S6.12.4.35 Find the volume of the parallelepiped

[karush]

$\tiny{s6.12.4.35}$

$\textrm{Find the volume of theparallelepiped with adjacent edges
$PQ, PR$, and $PS$.}$

\begin{align*}\displaystyle
P(2,0,-1),& \, Q(4,1,0), \, R(3.-1.1), S(2,-2,2)\\
\end{align*}
ok I know the problem is basically solved with

\begin{align*}\displaystyle
V&=\left|a\cdot(b\times c)\right|
\end{align*}

but was ? about the edges thing!

 

[Prove It]

$\tiny{s6.12.4.35}$

$\textrm{Find the volume of theparallelepiped with adjacent edges
$PQ, PR$, and $PS$.}$

\begin{align*}\displaystyle
P(2,0,-1),& \, Q(4,1,0), \, R(3.-1.1), S(2,-2,2)\\
\end{align*}
ok I know the problem is basically solved with

\begin{align*}\displaystyle
V&=\left|a\cdot(b\times c)\right|
\end{align*}

but was ? about the edges thing!

All that means is that the three edges touch each other...

 

[karush]

All that means is that the three edges touch each other...

I'm trying to plot this on:

https://technology.cpm.org/general/3dgraph/

but can't see to figure it

 

[HallsofIvy]

Yes, you can't graph the edges because you don't know what the edges are!

PQ is the line segment from P(2, 0, -1) to Q(4, 1, 0). That line can be written, parametrically, as x= (4- 2)t+ 2= 2t+ 2, y= (1- 0)t+ 0= t, and z= (0- (-1))t- 1= t- 1. That way, when t= 0, x= 2, y= 0, and z= -1 and, if t= 1, x= 4, y= -1, and z= 0.

It has "direction vector", which is the "a" in your formula, (4- 2)i+ (1- 0)j+ (0-(-1))k= 2i+ j+ k.

 

[skeeter]

$\vec{PQ} = 2i+j+k$

$\vec{PR} = i-j+2k$

$\vec{PS} = -2j+3k$

\(\displaystyle \vec{PQ} \times \vec{PR}=\begin{vmatrix}
i & j & k\\
2 & 1 & 1\\
1 & -1 & 2
\end{vmatrix}=3i-3j-3k\)

$|\vec{PS} \cdot (\vec{PQ} \times \vec{PR})| = |0 + 6 - 9| = 3$

graph shows edges translated to the origin ...