C(-14,0, –4), and D(-7,7,-4). See the following figure. Now, the vector n =AB x ÃD is perpendicular to the surface of the solar| panels. \text { Assume } \mathrm{s}=\frac{1}{\sqrt{3}} \hat{\imath}+\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k} \text { points towards the sun at a particular time } of the day and the flow of solar energy is F = 900s.
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