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

Mar 4, 2019

#1
+147
+2

Please help me soon. I don't know how to find the shortest distance. Would it be the line from T to the 'center' of ABC? If it is, I don't know how to find the center of ABC.

Mar 4, 2019
#2
+147
+2

I got 3sqrt5 but I think that is wrong.

Mar 4, 2019
#3
+106532
+3

This is a two-part problem

Let T  (0, 0, 0)      A= (12, 0, 0)   B = (0, 12, 0)     C =  (0, 0, 6)

The first thing we need to do is to find an equation for the plane containing  A, B and C

We can form vectors   AB =  (-12, 12, 0)    and BC = ( 0, -12, 6)

Now....we can find the normal vector to this plane by taking the cross-product of these two vectors

n =  AB x BC  =        i           j         k         i         j

-12       12      0        -12    12

0         -12     6         0       -12

[ 72i + 0j + 144k ]  - [ 0k + 0i  - 72j]  =

72i + 72j  + 144k

Using  C, the equation of the plane is

72(x - 0)^2 + 72(y - 0)^2 + 144(z - 6)^2  = 0     simplify

72x^2   + 72y^2 + 144z  - 864 =  0

The distance  from T to the face  is given by :

l 72(0) + 72(0) +   144(0) - 864  l                      864           2^5 * 3^3

__________________________      =            _____  =  _________    =

√ [ 72^2 + 72^2 +144^2 ]                               √31104      √[2^7*3^5]

2^5 * 3^3                          4*3           12√6

_______________  =  ______  =  _______  =    2√6  units

2^3 * 3^2 √ [2* 3]            √6                6

Mar 4, 2019
#4
+147
+2

Thank you!

DanielCai  Mar 4, 2019