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volume of a tetrahedron using vectors

Hi,

I was told that:

A(−1,−1,1),B(4,2,1),C(2,1,0)A(-1, -1, 1), B(4,2,1), C(2,1,0) and D(1,2,3)D(1,2,3)

I'm asked to find the volume of the tetrahedron ABCD. I did this by solving:

∣16AD⃗.(AC⃗×AB⃗)∣|\frac{1}{6} \vec{AD}.(\vec{AC} \times \vec{AB})|

My resulting volume was 76\frac{7}{6}

However, the mark scheme I was looking at did:

∣16AD⃗.(AB⃗×AC⃗)∣|\frac{1}{6} \vec{AD}.(\vec{AB} \times \vec{AC})|

and got the volume to be 116\frac{11}{6}

I thought that the triple scalar product was commutative? Why are these different?

If anyone wants to see my working to make sure I didn't make a calculation error I'd be happy to supply it :smile:

Thank you!
Reply 1
Avatar for seohyun
seohyun
8
I worked it out your way and got 11/6
My teacher said that it should work as long as all vectors have the same origin, in this case "A"
(edited 10 years ago)
Original post by so it goes

If anyone wants to see my working to make sure I didn't make a calculation error I'd be happy to supply it :smile:

Thank you!


I got 11/6 by your method.

I suspect a sign error on the k component of the cross product.
Reply 3
Avatar for so it goes
so it goes
OP
9
Original post by ghostwalker
I got 11/6 by your method.

I suspect a sign error on the k component of the cross product.


Thanks :smile: that's reassuring. I'll show you what I did, I made my j component negative, but not my k component when doing the cross product, is that wrong?

A⃗D=(232)\vec AD = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix}

A⃗C=(32−1)\vec AC = \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}

A⃗B=(530)\vec AB = \begin{pmatrix} 5 \\ 3 \\ 0 \end{pmatrix}

A⃗C×A⃗B=∣i‾j‾k‾32−1530∣.\vec AC \times \vec AB = \begin{vmatrix} \underline{i} & \underline{j} & \underline{k} \\ 3 & 2 & -1 \\ 5 & 3 & 0 \end{vmatrix} .

∴A⃗C×A⃗B=i‾∣2−130∣−j‾∣3−150∣+k‾∣3253∣.\therefore \vec AC \times \vec AB = \underline{i} \begin{vmatrix} 2 & -1 \\ 3 & 0 \end{vmatrix} - \underline{j}\begin{vmatrix} 3 & -1 \\ 5 & 0 \end{vmatrix} + \underline{k} \begin{vmatrix} 3 & 2 \\ 5 & 3 \end{vmatrix} .

∴A⃗C×A⃗B=(−35−1)\therefore \vec AC \times \vec AB = \begin{pmatrix} -3 \\ 5 \\ -1 \end{pmatrix}

∴A⃗D.(A⃗C×A⃗B)=(232).(−35−1)=7 \therefore \vec AD . (\vec AC \times \vec AB) = \begin{pmatrix} 2\\ 3 \\2 \end{pmatrix} . \begin{pmatrix} -3 \\ 5 \\ -1 \end{pmatrix} = 7
∴volume=76 \therefore volume = \frac{7}{6}

Can you see where I've gone wrong? Thanks :biggrin:
(edited 10 years ago)
Original post by so it goes


∴A⃗C×A⃗B=(−35−1)\therefore \vec AC \times \vec AB = \begin{pmatrix} -3 \\ 5 \\ -1 \end{pmatrix}


Can you see where I've gone wrong? Thanks :biggrin:


Ironically, the k component is correct, it's both the i,j components that are wrong. Should be:

A⃗C×A⃗B=(3−5−1)\vec AC \times \vec AB = \begin{pmatrix} 3 \\ -5 \\ -1 \end{pmatrix}

From the 3 determinants:

i component is (2)(0)−(−1)(3)(2)(0)-(-1)(3)

j component is −[(3)(0)−(−1)(5)]-[(3)(0)-(-1)(5)]
(edited 10 years ago)
Reply 5
Avatar for so it goes
so it goes
OP
9
Original post by ghostwalker
Ironically, the k component is correct, it's both the i,j components that are wrong. Should be:

A⃗C×A⃗B=(3−5−1)\vec AC \times \vec AB = \begin{pmatrix} 3 \\ -5 \\ -1 \end{pmatrix}

From the 3 determinants:

i component is (2)(0)−(−1)(3)(2)(0)-(-1)(3)

j component is −[(3)(0)−(−1)(5)]-[(3)(0)-(-1)(5)]


Ahh, I could kick myself :P thank you! I was doing them in my head, I'll make sure to write them out in full next time!
Thank you :biggrin:
Original post by so it goes
Thank you :biggrin:


You're welcome.

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