A) 45
B) 75
C) 20
D) 10
Correct Answer: A
Solution :
Key Idea: Power is equal to the scalar product of force with velocity. Power of the engine, \[P=\vec{F}\,.\,\vec{v}......(i)\] \[Given,\vec{F}=(20\hat{i}-3\hat{j}+5\hat{k})\,N\] \[\vec{v}=(6\hat{i}+20\hat{j}-3\hat{k})\,m/s\] Thus, after substituting for \[\vec{F}\,and\,\vec{v}\]in Eq. (i), it becomes, \[P=(20\hat{i}-3\hat{j}+5\hat{k})\,.\,(6\hat{i}+20\hat{j}-3\hat{k})\] \[=(20\times 6)\,(\hat{i}\,.\,\hat{i})+(-3\times 20)\,(\hat{j}\,.\,\hat{j})\,\]\[+(5\,x-3)\,(\hat{k}.\,\hat{k})\] \[=120-60-15\] = 45 Note: In the simplification for power, the dot product of a unit vector with same unit vector give 1. The dot product of a unit vector with its orthogonal gives zero. Thus, \[\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\] \[\hat{i}\cdot \hat{j}=\hat{i}\cdot \hat{k}=\hat{j}\cdot \hat{k}=0\] So, in above simplification second type of dot products are not shown.
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