Area of A Triangle in 3D

[DODGEVIPER13]

Homework Statement

Uploaded

Homework Equations

Uploaded

The Attempt at a Solution

I actually need someone to check my work for 1.1 and 1.2. Is what I have done in 1.3 correct I mean it does not seem right?

 

[gneill]

For 1.1 (b) it looks like they want you to take the magnitude of B in the denominator... otherwise the expression wouldn't make sense.

For 1.1 (c), note that the items u_x, u_y, and u_z are the unit vectors of the vector space. So when you take the cross product, don't introduce i, j, or k. The top row of your determinant should be |u_x u_y u_z|. The same applies to 1.1 (d)

For 1.2, consider what vectors being parallel or perpendicular implies for their dot and cross products.

For 1.3, form two vectors that define the sides of the triangle and carry out the cross product. Then $A = \frac{1}{2}|V_1 \times V_2|$.

 

[DODGEVIPER13]

Ok so the denominator for (b) should be 23.065. For (c) 8UxUyUz. For (d) -26UxUz. on 1.2 that is precisley what had me lost. I know that if they are parallel, then one vector must be a multiple of another. If perpendicular the dot product must result in a 0 and with alpha and beta only being tied to two terms. I struggle with how I can make this happen on the Z value for the parallel part. For part 1.3 A=48.489

 

[DODGEVIPER13]

AB dot AC = AB dot AC cos(theta) is how I find the angle right?

 

[gneill]

Ok so the denominator for (b) should be 23.065.

Sure. So what then are the components of the resulting vector?

For (c) 8UxUyUz.

I don't understand that. The result of a cross product should be a vector.

For (d) -26UxUz.

Part d contains a cross product followed by a dot product. The result should be a single scalar value.

on 1.2 that is precisley what had me lost. I know that if they are parallel, then one vector must be a multiple of another. If perpendicular the dot product must result in a 0 and with alpha and beta only being tied to two terms. I struggle with how I can make this happen on the Z value for the parallel part.

For the parallel vectors you can write x*A = B, for some scalar value x. Expand that and you'll see three equations. Solve for x (simple!). Use x to find $\alpha$ and $\beta$.

For part 1.3 A=48.489

Show your work; that's not the value I'm seeing.

EDIT: Sorry, I didn't see your attachments. Looking at your work for 1.3, I see that you've done something bizarre to find the vectors defining the triangle: you're multiplying the point components! A vector from point A to point B is given by V_AB = B - A.

For your cross products, the Ux, Uy, Uz stuff should ONLY appear in the top line of the determinant! Ux, Uy, and Uz designate the unit vectors of the vector space (or coordinate system axes). A vector written as: 3Ux + 5Uy -6Uz has Ux component 3, Uy component 5, and Uz component -6. It might also be written as: (3, 5, -6), with the understanding that those are the Ux, Uy, Uz components.

 

[gneill]

AB dot AC = AB dot AC cos(theta) is how I find the angle right?

For two vectors A and B, $A \cdot B = |A||B| cos(\theta)$

then $cos(\theta) = \frac{A \cdot B}{|A||B|}$

Pay attention to the vector directions: you want to take the angle between pairs such that the vectors are diverging from the triangle vertex. It may help to sketch the triangle "flat" using the vector lengths as the sides. Identify the "points" with the vertices and the vector directions along the triangle sides. You'll know something went wrong if the angle sum is not 180°

 

[DODGEVIPER13]

For (b) I get 2.948Ux+3.728Uy-1.213Uz. Did I do the cross product incorrect or something I am not understanding on part (c). Once again on part (d) I did the cross product and then used the dot product I am confused on what to do now?

 

[gneill]

For (b) I get 2.948Ux+3.728Uy-1.213Uz. Did I do the cross product incorrect or something I am not understanding on part (c). Once again on part (d) I did the cross product and then used the dot product I am confused on what to do now?

Your part (b) result looks fine.

For the cross products, take a look at the EDITs I added to post #5.

 

[DODGEVIPER13]

ok so X(Ux+3Uy-2Uz)=4Ux+Uy+8Uz then XUx+3XUy-2XUz = 4Ux+Uy+8Uz then Uz(-2x-8)=Ux(4-x)+(1-3x)Uy

 

[DODGEVIPER13]

How do I view the edits it still looks the same on my comp and BTW I can't read the code in #6

 

[gneill]

ok so X(Ux+3Uy-2Uz)=4Ux+Uy+8Uz then XUx+3XUy-2XUz = 4Ux+Uy+8Uz then Uz(-2x-8)=Ux(4-x)+(1-3x)Uy

What happened to the parameters $\alpha$ and $\beta$? You should have:

$X(\alpha, 3, -2) = (4, \beta, 8)$

from which you have three equations if you equate the like components.

