Given $$\vec x= \vec a+t\vec b$$ where $$ \vec a , \vec b $$ are non zero vector for all real number t .

The question asked about the minimum magnitude of x and the answer shown is $$ \frac {-\vec a \cdot \vec b}{\vec b\cdot \vec b}$$

How do i arrived at such conclusion?

I tried to take square but it didn't work out.

  • $\begingroup$ Minimum magnitude must surely be at $t = 0$, giving $\left\| a \right\|$. $\endgroup$ – user328032 Apr 22 '16 at 8:59
  • $\begingroup$ @Benedict Consider $a=(1,0)$, $b=(-1,0)$, or $a=(1,0)$, $b=(-1,1)$, or ... $\endgroup$ – amd Apr 22 '16 at 16:06
  • $\begingroup$ Did you mean to say that the minimum magnitude is at $t=-\frac{\vec a\cdot\vec b}{\vec b\cdot\vec b}$? The way you’ve written the question, it seems like you’re claiming that this is the minimum magnitude itself. $\endgroup$ – amd Apr 22 '16 at 16:09
  • $\begingroup$ Another poorly written question. $\endgroup$ – user328032 Apr 22 '16 at 16:20

We wish to minimise $\vec x\cdot\vec x=(\vec a+t\vec b)\cdot(\vec a+t\vec b)$. We can write that as $\vec b\cdot\vec b(t+\frac{\vec a\cdot\vec b}{\vec b\cdot\vec b})^2+\vec a\cdot\vec a-\frac{(\vec a\cdot\vec b)^2}{\vec b\cdot \vec b}$.

So we minimise by taking $t=-\frac{\vec a\cdot\vec b}{\vec b\cdot\vec b}$. The minimum value of $|\vec x|$ is thus $\sqrt{\vec a\cdot\vec a-\frac{(\vec a\cdot\vec b)^2}{\vec b\cdot \vec b}}$.

Or just draw a diagram and the answer is fairly immediate.


I can imagine that $\vec{x}$ is the diagonal vector formed by $\vec a$ and $ t\vec b$. Just look at $\vec b$ as a unit vector $\vec e_b$ since it is normalized as $\frac {\vec b}{\Vert\vec b\Vert}$.

Note: your answer is not the minimum distance, but what the parameter $t$ should be. If we recall Schwrz Inequality: $$0\leq\Vert\vec a+t\vec b\Vert=\Vert\vec a\Vert^2+2\Re{\langle\vec a,t\vec b\rangle}+\vert t\vert^2\Vert\vec b\Vert^2$$ If you plug $t$ in above, then $$\Vert\vec a\Vert^2-2\frac{\vert\langle\vec a,\vec b\rangle\vert^2}{\Vert\vec b\Vert^2}+\frac{\vert\langle\vec a,\vec b\rangle\vert^2}{\Vert\vec b\Vert^2}=\Vert\vec a\Vert^2-\frac{\vert\langle\vec a,\vec b\rangle\vert^2}{\Vert\vec b\Vert^2}\geq0$$ and we multiply $\Vert\vec b\Vert^2$ to get Schwardz Inequality: $$\Vert\vec a\Vert^2\cdot\Vert\vec b\Vert^2\geq\vert\langle\vec a,\vec b\rangle\vert^2$$

Hope it will be useful...


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