Finding the length of vectors

[RM5152]

Hello ? would someone be able to tell me if I have solved this question correctly? The last part of it 4(c) makes me think I must be missing something but I’m just not sure. Thanks in advance ..

 

[topsquark]

Hello ? would someone be able to tell me if I have solved this question correctly? The last part of it 4(c) makes me think I must be missing something but I’m just not sure. Thanks in advance ..

I presume you are trying to calculate the magnitude in part c). The magnitude of a vector is $|| \textbf{v} || = \sqrt{ \textbf{v} \cdot \textbf{v} }$,

-Dan

 

[Steven G]

y*y is not y^2
xy is not x*y
Use correct math symbols.

 

[topsquark]

y*y is not y^2
xy is not x*y
Use correct math symbols.

Actually, in Physics $\textbf{v} \cdot \textbf{v}$ is often denoted as $v^2$. As long as you have context you should be fine.

-Dan

 

[Deleted member 4993]

Actually, in Physics $\textbf{v} \cdot \textbf{v}$ is often denoted as $v^2$. As long as you have context you should be fine.

-Dan

Don't tell Steven about Physics - it is too uncertain for him.

 

[pka]

We are give that $\bf x~\&~y$ are unit vectors.
We also know that $\|2{\bf x}-9{\bf y}\|^2=(2{\bf x}-9{\bf y}){\Large\cdot}(2{\bf x}-9{\bf y})=4{\bf x}{\Large\cdot}{\bf x}-36{\bf x}{\Large\cdot}{\bf y}+81{\bf y}{\Large\cdot}{\bf y}=???$


$$$$$$

 

[Steven G]

Don't tell Steven about Physics - it is too uncertain for him.

Hey, stop talking about me!

 

[RM5152]

Thanks for all the help! Okay .. I think I got it .. Is this correct?

 

[topsquark]

Thanks for all the help! Okay .. I think I got it .. Is this correct?

You forgot to take the square root at the end.

-Dan