Finding Vector Equation for a Perpendicular Line Passing Through a Point

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Homework Statement


Find the vector equation with parameter T that passes through the point (6,0,1) and is perpendicular to the plane x+4y+6z = 5






The Attempt at a Solution


So I'm looking for the direction numbers n (a,b,c) for whatever direction is perpendicular to the plane given x+4y+6z = 5

What if I try to take the dot product of i+4j+6k with xi+yj+zk and solve for when it equals zero. The problem with doing that is there are more than one possible answers that solve that. Also since x+4y+6z = 5 is the linear equation of a plane I'm not even sure I'm thinking about this right at all X_x
 
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PsychonautQQ said:

Homework Statement


Find the vector equation with parameter T that passes through the point (6,0,1) and is perpendicular to the plane x+4y+6z = 5






The Attempt at a Solution


So I'm looking for the direction numbers n (a,b,c) for whatever direction is perpendicular to the plane given x+4y+6z = 5

What if I try to take the dot product of i+4j+6k with xi+yj+zk and solve for when it equals zero. The problem with doing that is there are more than one possible answers that solve that. Also since x+4y+6z = 5 is the linear equation of a plane I'm not even sure I'm thinking about this right at all X_x

One direction vector for the line is <1, 4, 6>, which I got by inspection. A normal to the plane Ax + By + Cz = D is the vector <A, B, C>. Once you know the direction of a line and a point on it, it's straightforward to get the equation of the line.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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