Vector

1. vector..................?

Using the suggestion A = A1 + A2 where A1 is the component in the direction of B, and A2 is the component perpendicular to B. Now bxA1 = 0, hence bxA =bxA1 + bxA2 = bxA2 That's a start, and I guess you finish it by taking the cross product -bx(bxA) = -bx(bxA2) and I would have known how to proceed from there, but don't remember now. Maybe you can show the righthand side is equal to A2. Or try a website such as Wolfram. Is there a formula for triple cross product something like ux(vxw) = (u.w)v - (u.v)w? If so, in this case the first term is (-b.A2)b which is zero since A2 is perpendicular to b, and the second is (b.b)A2, which is A2 since b.b = 1 for any unit vector.

2. Solving Vector Problems when they both start from the same origin?

When performing vector addition, you always place the start of one vector at the end of the other. Vector addition is commutative, so the order doesn't matter. If you put the start of B at the end of A, then the vector sum goes from the start of A to the end of B. Vectors, in the context in which you're working with them, do not have a defined start point--they only have magnitude and direction. So you can slide them around freely, as long as you don't change their length or rotate them.

3. How do you transfer a vector image from Illustrator to Publisher?

Export your Illustrator files as a wmf or emf (windows/enhanced meta file). They are both vector image formats and should work in publisher. Alternatively, you can directly copy (ctrl C) what you want from your illustrator file and past it (ctrl V) into your publisher document.

4. How do you find unit vectors that make an angle with another vector?

The important formula is the formula for dot products: For any two real vectors v and w, we have v dot w = |v|*|w|*cos(theta) where theta is the angle between v and w. We have theta=60 degrees, so cos(theta)=1/2. Let us call 3.4 v and we want to find two possible w. We know that |w| is 1, because it is a unit vector, and therefore we are left with 3, 4 dot x, y = 5/2 (|v|=5) Thus, we are left with the equation 3x + 4y = 5/2 With the added condition that x^2+y^2=1, because we want it to be a unit vector. Solving the first equation yields y=5/8 - 3x/4 Plugging this into the second yields (5/8-3x/4)^2+x^2=1 25/64 - 15x/16 + 9x^2/16 +x^2=1 25x^2/16 - 15x/16 - 39/64=0 100x^2 - 60x - 39 = 0 Solving for x via the quadratic formula will yield two possible values. Then you can use the other two equations we developed to solve for y, and get two values for w, as desired. =D

5. What would a vector orthogonal to three vectors not on the same plane look like?

It would have to be in the 4th dimension. Higher dimensions start to get pretty crazy... here's an awesome video to help you visualize them: http://www.tenthdimension.com/medialinks.php

6. What is the magnitude and direction of the resultant vector?

14.8N 15.7SE

7. How do you calculate an orbital vector?

Well i know this might sound pretty simple but you left out direction and time...now time is a no brain-er but direction for an orbit is segmented,,,,,in math we call those steps..in an engineering modeling calculator the step value is the increment that the new value is based on ..for example: step by 5 is all the integers from 1 to 100 pretty simple we just count by five there are about 20 steps not real accurate,,,,now step by .005 well you can see there is room for a much more accurate plot...so now with the planer coordinates of x y and z ...calculate in the locations of different spots versus the ever-changing affected planet you can see that there are a considerable number of calculations that have to be made...Here is an example : planet a moves from location 1 to 2 and has a uniform gravitational pull of x Weber's from planet B ,C and D all of which are fixed in plane around a central orbit loci (planet A)...now as the planet A gets closer to B it gets father from C and stays the same for D..well i think you might be seeing the problems associated with curve linear vector calcs..in fact the magnitude and vector angle are changed at each step,,.in closing draw the model as simple as possible ,only do a small number of calc and then try to predict the next outcome,add bells and whistles as you go ..pretty soon you'll have an active representation of what you need....From the E....

8. How do i get the simplified vector from 3 dimension plane?

for 3 dimension X,Y,Z you must also calculate the time to see real results.

