Question

The magnitude of vectors `vec(OA)`, `vec(OB)`, and `vec(OC)` in the given figure are equal. The direction of `vec(OA) + vec(OB) - vec(OC)` with the x-axis will be ______.

पर्याय

• `tan^-1  ((1 + sqrt3 - sqrt2))/((1 - sqrt3 - sqrt2)`

• `tan^-1  ((sqrt3 - 1 + sqrt2))/((1 - sqrt3 + sqrt2)`

• `tan^-1  ((sqrt3 - 1 + sqrt2))/((1 + sqrt3 - sqrt2)`

• `tan^-1  ((1 - sqrt3 - sqrt2))/((1 + sqrt3 + sqrt2)`

Answer 1

The magnitude of vectors `vec(OA)`, `vec(OB)`, and `vec(OC)` in the given figure are equal. The direction of `vec(OA) + vec(OB) -vec(OC)` with the x-axis will be `tan^-1  ((1 - sqrt3 - sqrt2))/((1 + sqrt3 + sqrt2)`.

Explanation:

We must determine the direction of

`vecR = vec(OA) + vec(OB) - vec(OC)`

= `vec(OA) + vec(OB) + (-vec(OC))`

Reverse the direction of `vec(OC)` to get `-vec(OC)`

Now, find `sumA_x` and `sumA_y`

`vecR = sumR_xhati + sumR_yhatj`

Direction, `theta = tan^-1  (sumR_y)/(sumR_x)`

addition of each's x-component,

`sumR_x = (1. cos30^circ + 1. cos45^circ + 1. cos60^circ)hati`

`sumR_y = (1. sin30^circ - 1. sin45^circ - 1. sin60^circ)hatj`

We considered the unit length of each value here.

`sumR_x = (sqrt3/2 + 1/sqrt2 + 1/2)hati`

`sumR_y = (1/2 - 1/sqrt2 - sqrt3/2)hatj`

`tantheta = (sumR_y)/(sumR_x) = (((sqrt2 - 2 - sqrt6)/(2sqrt2)))/(((sqrt6 + 2 + sqrt2)/(2sqrt2))`

= `(1 - sqrt2 - sqrt3)/((sqrt3 + sqrt2 + 1)`

= `tan^-1  ((1 - sqrt3 - sqrt2))/((1 + sqrt3 + sqrt2)`