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?
Sure. So what then are the components of the resulting vector?Ok so the denominator for (b) should be 23.065.
I don't understand that. The result of a cross product should be a vector.For (c) 8UxUyUz.
Part d contains a cross product followed by a dot product. The result should be a single scalar value.For (d) -26UxUz.
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##.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.
Show your work; that's not the value I'm seeing.For part 1.3 A=48.489
For two vectors A and B, ##A \cdot B = |A||B| cos(\theta)##AB dot AC = AB dot AC cos(theta) is how I find the angle right?
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?
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
How do I view the edits it still looks the same on my comp and BTW I cant read the code in #6
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