surface of a parallelogram - vectors

**harriette** · January 27th 2011, 12:16 PM

I have to determine the surface of a parallelogram whose vectors of diagonals are $\vec{c}=5\vec{m}+5\vec{n}$ , $\vec{d}=\vec{m}-\vec{n}$ , and $\vec{m}$ and $\vec{n}$ are unit vectors and $\angle (\vec{m}, \vec{n})=\frac{\pi}{6}$ .

Please help, I have to idea where to start.

**Sudharaka** · January 27th 2011, 05:04 PM

Originally Posted by harriette

I have to determine the surface of a parallelogram whose vectors of diagonals are $\vec{c}=5\vec{m}+5\vec{n}$ , $\vec{d}=\vec{m}-\vec{n}$ , and $\vec{m}$ and $\vec{n}$ are unit vectors and $\angle (\vec{m}, \vec{n})=\frac{\pi}{6}$ .

Please help, I have to idea where to start.

Dear harriette,

The vector equation of a plane is given by, $(\underline{a}-\underline{b}).\underline n=0$ where $\underline{a},\underline{b}$ are the position vectors of two points on the plane and $\underline n$ is a vector perpendicular to the plane.

Hope you would be able to continue along these lines.

**harriette** · January 27th 2011, 05:08 PM

I suppose I should express $\vec{a}, \vec{b}$ of the parallelogram (the base) using $\vec{m}, \vec{n}$ and then use cross product on that? Is that correct?

**Sudharaka** · January 27th 2011, 05:39 PM

Originally Posted by harriette

I suppose I should express $\vec{a}, \vec{b}$ of the parallelogram (the base) using $\vec{m}, \vec{n}$ and then use cross product on that? Is that correct?

Dear harriette,

We are given two vectors of the plane that we have to find out. Those are the vectors of the diagonals. What we dont know is a perpendicular vector to the plane. Can you think of a way to find a perpendicular vector using the two given vectors?

**harriette** · January 27th 2011, 06:48 PM

Their cross product will be perpendicular to the plane, but I don't see why do I even need a perpendicular vetor.

**harriette** · January 28th 2011, 01:29 AM

Anyone?

**Sudharaka** · January 28th 2011, 04:05 AM

Originally Posted by harriette

Their cross product will be perpendicular to the plane, but I don't see why do I even need a perpendicular vetor.

Correct. Therefore, you can take, $\vec{n}=\vec{c}\times\vec{d}$ . The equation of the plane will be, $\vec{c}.(\vec{c}\times\vec{d})=0$ because in this case $\vec{c}$ lies on the plane. Alternatively you can write the equation of the plane as $\vec{d}.(\vec{c}\times\vec{d})=0$ .

Hope you will be able to continue.

**harriette** · January 28th 2011, 04:20 AM

I'm so embarassed, but I really don't see where you're going with this.

Here's what I wanted to do. Let $ABCD$ be the parallelogram in question, $\vec{a}=AB, \vec{b}=AD$ . Surface area of that parallelogram is $S=|\vec{a}|h$ , where $h$ denotes the height. That equals to $|\vec{a}| |\vec{b}|\sin \alpha$ which is $| \vec{a} \times \vec{b}|$ .
So I was hoping I could express $\vec{a}, \vec{b}$ using $\vec{n}, \vec{m}$ and then calculate it using the above equation.

Does that make any sense?

**Sudharaka** · January 28th 2011, 04:29 AM

Originally Posted by harriette

I'm so embarassed, but I really don't see where you're going with this.

Here's what I wanted to do. Let $ABCD$ be the parallelogram in question, $\vec{a}=AB, \vec{b}=AD$ . Surface area of that parallelogram is $S=|\vec{a}|h$ , where $h$ denotes the height. That equals to $|\vec{a}| |\vec{b}|\sin \alpha$ which is $| \vec{a} \times \vec{b}|$ .
So I was hoping I could express $\vec{a}, \vec{b}$ using $\vec{n}, \vec{m}$ and then calculate it using the above equation.

Does that make any sense?

Ok. If you could find $\vec{a}$ and $\vec{b}$ using $\vec{m}$ and $\vec{n}$ that would be great (Can you please tell me how?). But I dont think that would be easy. Instead you could use the given vectors of the diagonals and get the equation of the plane as I had mentioned in my last post.

**harriette** · January 28th 2011, 04:30 AM

I can get the equation of the plane but then what? I have no idea how will I get the surface area of a parallelogram using the equation of a plane.

**alexmahone** · January 28th 2011, 04:48 AM

Originally Posted by harriette

I can get the equation of the plane but then what? I have no idea how will I get the surface area of a parallelogram using the equation of a plane.

Let $\vec{a}$ and $\vec{b}$ be two sides of the parallelogram.

$\vec{a}+\vec{b}=\vec{c}$

$\vec{a}-\vec{b}=\vec{d}$

We get $\vec{a}=\frac{1}{2}(\vec{c}+\vec{d})$ and $\vec{b}=\frac{1}{2}(\vec{c}-\vec{d})$ .

Find $\vec{a}$ and $\vec{b}$ in terms of $\vec{m}$ and $\vec{n}$ . Compute $\vec{a}\times\vec{b}$ .

**harriette** · January 28th 2011, 04:53 AM

Originally Posted by alexmahone

Let $\vec{a}$ and $\vec{b}$ be two sides of the parallelogram.

$\vec{a}+\vec{b}=\vec{c}$

$\vec{a}-\vec{b}=\vec{d}$

We get $\vec{a}=\frac{1}{2}(\vec{c}+\vec{d})$ and $\vec{b}=\frac{1}{2}(\vec{c}-\vec{d})$ .

Find $\vec{a}$ and $\vec{b}$ in terms of $\vec{m}$ and $\vec{n}$ . Compute $\vec{a}\times\vec{b}$ .

Thank you, both of you for your replies. This is what I wanted to do (I'm glad to see the idea was okay.)

I still don't understand the part with the perpendicular vector Sudharaka mentioned.

**alexmahone** · January 28th 2011, 04:55 AM

Originally Posted by harriette

I still don't understand the part with the perpendicular vector Sudharaka mentioned.

Nor do I.

**Sudharaka** · January 28th 2011, 05:05 AM

Originally Posted by harriette

I can get the equation of the plane but then what? I have no idea how will I get the surface area of a parallelogram using the equation of a plane.

In your first post.....

I have to determine the surface of a parallelogram.............

So I did not know you have to find the area of the paralleogram.........I told you the method of finding the plane (surface) which the parallelogram lies. Anyway if you want to find the area of the parallelogram you will have to find the sides of the parallelogram (as vectors). See the attached figure.

$\vec{x}+\vec{y}=5\vec{m}+5\vec{n}$

$\vec{x}-\vec{y}=\vec{m}-\vec{n}$

Solving the above equations you can obtain the vectors of the sides of the parallelogram. Then $\vec{x}\times\vec{y}$ would give you the area.

Sorry if I confused you in my previous posts..........I hope this helps to clarify things.....just forget what I told before.

**harriette** · January 30th 2011, 01:56 PM

Oh, now I see where the mixup was.

Thank you.

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