OpenStudy (anonymous):

What is the magnitude of the vector with an initial point of (1,-2) and a terminal point of (2,-5)? So basically I've no idea how to find the magnitude? I know you're supposed to plot the points, I'm just not sure what to do next.

3 years ago
OpenStudy (michele_laino):

please, you have to find the length of the segment joining your two points. So, first calculate the distance between your points, using the forula of the distance, namely: \[d=\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{1})^{2}}\],where x_1,x_2, y_1, y_2, are the coordinates of your points, try please, and write your result

3 years ago
OpenStudy (michele_laino):

@oswin \[d=\sqrt{(2-1)^{2}+(-5-(-2))^{2}}=...\] d is your magnitude

3 years ago
OpenStudy (anonymous):

So d =-2? @Michele_Laino

3 years ago
OpenStudy (michele_laino):

please, note that d is a positive number, retry please!

3 years ago
OpenStudy (anonymous):

Every way I try it, I get a negative number... All I'm doing is the operations... What am I missing?

3 years ago
OpenStudy (michele_laino):

(2-1)^2=1^1=1 is right?

3 years ago
OpenStudy (michele_laino):

\[(2-1)^{2}=1^{2}=1\]

3 years ago
OpenStudy (anonymous):

Hold on, I messed up. The answer would be \[2isqrt{2}\]

3 years ago
OpenStudy (anonymous):

\[2i \sqrt{2}\]

3 years ago
OpenStudy (anonymous):

There we go :P

3 years ago
OpenStudy (michele_laino):

please note d must be real, because you can measure it, \[(-5-(2))^{2}=(-5+2)^{2}=3^{2}=9\]

3 years ago
OpenStudy (michele_laino):

@oswin so d=sqrt(1+9)=...

3 years ago
OpenStudy (anonymous):

So D=10?

3 years ago
OpenStudy (michele_laino):

please, note taht there is a square root

3 years ago
OpenStudy (anonymous):

\[\sqrt{10}\]

3 years ago
OpenStudy (michele_laino):

that's right!

3 years ago
OpenStudy (anonymous):

Yaaaayyy! Thank you :)

3 years ago
OpenStudy (michele_laino):

thank you!

3 years ago
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