# Vector Geometry

• Mar 29th 2008, 02:36 AM
ah-bee
Vector Geometry
I haven't done any vector geometry in close to 2 years now and I got this question for one of my exercises and I've completely forgotten what to do.

Question: Show that the points A(1, 0, 1), B(2,−1, 3), C(5, 1,−1) and D(6, 0, 1) are vertices of a parallelogram. Is the parallelogram a square?

A full solution along with a step by step explanation would be great.. Just can't seem to remember how to do questions like this.

• Mar 29th 2008, 02:53 AM
Moo
Hello,

Try to sketch a figure.

If ABCD is a parallelogram, $\displaystyle \vec{AB}=\vec{DC}$

The coordinates of a vector, when given the coordinates of points is :

$\displaystyle \vec{AB} \ : \begin{Bmatrix} x_B-x_A \\ y_B-y_A \\ z_B-z_A \end{Bmatrix}$

Then, if it's a cube, diagonals may be equal and perpendicular (i don't remember if this condition is necesary).

If 2 vectors are perpendicular, their scalar product is 0.
The scalar product is the sum of the product of each coordinate (x with x, y with y etc..)

Try to work with these elements :D
• Mar 29th 2008, 03:47 AM
ah-bee
by expressing a vector like (x,y,z) i keep thinking of it as a point instead of a vector, is there an explanation that could help clarify things here?
• Mar 29th 2008, 03:50 AM
Moo
Hm... The coordinates of a vector correspond to a kind of slope of the line represented by the vector in the direction of x,y and z-axis. It may be similar to points, but not the same properties.

Am sorry, it's the best i can explain, this is the wall between two languages ^^'
• Mar 29th 2008, 03:58 AM
ah-bee
So basically, when the vector is represented in the form (x,y,z), that just specifies the magnitude and direction of the vector, not the position at which it is located at. That sounds about right... Right? lol
• Mar 29th 2008, 04:06 AM
Moo
Exactly !

The vector only determines directions, not positions (Clapping)

There is a simple illustration : two parallel lines have the same direction vector, but they are not the same.
• Mar 29th 2008, 07:55 AM
earboth
Quote:

Originally Posted by ah-bee
I haven't done any vector geometry in close to 2 years now and I got this question for one of my exercises and I've completely forgotten what to do.

Question: Show that the points A(1, 0, 1), B(2,−1, 3), C(5, 1,−1) and D(6, 0, 1) are vertices of a parallelogram. Is the parallelogram a square?

A full solution along with a step by step explanation would be great.. Just can't seem to remember how to do questions like this.

A parallelogram has
- 2 pairs of parallel sides
- the parallel sides have equal length
- the 2 diagonals have a common midpoint.

$\displaystyle \overrightarrow{AB} = \left(\begin{array}{c}2 \\-1 \\3 \end{array}\right) - \left(\begin{array}{c}1 \\0 \\ 1\end{array}\right) = \left(\begin{array}{c}1 \\-1 \\ 2\end{array}\right)$

$\displaystyle \overrightarrow{CD} = \left(\begin{array}{c}6 \\0 \\1 \end{array}\right) - \left(\begin{array}{c}5 \\1 \\ -1\end{array}\right) = \left(\begin{array}{c}1 \\-1 \\ 2\end{array}\right)$

$\displaystyle \overrightarrow{AC} = \left(\begin{array}{c}5 \\1 \\-1 \end{array}\right) - \left(\begin{array}{c}1 \\0 \\ 1\end{array}\right) = \left(\begin{array}{c}4 \\1 \\ -2\end{array}\right)$

$\displaystyle \overrightarrow{BD} = \left(\begin{array}{c}6 \\0 \\ 1\end{array}\right) - \left(\begin{array}{c}2 \\-1 \\3 \end{array}\right) = \left(\begin{array}{c}4 \\1 \\ -2\end{array}\right)$

So the quadrilateral is a parallelogram but not a square.

