A little problem

[matsu]

Hello! I am having trouble with this problem and I cannot seem to find materials on the internet to solve it. (hopefully it's the right subforum)

Find the surface area of the parallelogram built on vectors AC-> and AD->.
AC-> = a-> - b->
AD-> = 3a-> + 10b->
|a->| = 3
|b->| = 5
60 degrees between a-> and b->

I am sorry for the way I wrote it up, hopefully it is understandable. I am not looking for someone to solve it for me, just some tips on how to do it would suffice.

Thanks

 

[Ishuda]

Hello! I am having trouble with this problem and I cannot seem to find materials on the internet to solve it. (hopefully it's the right subforum)

Find the surface area of the parallelogram built on vectors AC-> and AD->.
AC-> = a-> - b->
AD-> = 3a-> + 10b->
|a->| = 3
|b->| = 5
60 degrees between a-> and b->

I am sorry for the way I wrote it up, hopefully it is understandable. I am not looking for someone to solve it for me, just some tips on how to do it would suffice.

Thanks

An outline:
Start at a point, and for convenience make that point the origin (0,0), and two non-linear vectors c and d. Then the parallelogram for those values can be described by (0,0), c, d, c+d. Let
AC-> = c = (x_c, y_c)
AD-> = d = (x_d, y_d)
Then corners of the parallelogram are (0,0), (x_c, y_c), (x_d, y_d), and (x_c + x_d, y_c + y_d). One set of the parallel sides is those lines connecting the pair of points (0,0) and (x_d, y_d) and the pair of points (x_c, y_c) and (x_c + x_d, y_c + y_d). The other set of the parallel sides is those lines connecting the pair of points (0,0) and (x_c, y_c) and the pair of points (x_d, y_d) and (x_c + x_d, y_c + y_d).

Again, for convenience, let a [= a->] line along the positive x axis, i.e. a = (x_a, 0). Since the length of a is 3, what is x_a? Since the line along which b lies is at 60 degrees, what is the formula for that line and, given that formula and that b is 5 units along that line [the length of b is 5], what are the values for x_b and y_b where b = (x_b, y_b).

Note: Depending on the class (and the instructor) and/or how you feel about it, you should justify those two statements about "for convenience". Sometimes the most obvious truths aren't.

Now that you know a and b, compute c and d. Given the formula for a parallelogram, how are c and d related to those quantities?

 

[HallsofIvy]

Given any parallelogram, draw a line from one vertex perpendicular to the opposite side. That cuts off a right triangle having one side of the parallelogram as hypotenuse. Given that \(\displaystyle \theta\) is the angle between the two sides in that right triangle, then the opposites side, the height of the parallogram, is given by \(\displaystyle h= Lsin(\theta)\) where h is the height of the parallelogram and L is the length of the side. The area of a parallelogram, I assume you know, is the length of the base times the height, perpendicular to the base. Putting those together, calling the length of the other side of the parallelogram M, the area is \(\displaystyle LM sin(\theta)\).

 

[matsu]

Thank you guys for your input, however, I'm sorry for not explaining how I am expected to solve this. As I am sure you know, the cross product of 2 vectos is equal to the surface area.of the parallelogram. So it goes something like this:

(a⃗ - b⃗)×(3a⃗ + 10b⃗) = a⃗×3a⃗ + a⃗×10b⃗ - b⃗×3a⃗ - b⃗×10b⃗ = a⃗×10b⃗ - b⃗×3a⃗

But I am stuck here, I don't know how to simplify it further and how the lengths of the vectors are involved. Halp :-(

 

[Ishuda]

Thank you guys for your input, however, I'm sorry for not explaining how I am expected to solve this. As I am sure you know, the cross product of 2 vectos is equal to the surface area.of the parallelogram. So it goes something like this:

(a⃗ - b⃗)×(3a⃗ + 10b⃗) = a⃗×3a⃗ + a⃗×10b⃗ - b⃗×3a⃗ - b⃗×10b⃗ = a⃗×10b⃗ - b⃗×3a⃗

But I am stuck here, I don't know how to simplify it further and how the lengths of the vectors are involved. Halp :-(

Actually, that is what HallsofIvy said in different words. Take a look at
http://en.wikipedia.org/wiki/Cross_product#Geometric_meaning
for more information.

 

[matsu]

This is really confusing.. I can find the cross product using coordinates, but I cannot find anything on how to solve a problem written up like this is. Anyway, I made a couple of schematics and found the lengths of vectors AC (sqrt(19)) and AD (sqrt(3031)), but I still don't have an idea of what the angle between them is to actually find the area of the parallelogram.. It is quite hard as english isn't my mother tongue as well. Would someone please show me through how this problem is solved? Ishuda, your explanation was very detailed, but sadly probably isn't what my lector expects.

 

[Ishuda]

This is really confusing.. I can find the cross product using coordinates, but I cannot find anything on how to solve a problem written up like this is. Anyway, I made a couple of schematics and found the lengths of vectors AC (sqrt(19)) and AD (sqrt(3031)), but I still don't have an idea of what the angle between them is to actually find the area of the parallelogram..

The method above outlines a process for finding the (relative) co-ordinates but, could you do anything with dot products, i.e.
cos(\(\displaystyle \theta\)) = \(\displaystyle \frac{a \bullet b}{|a| |b|}\)
where \(\displaystyle \theta\) is the angle between the vectors a and b. Especially considering the distributive and associative laws for dot products?

 

[matsu]

The method above outlines a process for finding the (relative) co-ordinates but, could you do anything with dot products, i.e.
cos(\(\displaystyle \theta\)) = \(\displaystyle \frac{a \bullet b}{|a| |b|}\)
where \(\displaystyle \theta\) is the angle between the vectors a and b. Especially considering the distributive and associative laws for dot products?

I don't know where to utilize it.

 

[Ishuda]

I don't know where to utilize it.

AC = a - b
AD = 3a + 10b

What is the dot product of AC and AD given that the dot product of a and b is |a||b|cos(\(\displaystyle \frac{\pi}{3})\)
and |a| and |b| are given?

 

[matsu]

AC = a - b
AD = 3a + 10b

What is the dot product of AC and AD given that the dot product of a and b is |a||b|cos(\(\displaystyle \frac{\pi}{3})\)
and |a| and |b| are given?

I can find find the dot product of a and b (7,5)
But I can't find the dot product of AC and AD, because I don't know the angle between them. What can I use the dot product of AC and AD for anyway?

I am fairly certain the solution starts with the cross product.

S = (a⃗ - b⃗)×(3a⃗ + 10b⃗)

I just don't know where to take it from here.

 

[Ishuda]

I can find find the dot product of a and b (7,5)
But I can't find the dot product of AC and AD, because I don't know the angle between them. What can I use the dot product of AC and AD for anyway?

I am fairly certain the solution starts with the cross product.

S = (a⃗ - b⃗)×(3a⃗ + 10b⃗)

I just don't know where to take it from here.

Read the Algebraic properties of cross products at
http://en.wikipedia.org/wiki/Cross_product