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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..

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    $\begingroup$ Compute the square of the norm. You will find a quadratic expression in $k$. From there you could for instance complete the square to find the minimum. Then take the square root. $\endgroup$
    – Julien
    Feb 18, 2013 at 2:33
  • $\begingroup$ actyally the answer depends on the domain of $k$. if $k$ belongs to $\mathbb{N}$ then $\sqrt{6}$ is correct. Else if, $\mathbb{R}$ then the answer will be in $\mathbb{R}$ $\endgroup$
    – tvamsisai
    Feb 18, 2013 at 2:43

3 Answers 3

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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?

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    $\begingroup$ all questions as an answer? $\endgroup$
    – tvamsisai
    Feb 18, 2013 at 2:47
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    $\begingroup$ It's a series of leading hints. The question appears to be related to coursework - and I strongly feel it rather defeats the purpose of giving students such questions to ponder if we are to just give explicit solutions. On the other hand if the original poster asks followup questions of a comment, it usually ends up as a sequence of clarifications which could be better dealt with by improving the answer as needed. If a person can end up providing intermediate answers themselves, it gives them more ownership of the solution. If this somehow contravenes the local community norms I can change it. $\endgroup$
    – Glen_b
    Feb 18, 2013 at 2:54
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    $\begingroup$ I think this is a good approach to an answer. I have gone through this a number of times. Sometimes this is enough for OP. Sometimes OP asks followup, and usually it shows s/he is thinking about it. You can then be more explicit where it seems needed. $\endgroup$ Feb 18, 2013 at 3:10
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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

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  • $\begingroup$ how did you find that equation..? $\endgroup$ Feb 19, 2013 at 7:20
  • $\begingroup$ using the distance formula between point one and point 2 in the euclidean metric. Which is distamce in $\mathbb{R}^3= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}.$ Where point 1 is equal to $(x_1,y_1,z_1) $and point 2 is $(x_2,y_2,z_2)$ in your case point 1 is (0,0,0) and point 2 is (k,k,3-k) $\endgroup$
    – Asinomás
    Feb 19, 2013 at 17:07
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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.

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  • $\begingroup$ @Glen_b: I disagree with the edit. The function $g(k)$ is supposed to be the length of the vector, but it is not what is shown here. $\endgroup$ Feb 18, 2013 at 4:25
  • $\begingroup$ Now we have what was intended. $\endgroup$ Feb 18, 2013 at 4:43
  • $\begingroup$ My apologies Ittay, @RossMillikan $\endgroup$
    – Glen_b
    Feb 18, 2013 at 6:22

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