt{3},\quad y=1,\quad \theta=45^\circ,$ and $\cos 45^\circ=\sin 45^\circ=\frac{1}{\sqrt2}.$

So,

=\frac{\sqrt3-1}{\sqrt2},$$ $$\beta=\sqrt3\cdot \frac{1}{\sqrt2}+1\cdot \frac{1}{\sqrt2} =\frac{\sqrt3+1}{\sqrt2}.$$ 2. **Area of the triangle** The triangle has vertices: $$(\alpha,\beta),\quad (0,\beta),\quad (0,0).$$ This is a right triangle with: - base = distance from $(0,\beta)$ to $(0,0)$ = $\beta$ - height = distance from $(0,\beta)$ to $(\alpha,\beta)$ = $\alpha$ Therefore, $$\text{Area}=\frac12\alpha\beta.$$ Now compute $\alpha\beta$: $$\alpha\beta=\frac{\sqrt3-1}{\sqrt2}\cdot \frac{\sqrt3+1}{\sqrt2} =\frac{(\sqrt3-1)(\sqrt3+1)}{2} =\frac{3-1}{2}=1.$$ Hence, $$\text{Area}=\frac12\cdot 1=\frac12.$$ 3. **Correct option** $$\boxed{\frac12}$$ So the correct option is **B**.