The magnitude of vectors OA OB, and OC in the given figure are equal. The direction of OA + OB - OC with x-axis will be: (a) tan ⁻¹((1-√(3)-√(2)))/((1+√(3)+√(2))) (b) tan ⁻¹((√(3)-1+√(2)))/((1+√(3)-√(2))) (c) tan ⁻¹((√(3)-1+√(2)))/((1-√(3)+√(2))) (d) tan ⁻¹((1+√(3)-√(2)))/((1-√(3)-√(2)))

Answer: (a). Take each magnitude as 1. OA = √(3)/2i + 1/2j, OB = 1/2i - √(3)/2j, OC = -1/√(2)i + 1/√(2)j x: √(3)/2 + 1/2 + 1/√(2) = (1 + √(3) + √(2))/(2) y: 1/2 - √(3)/2 - 1/√(2) = (1 - √(3) - √(2))/(2) tan θ = y/x = (1 - √(3) - √(2))/(1 + √(3) + √(2)) So the answer is (a).

The magnitude of vectors OA OB, and OC in the given figure are equal… | JEE Main 2021 (26 Aug, Shift 1)

Q15JEE Main 2021 · 26 AUG · SHIFT 1Medium

Vector Basics, Unit Vectors & Components

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:

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Answer: (a)

Take each magnitude as 1. $\vec{OA} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$ , $\vec{OB} = \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}$ , $\vec{OC} = -\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$
$x$ : $\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}} = \frac{1 + \sqrt{3} + \sqrt{2}}{2}$
$y$ : $\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} = \frac{1 - \sqrt{3} - \sqrt{2}}{2}$
$\tan\theta = \frac{y}{x} = \frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}}$
So the answer is (a).

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