MHB S6.793.12.4.33 Find the volume of the parallelepiped

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The volume of the parallelepiped formed by vectors a, b, and c is calculated using the scalar triple product, resulting in a volume of 82. The calculation involves finding the determinant of a matrix composed of the vectors. Participants discussed minor errors in transcription and suggested improvements for the LaTeX editing experience. There were also comments about the usability of the editing window on different devices, particularly tablets. Overall, the discussion focused on confirming the volume calculation and enhancing the user interface for better usability.
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$\tiny{s6.793.12.4.33}$
$\textsf{
Find the volume of the parallelepiped determined by the vectors, a b and c}$
$ a =\langle 6, 3, -1\rangle
\, b =\langle 0, 1, 2 \rangle
\, c =\langle 4, -2, 5 \rangle $

$\textsf{The volumn of the parallelepiped determined by the vectors }\\$
$\textsf{ $a, b$ and $c$ is the magnitude of their scalar triple product.}$

\begin{align}
\displaystyle
V&=|a \cdot(b \times c)|\\
\end{align}
then
\begin{align}
V=|a \cdot(b \times c)|&=
\begin{bmatrix}
6 & 3 & -1\\
0 &1 &2\\
4 &-2 &5
\end{bmatrix} \\
&=6\begin{bmatrix}=
1 &2\\
-2 &5
\end{bmatrix}
+3\begin{bmatrix}
0 &2\\
4 &5
\end{bmatrix}
-\begin{bmatrix}
0 &1\\
4 &-2
\end{bmatrix}\\
&= 6(9)-3(-8) +(4) \\
&\color{red}{V=82}
\end{align}
$\textit{ok think this is ok. but always suggestions! }\\$
$\textit{btw need more lines to expand to in latex window scrolling constantly not fun}$😰
 
Last edited:
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Looks good to me. A few errors in transferring your notes to here, but i think you have understood the process.

btw need more lines to expand to in latex window scrolling constantly not funbtw need more lines to expand to in latex window scrolling constantly not fun

I agree! There is a little tab down the bottom right of the text window that you can pull to increase the size.

EDIT: Ohh that window... I didn't even know it was there :). I usually just type in the big window and then hit preview to check my LaTeX.
 
\begin{align}
V=|a \cdot(b \times c)|&=
\begin{bmatrix}
6 & 3 & -1\\
0 &1 &2\\
4 &-2 &5
\end{bmatrix} \\
&=6\begin{bmatrix}
1 &2\\
-2 &5
\end{bmatrix}
+3\begin{bmatrix}
0 &2\\
4 &5
\end{bmatrix}
-\begin{bmatrix}
0 &1\\
4 &-2
\end{bmatrix}\\
&=6(9)-3(-8) +(4) \\
&=\color{red}{82}
\end{align}
$\textit{think this is it. well the expand tab doesn't work on a tablet! }$
 
Looks better :)

karush said:
$\textit{think this is it. well the expand tab doesn't work on a tablet! }$

True.. I just tried on my iPad and it doesn't work. I guess we'll just have to be more concise with our posts xD.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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