Question

The magnitude of vectors OA, OB and OC in figure below are equal .Find the direction of OA+OB-OC

Answer 1

Answer:

I Hope it is helpful

Answer 2

The direction of OA + OB - OC with the x-axis is equal to tan inverse (1-(√3+√2)) / (1+(√3+√2))

Given,

The magnitude of vectors OA, OB, and OC are equal

⇒ A = B = C

To Find,

the direction of OA + OB - OC

Solution,

An entity with both magnitude and direction is referred to be a vector. A vector can be visualized geometrically as a directed line segment, with an arrow pointing in the direction and a length equal to the magnitude of the vector. The vector points in a direction from its tail to its head.

We have been given 3 vectors OA, OB, and OC.

By observing the given figure, we can represent these vectors as follows:

vector OA = A cos(30°) i + A sin(30°) j = (√3/2A) i + (1/2A) j

vector OB = B cos(60°) i + -B sin(60°) j = (1/2B) i - (√3/2B) j

vector OC = -C cos(45°) i + C sin(45°) j = -(1/√2)C i + (1/√2)C j

We have been given that the magnitudes of OA, OB, and OC are equal

⇒ A = B = C

Therefore, vector OA + OB - OC could be expressed as the following:

⇒ OA + OB - OC = (√3/2)A i + (1/2)A j + (1/2)B i - (√3/2)B j - (-(1/√2)C i + (1/√2)C j)

⇒ OA + OB - OC = (√3/2)A i + (1/2)A j + (1/2)A i - (√3/2)A j + (1/√2)A i - (1/√2)A j

⇒ OA + OB - OC = A (√3/2 + 1/2 + 1/√2) i + A (1/2 - √3/2 - 1/√2) j

Now, the direction of this vector is θ°.

tan(θ°) could be found using the following formula:

$\implies tan\theta = \frac{y}{x}\\\\\implies tan\theta = \frac{(1 - \sqrt3 - \sqrt2)/2}{(\sqrt3 + 1 + \sqrt2)/2}\\\\\implies tan\theta = \frac{1 - \sqrt3 - \sqrt2}{\sqrt3 + 1 + \sqrt2}\\\\\implies tan\theta = \frac{1 - (\sqrt3 + \sqrt2)}{1 + (\sqrt3 + \sqrt2)}$

Therefore, the angle θ with the x-axis could be found as:

$\theta = tan^{-1}(\frac{1-(\sqrt3 + \sqrt2)}{1-(\sqrt3 + \sqrt2)})$

the direction of OA + OB - OC with x axis = tan inverse (1-(√3+√2))/(1+(√3+√2))

#SPJ3