+145 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?
+145 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.
+128407 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
