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MECE 2230U
Q1 Given; P = (−160𝑖̂+ 40𝑗̂+ 60𝑘̂)N Distance (d) between point O and A is 8m. a) Magnitude of vector P; |P| = √(−160) 2 + 402 + 602 |P| = 175.499 N From the direction cosines; P⃗ = Pcos𝜃𝑥𝑖̂+ Pcos𝜃𝑦𝑗̂+ Pcos𝜃𝑧𝑘̂ Pcos𝜃𝑥 = −160 cos𝜃𝑥 = − 160 175.499 = −0.912 𝜃𝑥 = 155.74° 𝑜𝑟 24.26° Pcos𝜃𝑦 = 40 cos𝜃𝑦 = 40 175.499 = 0.228 𝜃𝑦 = 76.83
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