# Thread: Hard vector parallelepped question

1. ## Hard vector parallelepped question

Three points A, B and C have coordinates (2,1,-2), (2,-1,-1) and (1,2,2) respectively. The vectors OA, OB and OC where O is theorigin form three concurrent edges of a parallelepiped OAPBCQSR in the diagram:

http://i45.tinypic.com/1ordkp.jpg

1. Find the coordinates of P, Q, R and S
2. Fina an equation for the plan OAPB
3. calculate the volume, B or the parallelepiped

Any help appreciated, im not sure where to start for Q1 and Q2

Thanks

2. Hello Jampop
Originally Posted by Jampop
Three points A, B and C have coordinates (2,1,-2), (2,-1,-1) and (1,2,2) respectively. The vectors OA, OB and OC where O is theorigin form three concurrent edges of a parallelepiped OAPBCQSR in the diagram:

http://i45.tinypic.com/1ordkp.jpg

1. Find the coordinates of P, Q, R and S
2. Fina an equation for the plan OAPB
3. calculate the volume, B or the parallelepiped

Any help appreciated, im not sure where to start for Q1 and Q2

Thanks
1. $\displaystyle \vec{OP} = \vec{OA} + \vec{AP}$
$\displaystyle =\vec{OA}+ \vec{OB}$

$\displaystyle =\begin{pmatrix}2\\1\\-2\end{pmatrix}+\begin{pmatrix}2\\-1\\-1\end{pmatrix}$

$\displaystyle =\begin{pmatrix}4\\0\\-3\end{pmatrix}$
So $\displaystyle P$ is the point $\displaystyle (4,0,-3)$.

Find the coordinates of $\displaystyle Q, R, S$ in a similar way.

2. The plane passes through the origin, so its equation is of the form:
$\displaystyle ax + by + cz = 0$
We may divide through by $\displaystyle a$, to obtain an equation of the form:
$\displaystyle x + py + qz =0$
The point $\displaystyle A$ lies in the plane. So:
$\displaystyle 2+p-2q=0$
Similarly for $\displaystyle B$. Solve for $\displaystyle p$ and $\displaystyle q$.

3. Use the triple scalar product to find the volume of the parallelopiped:
Volume $\displaystyle = |\vec{OA}.(\vec{OB} \times \vec{OC})|$
which is most easily calculated using the 3 x 3 determinant form. For reference see here.

3. Ohh i understand ! Thanks very much