Let $\vec{a} = i + 2 j + 3 k$ and $\vec{b} = 2 i + 5 k$. For which value of $t$ when $ −2 ≤ t ≤ 2 $ holds the length of the vector $\vec{c_t} = t\vec{a} + (1 − t) \vec{b}$ is as small as possible?
How should one approach this? The minimum would be when $\vec{c_t}$ is perpendicular to some vector $\vec{d} = \vec{b} - \vec{a}$ or is there something else Im not seeing?
On
Hint. The vector $c$ is a linear combination of the vectors $a$ and $b.$ Perform the substitution and find the length of $\vec c.$ This gives you a quadratic in $t,$ which you can minimize in the interval $[-2,2].$
On
Write $\vec{c_t} = t (\vec{a}- \vec{b} )+ \vec{b}$. Then,
$$\begin{align} |\vec{c}_t|^2 &= |\vec{a} - \vec{b}|^2t^2 + 2 (\vec{a}- \vec{b} )\cdot \vec{b}\> t+|\vec{b}|^2 \\ &= |\vec{a} - \vec{b}|^2 \left(t + \frac{(\vec{a}- \vec{b} )\cdot \vec{b} }{|\vec{a} - \vec{b}|^2}\right)^2 -\frac{((\vec{a}- \vec{b} )\cdot \vec{b} )^2}{|\vec{a} - \vec{b}|^2} +|\vec{b}|^2 \\ &\ge\frac{ (|\vec{a} - \vec{b}||\vec{b}|)^2 -((\vec{a}- \vec{b} )\cdot \vec{b} )^2 }{|\vec{a} - \vec{b}|^2} = |\vec{c}_t|_{min}^2\\ \end{align}$$
Hint: Write out the magnitude as a function of $t$ and then differentiate to find the maximum.