How can I find the component of a cross product along a given vector direction?

[Noone1982]

I'm using wangsness EM book. I keep doing ch 1 #5 and keep getting 19, despite the back of the book saying -14.4

A = <2,3,-4>
B = <-6,-4,1>

AxB = <-13, -22, 10> = D

Find the component of AXB along the direction of Vector C

C = <1,-1,1>

Now I am using

C•D = CDcos(theta)

Im getting 66.4º

Now, to get the component I used:

Magnitude = DCcos(66.4)

Does this method make sense?

 

[Doc Al]

To get the component of D in the direction of C, you need to take the dot product of D with the unit vector in the direction of C. (Divide \vec{D} \bullet \Vec{C} by the magnitude of C.)

(Also: Why mess around with angles when you have the components?)

 

[Noone1982]

Thanks for the help, I tend to do things the hard way. I have another question:

"A family of hyperbolas in the xy plane is given by u = xy. Find the gradient of u. Given vector A = 3i + 2j + 4k find the component of A in the direction of gradient u at the point u = 3 and x = 2"

Since we know that u = 3 and x = 2, we can gather that the point of the hyperbola is x =2 and y = 3/2

Lets call this vector E

so E = 2i + (3/2)j + 0k

This problem is similar to the previous one I asked, finding the component. So I take it since I want to find the component of A in the direction of Grad U, I will need to make E into a unit vector

U = (4/5)i + (3/5)j + 0k

Now I presume I dot U and A?

That would create

U•A = (8/5) +(9/10) = 5/2

 

[Noone1982]

Ignore the previous post. I was experiencing brain drain...