How do I find the gradient of a 3D vector?

[Saracen Rue]

Let's say I have point A(2, 6, 0) and B(3, -1, -2) and wanted to find the gradient of the vector joining these two points. I know how to find the vector representing the line joining these points:

OA = 2i + 6j , OB = 3i - j - 2k

AB = AO + OB
AB = -OA + OB
AB = -(2i + 6j) + 3i - j - 2k
AB = -2i - 6j + 3i - j - 2k
AB = i - 7j - 2k

But I don't know how I could find the gradient of this vector. Any help will be much appreciated.

 

[Raffaele]

You can get [itex]\vec{AB}[/itex] instantly as
[tex]\vec{AB}=\vec{OB}-\vec{OA}=(3-2)\vec{i}+(-1-6)\vec{j}+(-2-0)\vec{k}=\vec{i}-7\vec{j}-2\vec{k}[/tex]

gradient of [itex]\vec{AB}[/itex] means null vector, as [itex]\vec{AB}[/itex] is constant. Maybe you wanted the module, the norm
[tex]||\vec{AB}||=\sqrt{1^2+(-7)^2+(-2)^2}=\sqrt{54}=3\sqrt{6}[/tex]

 

[Saracen Rue]

You can get [itex]\vec{AB}[/itex] instantly as
[tex]\vec{AB}=\vec{OB}-\vec{OA}=(3-2)\vec{i}+(-1-6)\vec{j}+(-2-0)\vec{k}=\vec{i}-7\vec{j}-2\vec{k}[/tex]

gradient of [itex]\vec{AB}[/itex] means null vector, as [itex]\vec{AB}[/itex] is constant. Maybe you wanted the module, the norm
[tex]||\vec{AB}||=\sqrt{1^2+(-7)^2+(-2)^2}=\sqrt{54}=3\sqrt{6}[/tex]

Thanks for that first part, it should make things a little easier ^_^

I think I may have asked the question wrong. Let me try to rephrase it; it you have[tex]\vec{AB}[/tex], how would you find the gradient of the line AB.

 

[HallsofIvy]

What do you mean by "the gradient" of a 3D line? A line in 2 dimensions makes a single angle with the x-axis (and its angle with the y-axis is the conjugate of that) so we can take the tangent of that angle as the single number representing its direction, its "gradient".

But a line in 3 dimensions makes three different angle with the coordinate axes, the "direction cosines" for the line (and the sum of the squares of those cosines is 1) so we cannot have a single number that tells us the direction of the line. The best we can do is take those three direction cosines as components of a 3 d vector.

In particular, while a 3D vector may be a gradient vector for a line, a 3D vector does NOT "have" a gradient.

 

[Saracen Rue]

What do you mean by "the gradient" of a 3D line? A line in 2 dimensions makes a single angle with the x-axis (and its angle with the y-axis is the conjugate of that) so we can take the tangent of that angle as the single number representing its direction, its "gradient".

But a line in 3 dimensions makes three different angle with the coordinate axes, the "direction cosines" for the line (and the sum of the squares of those cosines is 1) so we cannot have a single number that tells us the direction of the line. The best we can do is take those three direction cosines as components of a 3 d vector.

In particular, while a 3D vector may be a gradient vector for a line, a 3D vector does NOT "have" a gradient.

Oh okay, thank you. If that's the case, is it possible to determine if two 3D vectors are parallel to each other or not?

 

[JonnyG]

Is one a scalar multiple of the other?

 

[HallsofIvy]

As JonnyG suggests, two vectors, in any dimension, are "parallel" if and only if one is a multiple of the other. (Sometimes the word "anti-parallel" is used if that multiple is negative.)