Finding distance using vector components

AI Thread Summary
To find the distance between two points represented by vector components r1 = (0.10)î + (-0.80)j and r2 = (-0.90)î + (0.10)j, one must first determine the vector connecting these points, which is r = r2 - r1. The distance is then calculated as the magnitude of this vector, using the formula distance = √(r·r). It's important to correctly add the i and j components separately to avoid errors. The discussion concludes with the user successfully finding the correct answer after clarifying the calculation method.
LadyTwi
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Homework Statement



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With reference to Fig. 3 (not accurate), where the origin is at the centre of the image, if r1 = (0.10)î+(-0.80)j hat, and r2 = (-0.90)î+(0.10)j hat, what is the distance between the dots?

Homework Equations



Unsure.

The Attempt at a Solution



I'm not so good with these i hat and j hat vector components. I've tried using (x2-x1) + (y2-y1) but that didn't give the correct answer. I'm not really sure what I can do with the two equations.
 
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here's some reading for you to do that should help!
http://www.physics.uoguelph.ca/tutorials/vectors/vectors.html
 
That is a helpful link, however I still can't get the answer. XD Do I have to find the angle between r1 and r2 and use that to find the distance or does it have to do with the addition of the i hat components and the j hat components?
 
the beauty of unit vectors is that it already takes the angle into consideration! all you have to do in this case is add the i vectors with i vectors and js with js.
 
I've tried that but BLS/CAPA is still telling me that my answer is wrong. Maybe I'm entering it wrong...?
 
The distance between two points is the magnitude of the vector connecting these two points. In your problem this vector is \vec{r} = \vec{r_{2}}-\vec{r_{1}}. To encounter the magnitude of this vector, all we have to do is calcute the scalar product of it with itself and take the square root. So,
distance = \sqrt{\vec{r}\dot\vec{r}}.
 
Thank you! I have the answer now. =D
 
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