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Find a vector equation for the line through the point $(3,-8,-8)$ perpendicular to these vectors $u=\langle 2,2,-1\rangle$ and $v=\langle -9,-8,-3\rangle$.

I'm fairly new to vector equations. Would I find a component vector of $u$ and $v$ and write the parametric equation for it and then use the formula $r(t) = r_0 + tv$?

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    $\begingroup$ The easiest way to find a vector in $\mathbb{R}^3$ that is perpendicular to two vectors $u$ and $v$, is to compute $u\times v$ (the cross product) $\endgroup$ – robjohn May 13 '15 at 23:32
  • $\begingroup$ @robjohn why the cross product? $\endgroup$ – Elsa May 13 '15 at 23:33
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    $\begingroup$ $u\times v$ is perpendicular to both $u$ and $v$. This can be seen from the triple products $u\times v\cdot u=0$, which is the determinant of a matrix with columns of $u$, $v$, and $u$, and $u\times v\cdot v=0$, which is the determinant of a matrix with columns of $u$, $v$, and $v$. Both of these determinants are $0$ since each matrix has two identical columns. $\endgroup$ – robjohn May 13 '15 at 23:36
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    $\begingroup$ @robjohn my textbook defines the cross product $u\times v$ to be a vector perpendicular to both $u$ and $v$ and with magnitude $|u||v|\sin (\widehat{u,v})$. $\endgroup$ – user26486 May 14 '15 at 0:39
  • $\begingroup$ @user31415: yes, but somehow, it seemed a bit artless to say, "because it's defined that way" :-) However, yes; it is defined that way. $\endgroup$ – robjohn May 14 '15 at 5:37
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Here's a sketch.

The best way to approach this is by thinking about what fundamentally defines a line: a vector and a point. In this case, we have the point, but instead of a vector in the direction of the line, we have two vectors perpendicular to the line. However, if we take the cross product of these two vectors, it will by definition be perpendicular to both of them and thus be in the direction of the line itself. Taking the cross product gives $\langle -14, 15, 2\rangle$. Then, the line can be parametrized using standard techniques as

$$\langle 3,-8,-8\rangle+t\cdot \langle-14, 15, 2\rangle=\langle 3-14t, -8+15t, -8+2t\rangle.$$

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  • $\begingroup$ Ohhh, ok so the cross product gives a vector perpendicular to BOTH u and v? So that makes it the "opposite" and makes it parallel to the line we want to find $\endgroup$ – Elsa May 13 '15 at 23:36
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    $\begingroup$ Yes, the cross product of two vectors is always perpendicular to both. $\endgroup$ – Samir Khan May 13 '15 at 23:37
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    $\begingroup$ < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. $\endgroup$ – Zev Chonoles May 13 '15 at 23:39
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    $\begingroup$ I wasn't aware of that - thanks for pointing it out. $\endgroup$ – Samir Khan May 14 '15 at 0:16
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Let $P = (3,-8,-8)$ be the point given above, $u$ and $v$ be defined as above, and $O = (0,0,0)$ (origin). Let $O \rightarrow P$ denote the distance from the origin to $P$. The cross product of two vectors results in a perpendicular vector. Hence $u \times v = (-14,15,2)$. Thus the equation for the line through the point $P$ perpendicular to $u$ and $v$ is

$r(t) = O\rightarrow P + t(-14,15,2) = (3,-8,-8) + t(-14,15,2)$

where $O \rightarrow P = P - O$.

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