Finding Vector Equation for a Perpendicular Line Passing Through a Point

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SUMMARY

The discussion focuses on finding the vector equation of a line that passes through the point (6,0,1) and is perpendicular to the plane defined by the equation x + 4y + 6z = 5. The normal vector to the plane is identified as <1, 4, 6>, derived from the coefficients of the plane equation. The solution involves using this normal vector as the direction vector for the line, allowing for the straightforward formulation of the line's vector equation.

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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.
 

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