d) The unit–vector notation of the two vectors a = 4.0i + 3.0j and b = −13.0i + 7.0j. Determine
the magnitude and direction of a + b [4mks
Sum of the vectors:
a+b=(4.0+(−13.0))i+(3.0+7.0)j=−9.0i+10.0ja + b = (4.0 + (-13.0))i + (3.0 + 7.0)j = -9.0i + 10.0ja+b=(4.0+(−13.0))i+(3.0+7.0)j=−9.0i+10.0j
Magnitude:
∣∣a+b∣∣=(−9.0)2+10.02≈13.45|| a + b || = \sqrt{(-9.0)^2 + 10.0^2} \approx 13.45∣∣a+b∣∣=(−9.0)2+10.02≈13.45
Direction:
tan(θ)=10.0−9.0θ=tan−1(10.0−9.0)≈132.0°\tan(\theta) = \cfrac{10.0}{-9.0} \\ \theta = \tan^{-1} (\cfrac{10.0}{-9.0}) \approx 132.0 \degreetan(θ)=−9.010.0θ=tan−1(−9.010.0)≈132.0°
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