# Missing apex on a parallelogram

• Apr 27th 2009, 02:23 PM
Undefdisfigure
Missing apex on a parallelogram
Let A=(1,0,-2), B=(-1,1,2),C=(3,4,0)

(a) Find the point D such that ABCD is a parallelogram where the apex D is opposite to A.

(b) Write parametric equations and symmetric equations of the line L passing through D and parallel to the segment BC.
• Apr 27th 2009, 03:07 PM
Soroban
Hello, Undefdisfigure!

Quote:

Let: .$\displaystyle A=(1,0,\text{-}2),\;\; B=(\text{-}1,1,2),\;\;C=(3,4,0)$

(a) Find point $\displaystyle D$ such that $\displaystyle ABCD$ is a parallelogram where vertex $\displaystyle D$ is opposite $\displaystyle A.$

We would find vector $\displaystyle AB$ like this . . .

. . $\displaystyle \overrightarrow{AB} \;=\;\underbrace{(\text{-}1,1,2)}_B - \underbrace{(1,0,\text{-}2)}_A \;=\;\langle\text{-}2,1,4\rangle$

Let $\displaystyle D \,=\,(x,y,z)$
We know that: .$\displaystyle \overrightarrow{AB} \,=\,\overrightarrow{CD}$

We would find vector CD like this:

. . $\displaystyle CD \;=\;(x,y,z) - (3,4,0) \:=\:\langle \text{-}2,1,4\rangle \quad\Rightarrow\quad \langle x-3,y-4,z-0\rangle \:=\:\langle \text{-}2,1,4\rangle$

So we have: .$\displaystyle \begin{array}{ccc}x-3 \:=\:\text{-}2 & \Rightarrow & x \:=\:1 \\ y-4\:=\:1 & \Rightarrow & y \:=\:5 \\ z - 0 \:=\:4 & \Rightarrow & z \:=\:4 \end{array}$

Therefore: .$\displaystyle D(1,5,4)$

Quote:

(b) Write parametric equations and symmetric equations of the line $\displaystyle L$
passing through $\displaystyle D$ and parallel to the segment $\displaystyle BC.$

The direction of $\displaystyle \overrightarrow{BC}$ is: .$\displaystyle \vec v \;=\;(3,4,0) - (\text{-}1,1,2) \;=\;\langle 4,3,\text{-}2\rangle$

The line through $\displaystyle D(1,5,4)$ with direction $\displaystyle \vec v \:=\:\langle 4,3,\text{-}2\rangle$ has equations:

. . . . $\displaystyle \begin{array}{ccc}x &=& 1 + 4t \\ y &=& 5 + 3t \\ z &=& 4 - 2t \end{array}$ . . and . . $\displaystyle \frac{x-1}{4} \:=\:\frac{y-5}{3} \:=\:\frac{z-4}{-2}$