Calculating the midpoints:

$\displaystyle M_{AD} \left(\frac{1+6}2\ , ~ \frac{0+(-1)}2\ , ~ \frac{1+3}2\ \right)$ and $\displaystyle M_{BC} \left(\frac{2+5}2\ , ~ \frac{(-1) +1}2\ , ~ \frac{3+(-1)}2\ \right)$

Both diagonals have the same midpoint. That means the 4 vertices are located in a plane and the quadrilateral is a parallelogram.
• Mar 29th 2008, 08:05 AM
earboth
Quote:

Originally Posted by Moo
...

Then, if it's a cube, diagonals may be equal and perpendicular (i don't remember if this condition is necesary).

...

That depends where the vertices are located. But if you are dealing with squares in 3-D these conditions are not sufficient.

You can construct a quadrilateral with 4 sides with equal length and the diagonals are perpendicular and have the same length - but this quadrilateral is neither a square nor a rhombus because the 4 vertices are not placed in a plane!
• Mar 29th 2008, 04:20 PM
ah-bee
Quote:

Originally Posted by earboth
A parallelogram has
- 2 pairs of parallel sides
- the parallel sides have equal length
- the 2 diagonals have a common midpoint.

$\displaystyle \overrightarrow{AB} = \left(\begin{array}{c}2 \\-1 \\3 \end{array}\right) - \left(\begin{array}{c}1 \\0 \\ 1\end{array}\right) = \left(\begin{array}{c}1 \\-1 \\ 2\end{array}\right)$

$\displaystyle \overrightarrow{CD} = \left(\begin{array}{c}6 \\0 \\1 \end{array}\right) - \left(\begin{array}{c}5 \\1 \\ -1\end{array}\right) = \left(\begin{array}{c}1 \\-1 \\ 2\end{array}\right)$

$\displaystyle \overrightarrow{AC} = \left(\begin{array}{c}5 \\1 \\-1 \end{array}\right) - \left(\begin{array}{c}1 \\0 \\ 1\end{array}\right) = \left(\begin{array}{c}4 \\1 \\ -2\end{array}\right)$

$\displaystyle \overrightarrow{BD} = \left(\begin{array}{c}6 \\0 \\ 1\end{array}\right) - \left(\begin{array}{c}2 \\-1 \\3 \end{array}\right) = \left(\begin{array}{c}4 \\1 \\ -2\end{array}\right)$

So the quadrilateral is a parallelogram but not a square.

Calculating the midpoints:

$\displaystyle M_{AD} \left(\frac{1+6}2\ , ~ \frac{0+(-1)}2\ , ~ \frac{1+3}2\ \right)$ and $\displaystyle M_{BC} \left(\frac{2+5}2\ , ~ \frac{(-1) +1}2\ , ~ \frac{3+(-1)}2\ \right)$

Both diagonals have the same midpoint. That means the 4 vertices are located in a plane and the quadrilateral is a parallelogram.

Thanks for the advice but now I'm stuck wondering how the midpoints have been calculated. Care to explain?
• Mar 29th 2008, 04:49 PM
Plato
Quote:

Originally Posted by ah-bee
Thanks for the advice but now I'm stuck wondering how the midpoints have been calculated. Care to explain?

Midpoints are just averages.
$\displaystyle \left( {a,b,c} \right)\,\& \,\left( {p,q,r} \right)\quad \Rightarrow \quad \left( {\frac{{a + p}}{2},\frac{{b + q}}{2},\frac{{c + r}}{2}} \right)$
• Mar 29th 2008, 05:04 PM
ah-bee
thats what i thought, so that just means that earboth has made an error in calculating the midpoint.. that confused me a bit.. Thx every1 for helping out!
• Mar 29th 2008, 09:33 PM
earboth
Quote:

Originally Posted by ah-bee
thats what i thought, so that just means that earboth has made an error in calculating the midpoint.. that confused me a bit.. Thx every1 for helping out!

You are right. Here is the correct midpoint:

$\displaystyle M_{AD} \left(\frac{1+6}2\ , ~ \frac{0+(-1)}2\ , ~ \frac{1+1}2\ \right)$