Volume of a tetrahedron

**tttcomrader** · January 27th 2008, 09:38 AM

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

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

Is that correct?

b) Find distance between A and face BCD:

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

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

I think is this one:

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

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

Is that right?

**earboth** · January 27th 2008, 10:50 PM

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

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

Is that correct?

b) Find distance between A and face BCD:

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

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

I think is this one:

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

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

Is that right?

Hello,

to #a) and #c): In my opinion your results are OK.

to #b) I have got $p_{BCD} : -x+3y+3 = 0$ as the equation of the plane BCD.
The distance formula you use is correct. But with my equation I got a distance of $d(A, p_{BCD})=\frac12 \cdot \sqrt{10}$

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