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Points A, B, C, and T are in space such that each of \(\overline{TA}\)\(\overline{TB}\), and \(\overline{TC}\) is perpendicular to the other two. If \(TA = TB = 12\) and \(TC = 6\),then what is the distance from T to face ABC?

 Jun 22, 2019
 #1
avatar+4325 
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We have the Vectors: \(AB=(-12, 12, 0), AB=<-12, 12,0>\) and \(BC=( 0, -12, 6), BC=<0, -12, 6>.\)

 

Using the cross-product, and the equation of the plane, we get(heft calculations), but this simplifies to \(\frac{12\sqrt{6}}{6}=\boxed{2\sqrt{6}}.\)

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 Jun 22, 2019
 #2
avatar+7725 
+1

Consider the volume of the triangular pyramid TABC.

\(\dfrac{6\cdot 12^2}{2\cdot 3} = \dfrac{\text{Area of }\triangle ABC\cdot \text{Distance from T to }\triangle ABC}{3}\\ \text{Distance} = \dfrac{432}{\text{Area of }\triangle ABC}\\ \text{Half perimeter of }\triangle ABC = \dfrac{2\sqrt{6^2+12^2} + \sqrt{12^2+12^2}}{2} = 6(\sqrt 5 + \sqrt 2)\\ \text{Area} = \sqrt{(6(\sqrt5+\sqrt2))\cdot (6(\sqrt5+\sqrt2) - 6\sqrt 5)^2 \cdot (6(\sqrt5+\sqrt2) - 12\sqrt 2)} = 36\sqrt6 \text{ unit}^2\\ \text{Distance} = \dfrac{432}{36\sqrt6} = 2\sqrt6 \text{ unit}\)

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 Jun 22, 2019

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