The magnitude of vectors `vec(OA), vec(OB) and vec (OC)` in figure are equal. Find the direction of `vec(OA)+vec(OB)-vec(OC)`.
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Options

A. ` tan^(-1) ((2 - sqrt 3 - 1)/(2 + sqrt 3 + 1))`
B. ` tan^(-1) ((1 - sqrt 3 +1)/(1 + sqrt 3 + sqrt 2))`
C. `tan ^(-1) (()1- sqrt 3 - sqrt2))/(1 + sqrt3 - sqrt 2))`
D. ` tan ^(-1) ((1 + sqrt 3 + sqrt 2)/(1- sqrt 3 + sqrt 2)`

Answer

B

Explanation

Let ` OA+ OB = OC = P`
Componet fo ` vec (OA)` along x-axis
` = P cos 30^@ = (P sqrt 3)/2 `
Component of ` vec (OA)` along y-xis `=P sin 30^@= P/2`
Component of ` vec (OB)` along x-axis`=P cos 60^@= P/2`.
Component of `OB` along y-axis`
`=- P sin 60^2 =( -P sqrt 3)/2 `
Component of ` vec (OP)` along x-axis
` =- P sin 45^@= (-P)/(sqrt 2)`
Component of ` vec (OC)` along y-axis
` P= cos 45^@ = P/(sqrt 2)`
:. x-component of ` vec (OA) + vec (OB) - vec (OC)`,
` | vec P_x| = (Psqrt 3)/2 + P/2 - ( (-P)/(sqrt 2)) = P/2 (sqrt 3 + 1 + sqrt2)`
y-component of ` vec (OA) +vec (OB) - vec (OC)`,
` | vec P_y| = P/2+ ((-P sqrt 3)/2) - P/(sqrt 2) = P/2 (-1 sqrt 2)`
Angle of ` vec (OA) + vec (OB)- vec (OC)` with x-axis is
` theta = tan^(-1) (P_y//P_x)= tan^(-1) [(P/2(1 - sqrt 3 - sqrt 2))/(P/2 (1+ sqrt + sqrt 2))]`
`= tan^(-1)` ((1-sqrt3 - sqrt 2)/(1+ sqrt3 + sqrt 2))`.