 

[gneill]

How do I view the edits it still looks the same on my comp and BTW I can't read the code in #6

If you refresh the page, the edits at the bottom of the post should appear.

By "code" I presume you mean the mathematical expressions rendered in LaTex? Perhaps your browser is not compatible with the Physics Forums' usage of LaTex? What type of machine/operating system are you using?

 

[DODGEVIPER13]

ok for c I get 6Uy+2Uz and for d I get -8

 

[DODGEVIPER13]

Ok after doing as you suggested if I did it right alpha=-1 and beta=-12 also x=-4

 

[DODGEVIPER13]

yah thanks the refresh worked

 

[gneill]

ok for c I get 6Uy+2Uz and for d I get -8

Check the signs on your expansion of the cross product for c. -8 for part d looks fine.

 

[gneill]

Ok after doing as you suggested if I did it right alpha=-1 and beta=-12 also x=-4

Right

 

[DODGEVIPER13]

ok my signs were off on (c) 6uy-2Uz

 

[gneill]

ok my signs were off on (c) 6uy-2Uz

I still see a sign problem with the first term...!

 

[DODGEVIPER13]

doh! so you are saying it is -6Uy-2Uz. Are you sure, I have checked it 3 times and feel confident in my initial answer?

 

[gneill]

doh! so you are saying it is -6Uy-2Uz. Are you sure, I have checked it 3 times and feel confident in my initial answer?

The order of the cross product matters! A x B = - B x A. Your attachment shows that you're computing A x Ux...

 

[DODGEVIPER13]

Ok good point it should be after reworking it again -6Uy-2Uz

 

[DODGEVIPER13]

to find magnitude (sqrt((-3)^2+(-30)^2+(-36)^2+(15)^2))/2 = 24.647

 

[DODGEVIPER13]

That is for the area of the triangle

 

[gneill]

to find magnitude (sqrt((-3)^2+(-30)^2+(-36)^2+(15)^2))/2 = 24.647

Ummm, why do you have two separate Uy terms? Your vector should have exactly three independent components.

 

[DODGEVIPER13]

Forgot to collect them up sorry now I get 25.719

 

[gneill]

Forgot to collect them up sorry now I get 25.719

Much better!

 

[DODGEVIPER13]

Calculate Gradient and Divergence

On 1.4 I think I am right but not sure? On 1.5 (a), (b), and (c) I am not sure if I did the partial derivatives correctly? On 1.6 Not sure if I computed the gradient right? On 1.7 not sure if i calculated the divergence right? On 1.8 I don't know what to do at all and need a bit of a push? I uploaded my work and the problem was in the first upload.

 

[gneill]

I think it would be better if you were to decompose the rest of these problems into individual threads. That will increase the chance that people with the appropriate skills for a given problem will take notice. Seeing a large batch of problems in one thread discourages some from jumping into what could be a lengthy and confusing fray.

Having said that, I will address the "latest" problem, 1.4.

Note that you wish the dl's to follow along the given curve, y = x². You correctly wrote that this means you can substitute y = x² in the integral. However, you also need to address dy. For the curve y = x², dy ≠ dx. What should it be?

 

[DODGEVIPER13]

Dy=2x

 

[DODGEVIPER13]

Now I get y^2-y

 

[gneill]

Dy=2x

$y = x^2$
$dy = 2x dx$

 

[gneill]

Now I get y^2-y

Show how. Your final result should be numeric.

 

[DODGEVIPER13]

What I did is take the integral from 1 to 0 of y dx and found it to be -y and then evaluated the integral from 0 to 1
Of 2xy^2 dx and got y^2 should I have turned y into x^2 and y^2 into x^4 hmm if I do that I get 0

 

[gneill]

What I did is take the integral from 1 to 0 of y dx and found it to be -y and then evaluated the integral from 0 to 1
Of 2xy^2 dx and got y^2 should I have turned y into x^2 and y^2 into x^4 hmm if I do that I get 0

I don't know why you'd do that first integral. The idea is to cast the dot product into terms of a single variable, and x is convenient. To do that you must replace the $y^2$ of the H function using $y=x^2$.

Now, the path you want L to follow is along the other function $y=x^2$. So then in terms of x, $dy = 2x dx$. Make the substitutions and carry out the dot product. Then integrate over x. By substituting y and dy, you need only integrate over the one variable, x.