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The distance between two vectors is the magnitude of their difference. Find the value of $t$ for which the vector ${v} = \begin{pmatrix} 2 \\ -3 \\ -3 \end{pmatrix} + \begin{pmatrix} 7 \\ 5 \\ -1 \end{pmatrix} t$ is closest to ${a} = \begin{pmatrix} 4 \\ 4 \\ 5 \end{pmatrix}.$

I'm kinda confused on the magnitude part. I got $\sqrt {901}$ for that but that doesn't sound right. I looked at the other links but didn't really get it.
Can someone please help me understand this?

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  • $\begingroup$ $\sqrt{9}01$ ? You mean $\sqrt{901}$, right ? $\endgroup$ – A---B Nov 1 '17 at 0:14
  • $\begingroup$ yea sorry, i''m not good at mathjax $\endgroup$ – ninjagirl Nov 1 '17 at 11:20
  • $\begingroup$ You need to use curly braces around the number so that it appears wholly inside the square root symbol. $\endgroup$ – A---B Nov 1 '17 at 11:43
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Consider the line $\ell$ in $\mathbb{R}^3$ parametrized by $$v(t) = \begin{pmatrix} 2 \\ -3 \\ -3 \end{pmatrix} + \begin{pmatrix} 7 \\ 5 \\ -1 \end{pmatrix} t, $$ and consider the point $$ {a} = \begin{pmatrix} 4 \\ 4 \\ 5 \end{pmatrix}. $$

Find the value of $t$ for which the point $v(t)$ on the line is closest to the point $a$.

$\textbf{Method 1}$. Let $p,q$ be two points in $\mathbb{R}^3$ and let $d(p,q) = || p-q||$ be the distance function between $p$ and $q$, where $||\cdot ||$ is the Euclidean norm.

Define $f(p,q)=d(p,q)^2$. Then thinking of $v(t)$, for a fixed $t$, as some point on the line $\ell$, $$ f(v(t),a)= \left(\sqrt{(2+7t-4)^2+(-3+5t-4)^2+(-3-t-5)^2}\right)^2 = 75t^2 - 82t + 117. $$ Since $t$ is arbitrary, now think of $t$ as a free parameter. So $f(v(t),a)$ is a function of $t$. Defining $g(t):= f(v(t),a)$, take the derivative of $g(t)=75t^2 - 82t + 117$ and set it equal to zero to obtain local min and local max (since $g$ is concave up, the point you get will be a global min): $$ g'(t) = 150 t-82=0 \mbox{ implies } t = \frac{41}{75} \approx \boxed{0.546667}. $$ We also see that the point on the line $\ell$ that is closest to the point $a$ is $$ v(0.546667) = \left( \frac{437}{75}, -\frac{4}{15}, -\frac{266}{75} \right) \approx \left( 5.82667, -0.266665, -3.54667 \right). $$

$\textbf{Method 2}$. Let $b=(2,-3,-3)$, a point on the line $\ell$, and let $w$ be the vector $$ w = a-b = (4,4,5)-(2,-3,-3) = \langle 2,7,8\rangle. $$ Then letting $u=\langle 7,5,-1\rangle$, a directional vector of the line $\ell$, the projection of the vector $w$ onto $u$ is: $$ \begin{align*} \text{proj}_{u}w &= \frac{u\cdot w}{u\cdot u}u \\ &= \frac{7(2)+5(7)+(-1)(8)}{7(7)+5(5)+(-1)(-1)} \langle 7,5,-1\rangle \\ &= \frac{14+35-8}{49+25+1} \langle 7,5,-1\rangle \\ &= \frac{41}{75}\langle 7,5,-1\rangle. \end{align*} $$ So the point on the line $\ell$ that is closest to the point $a$ is: $$ \begin{align*} v\left(\frac{41}{75}\right) &= ( 2,-3,-3) +\frac{41}{75} ( 7,5,-1 ) \\ &= \left( \frac{437}{75}, -\frac{4}{15}, -\frac{266}{75} \right) \\ &\approx \left( 5.82667, -0.266665, -3.54667 \right), \end{align*} $$ and $t$ must equal $$ t=\frac{41}{75} \approx \boxed{0.546667}. $$ $\textbf{Remark}$. It is worth noting that the distance between points $a$ and $v(0.546667)$ is $$ \begin{align*} ||\text{ortho}_{u}w || &= || w-\text{proj}_u w || \\ &= \Bigg|\Bigg| \langle 2,7,8\rangle -\frac{41}{75} \langle 7,5,-1 \rangle \Bigg|\Bigg| \\ &= \Bigg|\Bigg| \left( -\frac{137}{75} , \frac{64}{15}, \frac{641}{75} \right) \Bigg|\Bigg| \\ &= \sqrt{\frac{7094}{75}} \\ &\approx 9.72557. \end{align*} $$

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$$v-a = \begin{pmatrix} 2 \\ -3 \\ -3 \end{pmatrix} + \begin{pmatrix} 7 \\ 5 \\ -1 \end{pmatrix} t -\begin{pmatrix} 4 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 2+ 7t-4 \\-3+ 5t -4\\ -3-1t-5 \end{pmatrix} = \begin{pmatrix} -2+ 7t \\-7+ 5t\\ -8-1t \end{pmatrix} $$

The magnitude of $v-a$ is the square root of $(-2+ 7t)^2 + (-7+ 5t)^2 + (-8-1t)^2 $. Do you know how to minimise this expression?

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Write the vector form of the difference between the two vectors. Then find an expression in $t$ representing the square of the magnitude of the difference vector. If you minimize the square of the distance, you also minimize the distance. Find the value of $t$ which minimizes the square of the distance using the first derivative of the expression.

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Vector $v$ can be written as $$\begin{pmatrix} 2 +7t \\ -3 +5t\\ -3 -1t\end{pmatrix}$$

Then the distance between $a$ and $v$ is $$\sqrt{(2+7t-4)^2+(-3+5t-4)^2+(-3-t+5)^2} = \sqrt{(7t-2)^2+(5t-7)^2+(-t+2)^2} = \sqrt{49t^2-28t+4+25t^2-70t+49 + t^2-4t+4} = \sqrt{75t^2-102t+57}$$

We can't solve this radical explicitly, but we can find the zeros using the quadratic equation or graphing.

It turns out that this function has no zeros. Therefore, we look for the vertex, which is the minimum, and occurs at $t=\frac{--102}{2 \cdot75} = 102/150 = 0.68$

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  • $\begingroup$ wait, the t you wrote isn't correct... $\endgroup$ – ninjagirl Nov 1 '17 at 11:18
  • $\begingroup$ What value of t do you have as correct? $\endgroup$ – John Lou Nov 1 '17 at 14:31
  • $\begingroup$ Then how do you know I'm incorrect? I'm just asking because if I knew the right answer it might help me work backwards. $\endgroup$ – John Lou Nov 2 '17 at 1:42
  • $\begingroup$ i plugged the t you have into my original equation and got (6.76;0.4;-3.68) $\endgroup$ – ninjagirl Nov 2 '17 at 2:18
  • $\begingroup$ what is the correct vector? $\endgroup$ – John Lou Nov 2 '17 at 2:30

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