Question

You are given a vector in the xy plane that has a magnitude of 87. 0 units and a y component of -66. 0 units. Determine the direction of a vector

Answer 1

You are given a vector in the XY plane that has a magnitude of 87. 0 units and a y component of -66. 0 units. The direction of the vector V is;

How to know the direction of a vector?

We know that the formula for 2 vectors like this in the x and y directions is; A = xi^ + yj^

Where A is the magnitude of the resultant

x is the value of the x-component

y is the value of the y-component

We are given;

Magnitude of vector = 84 units

Y-component of the vector = -67 units

Thus,

$A = \sqrt(x^2 + y^2)\\\\87 = \sqrt(x^2+ (-66)^2)\\\\87^2= x^2+ 4356\\\\7569 = x^2+ 4356\\\\x= \sqrt(7569 - 4356)\\\\x = 56.68 units$

From A above, let us take the positive value of the x-component and as such our original vector will be;

A = 56.68i^ - 66j^

The direction of the vector V is;

$\theta = tan^{-1} \dfrac{y}{x} \\\\\theta = tan^{-1} \dfrac{-66}{56.68} \\\theta =-27.15°$

Since it points entirely to the negative x-axis, then the angle is;

180 - (-27.15) = 207.15°

Learn more about vectors;

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