# Vector Methods To Find An Area

• BOAS
In summary: Ok, I've got myself in a bit of a muddle here.Any side can be my base, so i'll say AB is. AC is therefore, the diagonal to the 'top'.The height is |AC|sinθ, where θ is the angle between AC and AB.
BOAS
Hello,

after searching around on the internet about this problem, it looks like it is Cramer's rule that I want to use, though it wasn't shown to us under that name.

My textbook doesn't cover the material required for this problem, so i'd really like to run what I have done past you guys :)

## Homework Statement

The triangle ABC has vertices A(-1,3,0), B(-3,0,7), C(-1,2,3). Find AB, AC, CB.

Find also the area of the triangle, using a vector method.

## The Attempt at a Solution

I found the vectors easily enough.

AB = -2i - 3j + 7k

AC = -j + 3k

CB = -2i - 2j + 4k

Area of a triangle = 1/2 base x height

$A$ = $( \frac{1}{2})$ (AB x AC)

I don't know how to make matrices in latex, but I use what looks to me like cramer's rule (we were just told how to do it, as we haven't covered determinants yet) and come to an answer that;

A = -i + 3j + K

I hope it's clear to you, but if not, if you can show me how to display matrices i'll gladly show the working.

Thanks!

Area is supposed to be a number no? You seem to get that area is a vector.

1 person
micromass said:
Area is supposed to be a number no? You seem to get that area is a vector.

Hm, yes.

Area = |AB x AC| <--- Which I didn't actually say, so thanks for pulling that up.

So I should give my answer as '3'?

BOAS said:
Hm, yes.

Area = |AB x AC|

That's the area of the parallellogram formed by AB and AC, no? You want the triangle, so the factor 1/2 is necessary.

1 person
micromass said:
That's the area of the parallellogram formed by AB and AC, no? You want the triangle, so the factor 1/2 is necessary.

Ack, yes.

The factor of a half was included in my calculations, but I forgot to show it in my reply to your comment.

A = 1/2 |AB x AC|

= 1/2( -2 + 6 + 2)
= 3

BOAS said:
Ack, yes.

The factor of a half was included in my calculations, but I forgot to show it in my reply to your comment.

A = 1/2 |AB x AC|

= 1/2( -2 + 6 + 2)
= 3

That's a weird way of calculating the length. Isn't the length of a vector given by

$$|(x,y,z)| = \sqrt{x^2 + y^2 + z^2}$$

micromass said:
That's a weird way of calculating the length. Isn't the length of a vector given by

$$|(x,y,z)| = \sqrt{x^2 + y^2 + z^2}$$

Yes...

Is the method I tried to use wrong? (excusing the mistakes we have already discussed)

When using the method described above, my answers don't agree.

BOAS said:
Yes...

Is the method I tried to use wrong? (excusing the mistakes we have already discussed)

When using the method described above, my answers don't agree.

You want to calculate the length of the vector ##(-1,3,1)##. You seem to do this by calculating

$$-1 + 3 + 1$$

I don't see where that comes from. Isn't the length rather equal to

$$\sqrt{(-1)^2 + 3^2 + 1^2}$$

micromass said:
You want to calculate the length of the vector ##(-1,3,1)##. You seem to do this by calculating

$$-1 + 3 + 1$$

I don't see where that comes from. Isn't the length rather equal to

$$\sqrt{(-1)^2 + 3^2 + 1^2}$$

I agree with you.

Area of a triangle = 1/2 base x height

|AB| = $\sqrt{(-2)^{2} + (-3)^{2} + (7)^{2}}$ = $\sqrt{62}$

|AC| = $\sqrt{(0)^{2} + (-1)^{2} + (3)^{2}}$ = $\sqrt{10}$

∴ A = 1/2($\sqrt{10}$ $\sqrt{62}$) = $\sqrt{155}$ ≈ 12.45

Can this problem be solved using the cross product, or am I barking up the wrong tree?

Thank you for the help by the way.

BOAS said:
I agree with you.

