# Problem with vector operator

moatasim23
Why do we use the coordinates of r in terms of x,y,z?Why dont we express coordinates of A in x,y,z?

#### Attachments

• Physics.png
19.6 KB · Views: 431

Homework Helper
A is expressed in terms of x y and z.

Homework Helper
It's a matter of notation: we're just giving names to the three components of both vectors.
You can replace them by ##r_1, r_2, r_3##, if that makes you feel any better. In general, if v is a vector, it is customary to denote its components by v1, v2, v3. However, if r is the position vector, then (x, y, z) is also quite common.

Also note that though the fact that one is named r hints that it comes from a physical application in which a position vector is involved, the mathematical identity actually holds for any two vectors u, v.

moatasim23
No its not..here at least

moatasim23
It's a matter of notation: we're just giving names to the three components of both vectors.
You can replace them by ##r_1, r_2, r_3##, if that makes you feel any better. In general, if v is a vector, it is customary to denote its components by v1, v2, v3. However, if r is the position vector, then (x, y, z) is also quite common.

Also note that though the fact that one is named r hints that it comes from a physical application in which a position vector is involved, the mathematical identity actually holds for any two vectors u, v.

If we use r1,r2,r3 then how would the vector operator operator operate on it?Like it didnt in A when we used A1,A2,A3.

Staff Emeritus
Homework Helper

Homework Helper
No its not..here at least
I'm sorry - the example in your attachment very clearly states that

A=A1i+A2j+A3k

That means that
- the x component of A is A1,
- the y component of A is A2,
- the z component of A is A3.

Therefore: A is resolved in terms of x, y, and z.

What did you think it meant?

$$\mathbf u \times \mathbf v = (u_2 v_3 - u_3 v_2) \mathbf i + (u_3 v_1 - u_1 v_3) \mathbf j + (u_1 v_3 - u_3 v_1) \mathbf k$$