Originally Posted by

**tttcomrader** Given a tetrahedron ABCD with A = (1,1,1), B = (3,2,1), C = (0,1,2) , D = (3,2,2)

a) Find volume of ABCD:

I have $\displaystyle V= \frac {|(A-D) \cdot (B-D) \times (C-D)|}{6} $

$\displaystyle = \frac {|(-2,-1,-1) \cdot (0,0,-1) \times (-3,-1,0)|} {6} = \frac {|-1|}{6} = \frac {1}{6}$

Is that correct?

b) Find distance between A and face BCD:

Face BCD is $\displaystyle ( \bar {x} - D) \cdot (B-D) \times (C-D) = -x-3y+9=0$

But I can't remember the distance formula, any help?

I think is this one:

dist(A,BCD) = $\displaystyle | \frac {-(1)-3(1)+9}{||(-1,-3,0)||} | = \frac {4}{10^{1/2}} $

Sorry, we don't have a textbook for this course, and the professor didn't have the formula clearly, I tried my best to take the notes but I just can't follow him.

c) Find the distance between the lines AC and BD:

AC = (1,1,1)+t(-1,0,1) , BD = (3,2,1)+s(0,0,1)

dist(AC,BD) = $\displaystyle | \frac {(B-A) \cdot (0,0,1) \times (-1,0,1)}{(0,0,1) \times (-1,0,1)}| = \frac {|(2,1,0) \cdot (0,-1,0)|}{|(0,-1,0)|} = 1 $

Is that right?