Finding the Equation of a Plane Perpendicular to a Given Line

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SUMMARY

The discussion focuses on deriving the equation of a plane that is perpendicular to a line defined by points A(0, 1, 1) and B(1, -1, -1) and passes through the point (2, 1, 5). The normal vector to the line, calculated as (0, 1, -1), serves as the normal vector for the plane. The equation of the plane is expressed as ax + by + cz = d, where d can be determined using the coordinates of the point (2, 1, 5). The steps outlined provide a clear method for finding the equation of the desired plane.

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brunette15
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I am trying to find an equation for a plane that passes through the point (2, 1, 5) however is also perpendicular to the line that passes through the points A(0, 1, 1) and B(1,-1,-1).

I am unsure how to begin with this. I have started by finding the normal vector to A and B = (0,1,-1), to find the direction of the plane and don't really know what to do from here.
 
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My suggestion would be:

Step 1: Use the two points $A$ and $B$ on the straight line to determine the direction of the line.
Step 2: Since the plane is perpendicular to the line, the direction of the line is a normal to the plane. If we let $(a,b,c)$ be your computed normal to the plane then the equation of the plane is given by: $ax+by+cz=d$, where $d$ is still unknown.
Step 3: Last step is to find $d$, but that is now straightforward since the plane passes through the point $(2,1,5)$ and thus $2a+b+5c=d$.
 
Siron said:
My suggestion would be:

Step 1: Use the two points $A$ and $B$ on the straight line to determine the direction of the line.
Step 2: Since the plane is perpendicular to the line, the direction of the line is a normal to the plane. If we let $(a,b,c)$ be your computed normal to the plane then the equation of the plane is given by: $ax+by+cz=d$, where $d$ is still unknown.
Step 3: Last step is to find $d$, but that is now straightforward since the plane passes through the point $(2,1,5)$ and thus $2a+b+5c=d$.

Thankyou so much!
 

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