9. How can I change the stroke of multiple vector shapes at once in Photoshop?

There is in Illustrator. You can use the magic wand in there and it selects all objects with the specific attributes you clicked on. In PS, I'm not sure. I don't care for the vector capabilities in PS. If you have Illustrator, you can export the paths to it from PS and edit them from there. Just a thought

10. How to solve the magnitude of a vector when it is unknown?

The easiest way to solve this is through a graphical solution. But to do this you're gonna need a good ruler and some sort of angle measurer. First draw vector A, in a certain scale, let's say 1 in = 10 units. That means you'll be drawing a straight vertical line which has a length of 18.8 in. Next you draw vector B from the edge of vector A 50 deg east to north. Draw a very very long line. Next you draw vector C from the other edge of vector A, 20 deg west to south. Again draw a very very long line. You'll then get a intersection where the line representing vector B C meet. Measure the line, from that intersection to the edge of vector A. Multiply the measurement with the scale you've already stated in the first step (1 in = 10 units). Et voila!! You got the magnitude of each vector

11. How do I convert a cartesian vector into spherical coordinates?

ro^2 = x^2 + y^2 + x^2 ro = sqrt(x^2 + y^2 + x^2) cos(phi) = z / ro cos(theta) = x / (ro * sin phi)

12. How do you use vector files in Photoshop CS?

To add to answers given above, starting with the CS2 package Adobe synced Photoshop and Illustrator nicely. If you have both, you can import vector layers into Photoshop as vector smart objects. By double clicking on them in the layer palette it will open them in Illustrator, and any saved changes will take their effect in Photoshop when you return to the program. Not quite the same, but in two totally different graphic worlds (vector and bitmap) its the closest you're going to get to joining them for now.

13. What is the average magnitude of the poynting vector?

The average value of the Poynting vector is zero for isotropic radiation.

14. Will a vector at 45° to the horizontal be larger or smaller than its horizontal and vertical components?

Larger. Just remember the Pythagorean theorem. When you draw a right triangle, the vector will be the hypotenuse. Since it's a 45 degree angle, the vertical and horizontal components are equal. The vector will be larger than them by a factor of square root of 2.

15. How do I find the matrix representation for the orthogonal projection onto vector v?

I'm not sure what you are asking. If you want to project a vector V onto a standard basis composed of E1 and E2, then: If E1 and E2 are orthonormal, then any vector V = (V.E1)E1 + (V.E2)E2. Notes: - E1 and E2 are orthonormal if they are perpendicular and have length 1 http://en.wikipedia.org/wiki/Orthonormal_basis - (A.B) means the dot product of the vectors A and B. The results is a scalar that gives the length of component of B parallel to A times the length of A (which is equal to the length of the component of A parallel to B times the length of B) http://en.wikipedia.org/wiki/Dot_product If, on the other hand, you have some linear transformation, then you can find its matrix representation by seeing what it does to E1 and E2. Let the transform be T: T11 T12 T21 T22 Then since E1 is (by definition) 1, 0, if T(E1) = X1, Y1, then: T11 = X1 and T21 = Y1 Similarly, since E2 = 0, 1, if T(E2) = X2, Y2 you get: T12 = X2 and T22 = Y2

16. How to find the vector with a given magnitude and same direction?

Treat it as a geometry problem. Any vector passing through (3, 3) will have a slope of 1. So, you need a line of slope 1 such that it's length is 5. Let the vector be (x, y). You know that x = y, so we can write the vector as (x, x). From the Pythagorean Theorem, 5 = sqrt(x^2 + x^2) = sqrt(2x^2) 2x^2 = 25 x^2 = 12.5 x ~ 3.54 So, 3.54, 3.54

17. why to use the normal vector for constructing the plane equation in 3-D Calculus?

In two dimensions, we can take the normal vector: a , b and turn it into the slope: - b / a However, if you have a normal vector in 3D: a , b , c You can't turn it into a fraction - there are three things! You can't have a parallel vector to a plane, since a plane goes many directions. It's also important to remember that the normal vector is more like the slope, whereas the tangent plane is like the tangent line. Alot of 2D calculus, in high level math, is done with vector methods instead, even though they aren't as necessary.

18. How to display a vector with out the disp function?

omg i learning that stuff at school right now and i have no idea what it is

19. How do you determine magnitude and direction of a force that completely cancels a resultant force vector?

same magnitude; opposite direction

20. where and how can I find cheap vector artists?

check out deviantart.com find a couple artists you like email them and try and make some contacts most are young up and coming artists Also, post an add on Craigslist, you will get responses. You can also try posting fliers at a local art school.