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A vector, starting at the origin and going to point (-2,0):
Since there is no y-component, the magnitude is the absolute value of the x component
magnitude = 2
magnitude of a vector = sqrt( X2 + Y2) = sqrt ((-2)2 + 02) = sqrt(4) = 2
where X & Y are the x-component & y-component of the vector.
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How can you find a unit vector in the same direction as the given vector?
Divide the vector by it's length (magnitude).
Why is it necessary to use the trig function cosine when working with vectors?
The cosine function is used to determine the x component of the vector. The sine function is used to determine the y component. Consider a vector drawn on an x-y plane with its initial point at (0,0). If L is the magnitude of the vector and theta is the angle from the positive x axis to the vector, then the x component of the vector is L * cos(theta) and the y component is L * sin(theta).
Prove that a constant vector always has a perpendicular derivative?
A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular
How do you solve equation by completing the square p2-2p-3 equals 0?
p^2 - 2p - 3 = 0 p^2 - 2p = 3 halve the linear term (-2), square it and add it to both sides p^2 - 2p + 1 = 3 + 1 factor left, gather terms right (p - 1)^2 = 4 (p - 1)^2 - 4 = 0 (1,-4) are the vector points of this parabola You can factor to solve for p (p + 1)(P - 3) p = - 1 p = 3
Solve 2x squared - 6x plus 4 equals 0?
2x2 - 6x + 4 = 0 2(x2 - 3x + 2) = 0 2(x - 1)(x - 2) = 0 x - 1 = 0 and x - 2 = 0 So, x = 1 and x = 2.
