Math Help - vector parallelogram

1. vector parallelogram

ABCD is a parallelogram. A, B and C are represented by the position vectors
i + 2j - k, 2i + j -2k and 4i - k respectively. Find:
a) A->D

2. Hi, have a look at the attached diagrams:

3. Hello, scorpion007!

$ABCD$ is a parallelogram.
$A, B$ and $C$ are represented by the position vectors:
. . $i + 2j - k,\;2i + j -2k\;4i - k$, respectively.

$(a)\text{ Find: }\overrightarrow{AD}$

We have: . $A = \langle1,2,\text{-}1\rangle,\;B = \langle2,1,\text{-}2\rangle,\;C = \langle4,0,\text{-}1\rangle,\;D = \langle x,y,z\rangle$

Then: . $\overrightarrow{BA} = \langle\text{-}1,\,1,
\,1\rangle$
and $\overrightarrow{CD} = \langle x-4,\,y-0,\,z+1\rangle$

Since $BA \parallel CD:\;\begin{array}{ccc}x - 4 \,=\,\text{-}1 \\ y\,=\,1 \\ z+1\,=\,1\end{array}$ . $\Rightarrow$ . $\begin{array}{ccc}x=3 \\ y=1 \\ z = 0\end{array}$

Therefore: . $D = \langle3,1,0\rangle\quad\Rightarrow\quad AD \:=\:\langle3-1,\,1-2,\,0-(\text{-}1)\rangle\:=\:\langle2,\text{-}1,1\rangle$

$(b)\text{ Find }\cos(\angle BAD)$

The angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by: . $\cos\theta \:=\:\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}| }$

We have: . $\begin{array}{cc}\overrightarrow{AB} = \langle1,\text{-}1,\text{-}1\rangle \\ \overrightarrow{AD} = \langle2,\text{-}1,1\rangle\end{array}$ . $\Rightarrow$ . $\begin{array}{cc}|AB| = \sqrt{3} \\ |AD| = \sqrt{6}\end{array}$

. . and: . $\overrightarrow{AB}\cdot\overrightarrow{AD} \:=\:(1)(2)+(\text{-}1)(\text{-}1)+(\text{-}1)(1) \:=\:2$

Therefore: . $\cos(\angle BAD) \;= \;\frac{2}{\sqrt{3}\sqrt{6}}\;=\;\frac{\sqrt{2}}{3 }$

. . $\left(\theta \:\approx\:61.87^o\right)$

4. thanks for the help guys

5. earboth, according to your diagram, isn't a + BC = d or OD? not AD as you have it written?
since AD || BC, and since its a parallelogram, shouldnt AD = BC? so why is AD = a + BC? Im thinking a + BC yields the vector from the origin to D, OD.

EDIT: i see [.math] notation is back.

6. Originally Posted by scorpion007
earboth, according to your diagram, isn't a + BC = d or OD? not AD as you have it written?
since AD || BC, and since its a parallelogram, shouldnt AD = BC? so why is AD = a + BC? Im thinking a + BC yields the vector from the origin to D, OD....
Hi,

you are right. Sorry for that typo.

EB