Volume of a Parallelepiped, Vectors

Question

I have designed three vectors that consist of: \begin{align} a &= -4i-10j-k\\ b &= 7i+9j-2k\\ c &= 3i+9j+4k\\ \end{align} And the task was to find the volume that was created from the above vectors when creating a parallelepiped. The current method I have used is modulus matrix of the scalar triple product ($a$.$b*c$) and also the distributive law. the results concluded from this was $88$ units cubed, but I was wondering whether there were any other methods I could use to verify my answer?

Answer 1

$$V=|-4\cdot9\cdot4+(-1)\cdot7\cdot9+3\cdot(-10)\cdot(-2)-3\cdot9\cdot(-1)-(-4)\cdot9\cdot(-2)-4\cdot7\cdot(-10)|=88$$ It's the best way, I think.

We can also just to prove this formula.

Id est, let $ABCDA'B'C'D'$ be our parallelepiped.

We can find $S_{ABCD}=AB\cdot AC\cdot\sin\measuredangle BAC$ and calculate an altitude $A'K$ to $ABCD$ of the parallelepiped.

By this way we calculate a module of the scalar triple product of our vectors.

Answer 2

To my knowledge, that is the best and easiest way of finding the volume of a parallelepiped. The area of the base is given by $b*c$. multiplying that by its height, $a$, we get the volume. Although, I suppose the product of the moduli of each vector might be an idea. I'm not quite sure.