Area of a triangle = 1/2 base x height

|AB| = $\sqrt{(-2)^{2} + (-3)^{2} + (7)^{2}}$ = $\sqrt{62}$

|AC| = $\sqrt{(0)^{2} + (-1)^{2} + (3)^{2}}$ = $\sqrt{10}$

∴ A = 1/2($\sqrt{10}$ $\sqrt{62}$) = $\sqrt{155}$ ≈ 12.45

Can this problem be solved using the cross product, or am I barking up the wrong tree?

Thank you for the help by the way.

Right now you're going wrong. You can't do this because you don't know that AC is actually the height. You just know it's a side from the triangle.

You were actually doing very well before. We established that the area of the triangle is given by

$$\frac{1}{2}|\mathbf{AC}\times\mathbf{AB}| = \frac{1}{2}|(-1,3,1)|$$

The only thing I disagreed with is how you practically calculated the length.

micromass said:
Right now you're going wrong. You can't do this because you don't know that AC is actually the height. You just know it's a side from the triangle.

You were actually doing very well before. We established that the area of the triangle is given by

$$\frac{1}{2}|\mathbf{AC}\times\mathbf{AB}| = \frac{1}{2}|(-1,3,1)|$$

The only thing I disagreed with is how you practically calculated the length.

Ok, I've got myself in a bit of a muddle here.

Any side can be my base, so i'll say AB is. AC is therefore, the diagonal to the 'top'.

The height is |AC|sinθ, where θ is the angle between AB and AC.

AB.AC = 24

AB.AC = |AB||AC| cosθ

|AB||AC| cosθ = 24

θ = arccos ($\frac{24}{\sqrt{10}\sqrt{62}}$)

= 15.45°

Height, |AC|sinθ = $\sqrt{10}$ sin(15.45)
≈ 0.84

Now I should be able to say that the area = 1/2 base x height

= 1/2 * $\sqrt{62}$ * 0.84
= 3.32

You're really making it difficult for yourself. You almost had the right answer in post 5, the only thing was that you calculated the distance wrongly.

micromass said:
You're really making it difficult for yourself. You almost had the right answer in post 5, the only thing was that you calculated the distance wrongly.

A parallelogram can be divided into 2 congruent triangles, so the area of the triangle should be half the area of the parallelogram as worked out by doing the cross product.

Area = $\frac{1}{2}$ |AB|x|AC| = $\frac{1}{2}$ |-2, -3, 7|x|0, -1, 3|

= $\frac{1}{2}$ (-2, -6, 2)

= $\frac{1}{2}$ $\sqrt{(-2)^{2} + (-6)^{2} + (2)^{2}}$

= $\sqrt{11}$

That was a lot simpler, but i'll take some solace from the fact that it agrees with my longwinded route. (fingers crossed that this is correct!)

Yep, you've got it! Congratulations!

1 person
micromass said:
Yep, you've got it! Congratulations!

Thanks so much for your patience.

The format here, where you allow the question askers to arrive at the solution themselves is very beneficial and definitely more conducive to understanding than simply providing a solution.

## 1. What are vector methods used for finding an area?

Vector methods are used to calculate the area of a 2-dimensional shape by breaking it up into smaller vectors and using mathematical formulas to determine the total area.

## 2. How do vector methods differ from traditional methods of finding an area?

Unlike traditional methods, which rely on measurements and calculations of length and width, vector methods use the magnitude and direction of vectors to determine the area. This allows for more accurate calculations, especially for irregularly shaped figures.

## 3. What types of shapes can be measured using vector methods?

Vector methods can be used for any 2-dimensional shape, including regular polygons, irregular polygons, and curved shapes such as circles and ellipses.

## 4. Can vector methods be used for finding the area of 3-dimensional shapes?

No, vector methods are only applicable for 2-dimensional shapes. For 3-dimensional shapes, other methods such as integration must be used to determine the surface area.

## 5. Are there any limitations to using vector methods for finding an area?

Vector methods may not be suitable for very complex or irregular shapes, as they require breaking the shape into smaller vectors. In these cases, other methods such as integration or approximations may be more appropriate.

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