Let R^3 have the Euclidean inner product, and suppose that u =(1,1,1) v =(6,7,-15.Find the value of k for which ||ku+v||=13

Question

Let R^3 have the Euclidean inner product, and suppose that  u =(1,1,1) v =(6,7,-15.Find the value of k for which ||ku+v||=13

anonymous · Answer

any idea, friends, I got negative value for k^2. stuck!!!

anonymous · Answer

Why don't you brute force it?

anonymous · Answer

what you mean?

anonymous · Answer

$$
||ku+v||=13\
||ku+v||^2=13^2\
<ku+v,ku+v>=13^2
$$

anonymous · Answer

so, it is not ||ku+v||^2 = ||ku||^2 +||v||^2= k^2||u||^2 +\v||^2 =13^2?

anonymous · Answer

I think they want you to use inner products here.

anonymous · Answer

ohoh, see, can you see my weak logic at breaking down the ||ku||^2

anonymous · Answer

Also: $$
||a+b|| \neq ||a||+||b||
$$

anonymous · Answer

yes, you are right, but square is another topic.

anonymous · Answer

Consider if $a=-b$

anonymous · Answer

yes, you are right, my bad. if they are orthogonal, then sqr is applied

anonymous · Answer

ku = (k,k,k) , right? break down follow that way, right?

anonymous · Answer

thanks a lot, friend, I think I can get the answer from your way. got it, k != negative value anymore.