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I am given the point $P(1,0,6)$ and I need to find a line that goes through $P$ and is also perpendicular to $x+3y+z=5$. Background info: I've gotten the help I needed now but when I started I was unaware about the method to find a vector perpendicular to the plane and how to find a line equation which uses that vector and goes through the point P.

ShaneBird
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    Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Fly by Night Sep 01 '14 at 18:22
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    You should change the end of the title from "line" to "plane", because: A. $x+3y+z=5$ is a plane. B. If it was indeed a line, then there were infinitely many solutions. – barak manos Sep 01 '14 at 18:23
  • Well, if it was a line, there would still only be one solution. But you are correct that it is a plane. @barakmanos – Thomas Andrews Sep 01 '14 at 18:25
  • @ThomasAndrews: Haha, just realized that, using one finger for a line and another for a point... I was probably thinking of the "opposite" more or less (given a line in a $3D$ space, there are infinitely many lines that are perpendicular to it)... Thanks :) – barak manos Sep 01 '14 at 18:28
  • Thanks Fly by Night, I'm new here. – ShaneBird Sep 01 '14 at 18:35

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Hint: the normal vector to the plane $ax+by+cz=d$ is $(a,b,c)$. To find a line you only need a point and a direction. Now you finish it.

Ivo Terek
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  • Thanks! I appreciate you now giving me the whole solution. – ShaneBird Sep 01 '14 at 18:33
  • I'm pretty sure the normal vector is perpendicular to the plane, so I'm just going to figure out a formula for a line which goes through P in the direction of the normal vector. – ShaneBird Sep 01 '14 at 18:35
  • And my final answer is r = <1,0,6> + t<1,3,1> – ShaneBird Sep 01 '14 at 18:40
  • Your answer is correct. As I said, you need only a point and a direction. The point is given, and it is straightforward to find the direction. In general, the line passing through $P$ with direction $v$ is given by $X = P + tv$, with $t \in \Bbb R$. – Ivo Terek Sep 01 '14 at 18:43
  • Thanks again, it's a little counter intuitive for me to see that <a,b,c> is the normal vector, I'd imagine it'd be something like <-1/a,-1/b,-1/c>, but I'll remember that. – ShaneBird Sep 01 '14 at 18:44
  • Let's think of planes through the origin, for simplicity, that is, $ax + by + cz = 0 $. The LHS can be written as the dot product $(a,b,c) \cdot (x,y,z)$. So, a point $(x,y,z)$ satisfies the equation iff it is orthogonal to $(a,b,c)$. – Ivo Terek Sep 01 '14 at 18:47
  • Nice! I really like seeing those connections, it helps me understand the material on a deeper level. What is our d is not equal to zero? Like in the problem above, how come that still produces an orthogonal vector? – ShaneBird Sep 01 '14 at 18:52
  • If $d \neq 0$, take a point $P$ in the plane. The point $(x,y,z)$ is in the plane iff the vector $((x,y,z) - P)$ is contained in the plane, and so, is orthogonal to $(a,b,c)$. You can determine $d$, by using that: $(a,b,c) \cdot ((x,y,z) - P) = 0 $ implies $(a,b,c) \cdot (x,y,z) = (a,b,c) \cdot P$. The intepretation for the LHS is my previous comment, and $(a,b,c) \cdot P $ is our $d$. – Ivo Terek Sep 01 '14 at 18:58