Equation of Normal: M(2,3,1) to Line l

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The discussion focuses on deriving the equation of a normal plane from the point M(2,3,1) to the line defined by the equation \(\frac{x+1}{2}=\frac{y-0}{-1}=\frac{z-2}{3}\). Participants emphasize the need to calculate the direction vector of the line, find the intersection point between the line and the plane, and utilize the cross product to determine the normal vector of the plane. The normal vector shares the same direction as the line's direction vector, which is essential for formulating the plane's equation.

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chmate
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Write equation of normal from point M(2,3,1) into line:l: \frac{x+1}{2}=\frac{y-0}{-1}=\frac{z-2}{3}.

I just have no idea about this problem. Any hint?

Thank you.
 
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What is the relationship between the line and the normal?
 
Calculate the plane normal to the given line and that passes through 2,3,1

Then calculate the point where the given line and the plane intersect.

This point will enable you, together with the (2,3,1), to calculate the direction of the normal to the given line

marlon
 
marlon said:
Calculate the plane normal to the given line and that passes through 2,3,1

Then calculate the point where the given line and the plane intersect.

This point will enable you, together with the (2,3,1), to calculate the direction of the normal to the given line

marlon

Let me understand this. I have to create a plane which passes through that point. Another point is point of intersection. But to find that point i must have a direction vector of that plane and I don't have so can you help me with point of intersection. How can i find it?

So, when I have that point I have to create a vector which passes through point (2,3,1) and point of intesection and cross product of vector direction of line and vector I created will create a normal vector of plane. Am I right?

Thank you.
 
Last edited:
chmate said:
Let me understand this. I have to create a plane which passes through that point. Another point is point of intersection. But to find that point i must have a direction vector of that plane and I don't have so can you help me with point of intersection. How can i find it?

So, when I have that point I have to create a vector which passes through point (2,3,1) and point of intesection and cross product of vector direction of line and vector I created will create a normal vector of plane. Am I right?

Thank you.

Let's take this one step at the time :

First you need to find a plane that contains the point (2,3,1) and that is normal to the given line (your first equation).

Let's start with that.

Do you know how to do this ?

marlon
 
No, can you help me with this and line of intersection?

Thank you.
 
chmate said:
No, can you help me with this and line of intersection?

Thank you.

let's say that the plane has equatiuon ux+vy+wz+t=0 where (u,v,w) is the normal vector of the plane and t is a constant. We need to find u,v,w and t.

A) find u,v,w
B) find t

For A) i say that (u,v,w) denotes the same direction vector as the line.
Do you know why and if so, what's the direction vector of the line ?

If you have u,v,w then you can use the fact that (2,3,1) lies on the plane to find t

marlon
 
So the normal vector has the same direction as line. So how to find that vector and t?
 
chmate said:
So the normal vector has the same direction as line. So how to find that vector and t?

Well, you know the normal vector (u,v,w) and you know one point on the plane

Substitute these two variables in the ux+vy+wz+t=0 and solve for t

marlon
 

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