# Thread: Missing apex on a parallelogram

1. ## 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.

2. Hello, Undefdisfigure!

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

(a) Find point $D$ such that $ABCD$ is a parallelogram where vertex $D$ is opposite $A.$
We would find vector $AB$ like this . . .

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

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

We would find vector CD like this:

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

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

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

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

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

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

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