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Vector Algebra Part 1

Section: Mathematics

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Question : 13 of 100

Marks: +1, -0

Let a vector

\alpha \hat{i} + \beta \hat{j}

be obtained by rotating the vector

\sqrt{3} \hat{i} + \hat{j}

45^{\circ}

about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices

(\alpha, \beta), (0, \beta)

(0,0)

[16 Mar 2021 Shift 1]

$\; \frac{1}{2}$
1
$\; \frac{1}{\sqrt{2}}$
$2 \sqrt{2}$

Solution:

Let OA be

\sqrt{3} \hat{i} + \hat{j}

and

O B

be

\alpha \hat{i} + \beta \hat{j}

.

As, we can notice in

O A, \; \frac{1}{\sqrt{3}} = \tan 30^{\circ}

. So, it makes an angle of

30^{\circ}

with the

X

-axis.

Now, when

O A

is rotated further by

45^{\circ}

anticlockwise, the resultant vector

O B

makes an angle of

75^{\circ}

with the

X

-axis.

So, OB

= |OA| (\cos 75^{\circ} \hat{i} + \sin 75^{\circ} \hat{j})

Let

riangle O B C

be the required triangle whose area we have to determine.

Area of

\triangle O B C = \frac{1}{2} \times (

Base

) imes (

Height

)

= \frac{1}{2} \times \beta \times \alpha

= \frac{1}{2} (2 \sin 75^{\circ}) (2 \cos 75^{\circ}) = 2 \sin 75^{\circ} \cos 75^{\circ}

= \sin 150^{\circ} = \sin 30^{\circ}

= \frac{1}{2}

Hence, the area is

rac{1}{2} ext{sq}

. unit.

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