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

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In summary, to find the equation of the normal from point M(2,3,1) to line l: \frac{x+1}{2}=\frac{y-0}{-1}=\frac{z-2}{3}, you need to first find the plane that contains point M and is normal to the given line. To do this, you can use the direction vector of the line and the given point to find the normal vector (u,v,w) and then use this to find the constant t in the equation ux+vy+wz+t=0. Once you have the equation of the plane, you can find the point of intersection between the line and the plane, which will give you the direction of the normal
  • #1
chmate
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Write equation of normal from point [tex]M(2,3,1)[/tex] into line:[tex]l: \frac{x+1}{2}=\frac{y-0}{-1}=\frac{z-2}{3}[/tex].

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

Thank you.
 
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  • #2
What is the relationship between the line and the normal?
 
  • #3
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
 
  • #4
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.
 
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  • #5
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
 
  • #6
No, can you help me with this and line of intersection?

Thank you.
 
  • #7
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
 
  • #8
So the normal vector has the same direction as line. So how to find that vector and t?
 
  • #9
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
 

1. What is the equation of normal for a point M(2,3,1) to a line l?

The equation of normal for a point M(2,3,1) to a line l is a mathematical expression that represents the shortest distance from the point to the line, and the direction perpendicular to the line.

2. How do you calculate the equation of normal for a point to a line?

To calculate the equation of normal, you first need to find the vector representing the direction of the line. Then, construct a vector from the point to any point on the line. Finally, take the dot product of these two vectors to get the equation of normal.

3. Can the equation of normal be negative?

Yes, the equation of normal can be negative. The equation represents the distance between the point and the line, which can be positive or negative depending on the positioning of the point relative to the line.

4. What is the significance of the equation of normal?

The equation of normal is important in geometry and physics, as it allows us to find the shortest distance from a point to a line. It is also used in calculating the angle of incidence and reflection in optics.

5. How is the equation of normal related to the normal vector?

The equation of normal is related to the normal vector as it is essentially the equation of a line perpendicular to the given line, passing through the given point. The normal vector represents the direction of this line, which is used in calculating the equation of normal.

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