Shortest length that a vector can have I came to the following question from a past exam: The vector $v = (k, k, 3 − k)$ depends on a variable $k$.
What is the shortest length of the vector $v$ can have? I know that the answer is $\sqrt{6}$, but why...? How do I proceed here to find the result..
 A: The length is going to be $\sqrt{k^2+k^2+(3-k)^2}= \sqrt{3k^2-6k+9}$ now use calculus to find the minimum of the function $3k^2-6k+9$ find the derivative is $6k-6$ so if $6k-6=0$ then $k=1$ substitute to find the minimum distance is 
$\sqrt 6$ as desired. Regards
A: Assuming that $k$ may range over $\mathbb R$, and let $v_k=(k,k,3-k)$, start out by considering the function $f(k)=\|v_k\|$. What you want is to locate the minimum of $f$. Now, work with the definition of the length of a vector. You will see that it contains a square root. Make your life a lot easier by noticing that it is actually the same problem to find the minimum of the function $g(k)=\|v_k\|^2$, so now the square root is gone. Now you just have a rather simple single variable real valued function. Find its minimum using standard calculus techniques. 
A: Are you able to write the formula for the length of the vector? 
What about its square? How does minimizing the square of the length relate to minimizing the length? (How are the arguments related, if at all?)
What kind of a function of $k$ is that square?
