some vector problem (1 Viewer)

[fatmuscle]

1.
A is a vector
C is a scalar

Ho is CA related to A geometically
How do you find CA algebraically


2.
Prove that when you divide a vector by it's own length, the result is a unit vector



no, i don't have the answers
nor is this homework

 

[Calculon]

CA relates to A by having the same principle argument

 

[fatmuscle]

ta, next part and next Q?

or give me a hint on Q2
i can't think of a starting point from the top of my head

 

[McLake]

Moving ...

 

[Affinity]

1)

cA is a scalar multiple of A, so geometrically, it's parallel to A (except in the degenerate case where c = 0), cA's magnitude (length) is equal to |c|*length of A

to find cA algebraically, just multiply each element of A by c
eg A= (w,x,y,z) then cA = (cw,cx,cy,cz)

2.) there may be a simpler answer depending on the results you can assume/use

let the vector be (x_1,x_2,...,x_n)

and Length of vector = L = sqrt((x_1)^2 + ... + (x_n)^2)

if you divide the the vector by it's length (or multiply by L^-1)

it becomes

(x_1/L , x_2/L,....,x_n/L)

calculate it's length

= sqrt[ (x_1)^2/L^2 + ... + (x_n)^2/L^2 ]

= sqrt[ L^2 / L^2 ]

= 1

and therefore it's a unit vector