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Represent the 400N Force in Cartesian vector format. Use x=20m, y=15m.



D (0,0,24)m
A (20,15,0)m

position vector (from D to A) rDA = [20i+15j-24k]m

magnitude of rDA = length of DA = distance btwn DA = sqrt(202+152 + (24)2)sqrt\left(20^{2}+15^{2}\ +\ \left(-24\right)^{2}\right) = 34.7

uDA = position vector rDA / magnitude rDA = 0.576i + 0.432j - 0.962k


F400 = 400 uDA = [231i + 173j - 277k]N


F400 = <230.8,173.1,-277>N
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Determine the magnitude and coordinate direction angles of the resultant force