The magnitude of vectors $\overrightarrow{OA}$, $\overrightarrow{OB}$ and $\overrightarrow{OC}$ in the given figure are equal. The direction of $\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}$ with $x$-axis will be :-

(C) $\tan ^{-1} \frac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}$

(A) $\tan ^{-1} \frac{(\sqrt{3}-1+\sqrt{2})}{(1-\sqrt{3}+\sqrt{2})}$

(B) $\tan ^{-1} \frac{(1+\sqrt{3}-\sqrt{2})}{(1-\sqrt{3}-\sqrt{2})}$

(D) $\tan ^{-1} \frac{(\sqrt{3}-1+\sqrt{2})}{(1+\sqrt{3}-\sqrt{2})}$

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Vector

The magnitude of vectors

O A

O B

and

O C

in the given figure are equal. The direction of

O A + O B - O C

with

x

-axis will be :-

Answer: (C)

Each vector has the same magnitude

a

, and the figure places

O A

3 0^{\circ}

above the

x

-axis,

O B

6 0^{\circ}

below it and

O C

4 5^{\circ}

above the negative

x

-axis:

O A = a (\frac{3}{2} \hat{i} + \frac{1}{2} \hat{j}), O B = a (\frac{1}{2} \hat{i} - \frac{3}{2} \hat{j}), O C = a (- \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j})

Adding the first two and subtracting the third,

O A + O B - O C = \frac{a}{2} [(1 + 3 + 2) \hat{i} + (1 - 3 - 2) \hat{j}]

The direction with the

x

-axis follows from

tan θ = \frac{1 - 3 - 2}{1 + 3 + 2}

so the resultant makes an angle

tan^{- 1} \frac{( 1 - 3 - 2 )}{( 1 + 3 + 2 )}

with the

x

-axis.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option