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Sagot :
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,
[tex]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[/tex]
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;
[tex]\theta = tan^{-1} \dfrac{y}{x} \\\\\theta = tan^{-1} \dfrac{-66}{56.68} \\\theta =-27.15°[/tex]
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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