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MiscII · Q148

Q.Find a unit vector perpendicular to the plane containing the points (a,0,0)(a,0,0), (0,b,0)(0,b,0), and (0,0,c)(0,0,c). What is the area of the triangle with these vertices?

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Let P=(a,0,0),Q=(0,b,0),R=(0,0,c)P=(a,0,0),Q=(0,b,0),R=(0,0,c). Form two sides from PP:

PQ→=(−a,b,0),PR→=(−a,0,c).\overrightarrow{PQ}=(-a,b,0),\qquad\overrightarrow{PR}=(-a,0,c).

PQ→×PR→=∣i^j^k^−ab0−a0c∣=i^(bc−0)−j^(−ac−0)+k^(0+ab)=bc i^+ac j^+ab k^.\overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix}\hat i&\hat j&\hat k\\-a&b&0\\-a&0&c\end{vmatrix} =\hat i(bc-0)-\hat j(-ac-0)+\hat k(0+ab)=bc\,\hat i+ac\,\hat j+ab\,\hat k.

Its magnitude is b2c2+a2c2+a2b2\sqrt{b^2c^2+a^2c^2+a^2b^2}. The unit vector perpendicular to the plane is

n^=bc i^+ac j^+ab k^a2b2+b2c2+c2a2.\hat n=\frac{bc\,\hat i+ac\,\hat j+ab\,\hat k}{\sqrt{a^2b^2+b^2c^2+c^2a^2}}.

The area of the triangle is …

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