Given that P = (-4, -5) and Q = (2,3), express →PQ in the form (k,θ). where k is the magnitude and θ the bearing.
The correct answer is A. (10 units, 053º)
|âPQ| = â[(2-(-4))\(^2\) + (3-(-5))\(^2\)]
|âPQ| = â\(6^2 + 8^2\)
|âPQ| = â100
|âPQ| = 10units
tanθ = \(\frac{3--5}{2--4}\)
tanθ = \(\frac{4}{3}\)
θ = \(tan^{-1}\frac{4}{3}\)
= 53º
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