How to Resolve a Vector into Parallel and Perpendicular Components?

[songoku]

Homework Statement

Resolve the vector 6i+2j-2k into two vectors, one parallel and another perpendicular to i+j+k

Homework Equations

$$a\cdot b = 0 \; \text{for two perpendicular vectors}$$

$$a=\lambda \;b\;\text{for parallel vectors}$$ , $$\lambda = \text{parameter}$$

The Attempt at a Solution

I have no idea to start. How to resolve one component of vector into two components ?

Thanks

 

[CompuChip]

Have you learned about the geometrical meaning of dot product?
I.e. if I drew v = 6i+2j-2k and w = i+j+k for you, could you explain to me how v · w appears in the picture?

 

[Дьявол]

Have you learned about the geometrical meaning of dot product?
I.e. if I drew v = 6i+2j-2k and w = i+j+k for you, could you explain to me how v · w appears in the picture?

I don't think that he can give you geometrical interpretation because dot product is just a scalar and not vector.
Although, he can give you the scalar projection of v onto w.

And do you mean to resolve 6i+2j-2k = c + d ?

If so, let c be the vector parallel to and d perpendicular to the vector (1,1,1) i.e i+j+k

(c_x,c_y,c_z)=$\lambda$(1,1,1)

and do you know what d will equal to? What are the conditions of the task?Regards.

 

[HallsofIvy]

Homework Statement

Resolve the vector 6i+2j-2k into two vectors, one parallel and another perpendicular to i+j+k

Homework Equations

$$a\cdot b = 0 \; \text{for two perpendicular vectors}$$

$$a=\lambda \;b\;\text{for parallel vectors}$$ , $$\lambda = \text{parameter}$$

The Attempt at a Solution

I have no idea to start. How to resolve one component of vector into two components ?

Thanks

Okay, so a vector parallel to i+ j+ k must be $\lambda i+ \lambda j+ \lambda k$. Suppose ai+ bj+ ck is the vector perpendicular to that. Then you have $(ai+ bj+ ck)\cdot(\lamba i+ \lambda j+ \lambda k)= a\lambda+ b\lambda+ c\lambda$$= \lambda(a+ b+ c)= 0$ and $(ai+ bj+ ck)+ (\lambda i+ \lambda j+ \lambda k)$$= (a+\lambda)i+ (b+\lambda)j+ (c+ \lambda)k$$= 6i+ 2j- 2k$. That gives you four equations to solve for a, b, c, and $\lambda$.

 

[songoku]

Hi CompuChip, Дьявол, and Mr. HallsofIvy

I get it now. Sorry, but I have another simple question. What is the meaning of dot product ?

It's easier for me to imagine cross product. If we cross two vectors, we will get third vector that is perpendicular to the previous two vectors.

But, what about dot product ? If we dot 2 vectors, we get a numerical value, What does the numerical value represent?

Thanks

 

[mg0stisha]

As I was taught, the dot product is where you only consider the part of the second vector being multiplied that is parallel to the first vector.

 

[Дьявол]

Hi CompuChip, Дьявол, and Mr. HallsofIvy

I get it now. Sorry, but I have another simple question. What is the meaning of dot product ?

It's easier for me to imagine cross product. If we cross two vectors, we will get third vector that is perpendicular to the previous two vectors.

But, what about dot product ? If we dot 2 vectors, we get a numerical value, What does the numerical value represent?

Thanks

http://upload.wikimedia.org/wikipedia/commons/thumb/3/3e/Dot_Product.svg/300px-Dot_Product.png

And the dot product is A • B = |A| cos(θ) |B|

|A| cos(θ) is the scalar projection of A onto B.

So you got the part down there, just you multiply it with the magnitude of B.

Regards.

 

[songoku]

Hi Дьявол and mg0stisha

Wow, now I get the meaning of dot product. Thanks a lot to all of you ! (CompuChip, Дьявол, MR. HallsofIvy, mg0stisha) ^^