0
$\begingroup$

Let $S = \textbf{x}\cdot\textbf{y}+\textbf{y} \cdot \textbf{z}+\textbf{z} \cdot \textbf{x}$. Show the minimum of $S$ over $\textbf{z}$ given that $\textbf{x}$ and $\textbf{y}$ are fixed and linearly independent is attained when $\textbf{z} = \lambda(\textbf{x}+\textbf{y})$. Show also that for this value of $\lambda$

i) $\lambda\leq-\frac{1}{2}$ for any choice of $\textbf{x}$ and $\textbf{y}$

ii) $\lambda = -1$ and $S=-\frac{3}{2}$ when $\textbf{x}\cdot\textbf{y} = -\frac{1}{2}$

Since $\textbf{x}$ and $\textbf{y}$ are fixed, so is $\textbf{x}\cdot\textbf{y}$, so we focus on $\textbf{y} \cdot \textbf{z}+\textbf{z} \cdot \textbf{x} = \textbf{z}(\textbf{x}+\textbf{y}) = \lvert \textbf{z}\rvert \lvert \textbf{x}+\textbf{y}\rvert \cos(\theta)$, which is minimised when $\cos(\theta) = -1$, i.e. when $\textbf{z}$ is in the opposite direction to $\textbf{x}+\textbf{y}$, so when $\textbf{z} = \lambda(\textbf{x}+\textbf{y})$ with $\lambda \leq 0$. I don't see where the $-\frac{1}{2}$ comes in, or how to do ii).

$\endgroup$
  • $\begingroup$ Hint: consider what value of lambda minimizes S given the restricted form of z you have derived. $\endgroup$ – John Polcari Apr 7 at 15:52
  • $\begingroup$ @JohnPolcari why can't $\lambda$ be as small as possible? $\endgroup$ – user112358 Apr 7 at 16:09
  • $\begingroup$ Plug and chug and I believe you will find the result has a minimum value because of how the length of x+y and the dot product x*y interact. $\endgroup$ – John Polcari Apr 7 at 16:12
  • $\begingroup$ Or, more likely, not an actual minimum, but at least some obvious limits (like lambda <= -1/2). $\endgroup$ – John Polcari Apr 7 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.