Finding the Magnitude of a Diplacement Vector

[Love_to_Learn]

1. Cavers spelunked 2.6 km westward, 3.9 southward, and 25 meters upward. What is their displacement vector?

Homework Equations

The Attempt at a Solution

- The formula for the magnitude of a vector which is

sqrt of [(3.6)^2+(2.9)^2+(0.025)^2] ≈4.623 km

 

[jegues]

Let the point P1 lie at (-2600, -3900, 25).


Then OP1 will be a line segement whos magnitude is denoted as follows

sqr[ (-2600)^2 + (-3900)^2 + (25)^2 ] = 4687.28m = 4.69km

Looks about right, assuming it's 3.9km southward.

EDIT: I failed to read the question, The displacement vector would simply be r = -2600i -3900j + 25k, where i, j, k are the unit vectors of x, y, and z.

2nd EDIT: I reread the title and if you're looking for the magnitude of that displacement vector it is indeed 4.69km, sorry for the confusion.

 

[Love_to_Learn]

Alright, then I am not understanding how to find a displacement vector. I thought that the displacement of a vector was the same as its magnitude. How is finding the displacement of a vector somehow different than finding the magnitude of a vector?

 

[denverdoc]

The displacement vector conveys both magnitude and direction. You have the distance, does the problem want direction as well?

 

[jegues]

How is finding the displacement of a vector somehow different than finding the magnitude of a vector?

I think you are mixing up terms, its not a vector with a displacement, its a displacement vector, this is a vector that defines displacement of an object based on another parameter(usually time).

 

[Love_to_Learn]

I think you are mixing up terms, its not a vector with a displacement, its a displacement vector, this is a vector that defines displacement of an object based on another parameter(usually time).

You were exactly correct. That is what I was doing. Thanks for the replies everyone.