Find the distance from the point P to a line - linear algebra

In summary: The minimum occurs when the derivative is zero. You could also use vector projection to get a similar equation. In summary, there are two methods to find the distance from a point to a line: using calculus or using vector projection.
  • #1
Oliviacarone
18
1

Homework Statement


Find the distance from point P (1,7,3) to the line
(x,y,z) = (-2,1,4) + s(1,-3,4),
s is a free variable

Homework Equations


projnQP = ( QP⋅n/(lengthQP)(lengthn) )(n)

The Attempt at a Solution


I'm not quite sure about how to find the normal (n) here, but if I make s=0, I'm guessing it's just (-2,1,4)?
QP would be Q= a point on the line, so I'm not sure which point to use since all I have here is a point, it feels like, and I used this point (-2,1,4) for the normal...
Then if I found QP I would just plug in everything.
 
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  • #2
Oliviacarone said:
I'm not quite sure about how to find the normal (n) here, but if I make s=0, I'm guessing it's just (-2,1,4)?
I think the problem is that you are guessing. Try sketching the line and the point. When you understand how parametric representations of lines work, you can start to solve problems with them.
 
  • #3
Oliviacarone said:

Homework Statement


Find the distance from point P (1,7,3) to the line
(x,y,z) = (-2,1,4) + s(1,-3,4),
s is a free variable

Homework Equations


projnQP = ( QP⋅n/(lengthQP)(lengthn) )(n)

The Attempt at a Solution


I'm not quite sure about how to find the normal (n) here, but if I make s=0, I'm guessing it's just (-2,1,4)?
QP would be Q= a point on the line, so I'm not sure which point to use since all I have here is a point, it feels like, and I used this point (-2,1,4) for the normal...
Then if I found QP I would just plug in everything.

You could use calculus to minimize the squared distance between P and a point Q in the line; that squared-distance is a quadratic form in the variable ##s##.
 

1. What is the formula for finding the distance from a point to a line?

The formula for finding the distance from a point P(x1, y1) to a line ax + by + c = 0 is given by: d = |ax1 + by1 + c| / √(a2 + b2).

2. How do you interpret the result of the distance formula?

The result of the distance formula represents the shortest distance between the point P and the line. It can be positive or negative depending on which side of the line the point is located on. A positive distance means the point is on one side of the line, while a negative distance means the point is on the other side of the line.

3. Can the distance from a point to a line be negative?

Yes, the distance from a point to a line can be negative. A negative distance indicates that the point is on the opposite side of the line from where the positive distance is measured.

4. What is the significance of the absolute value in the distance formula?

The absolute value in the distance formula ensures that the result is always positive, regardless of which side of the line the point is on. This allows for a consistent and standardized way of measuring distance from a point to a line.

5. Can the distance from a point to a line be 0?

Yes, the distance from a point to a line can be 0. This occurs when the point lies directly on the line, so there is no distance between them. In this case, the point is considered to be collinear with the line.

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