# A Dot product analysis proof (might be simple)

• Andrax
In summary: In fact, it follows from the Pythagorean theorem. If we measure from point P to line l along a NON perpendicular line, we can drop a perpendicular from P to l giving us a right triangle in which the original, non-perpendicular, distance is the hypotenuse. By c^2= a^2+ b^2, c, the length of the hypotenuse is longer than the length of either leg.Alternatively, you can define the "distance between point P and line l" to be measured along the perpendicular and use that theorem to show that it is the shortest distance.
Andrax

## Homework Statement

so we have a (D) Line (geometry) it's Cartesian equation is ax+by+c=0
we have an $A$($\alpha$,$\beta$)
prove that the distance between the line(D) and the point A is
d=$\frac{la\alpha+b\beta+cl}{\sqrt{a^2+b^2}}$

## The Attempt at a Solution

let every distance be a vector (sorry didn't find vectors in the latex reference)
AH h belongs to (D) and AH factors (D) so AH(a,b)
we also have n vector (-b,a)
we can say that the catersian equation of ah is -b(x-$\alpha$)+a(y-$\beta$)+c I am pretty musch stuck here now to get this \sqrt{a^2+b^2} we will probably use cos but ireally can't find how r we supposed to use it :( please help guys:

What are H and h? What is l?
Do you know the cross-product?
Alternatively, can you transform your line to get ##\sqrt{a^2+b^2}=1##? This is always possible (as a!=0 or b!=0 for a meaningful line).

The distance from a point to a line is, by definition, measured along a line through the given point perpendicular to the given line.

We can write ax+ by= c as y= -(a/b)x+ c/b showing that the line has slope -(a/b). A line perpendicular to it must have slope b/a. A line with that slope and passing through the point $(\alpha, \beta)$ has equation $y= (b/a)(x- \alpha)+ \beta$ or $bx- ay= b\alpha- a\beta$. Determine where the two lines intersect, by solving ax+ by= c and $bx- ay= b\alpha- a\beta$ simultaneously and then determine the distance between that point of intersection and $(\alpha, \beta)$.

That should get the formula you want.

HallsofIvy said:
The distance from a point to a line is, by definition, measured along a line through the given point perpendicular to the given line.

We can write ax+ by= c as y= -(a/b)x+ c/b showing that the line has slope -(a/b). A line perpendicular to it must have slope b/a. A line with that slope and passing through the point $(\alpha, \beta)$ has equation $y= (b/a)(x- \alpha)+ \beta$ or $bx- ay= b\alpha- a\beta$. Determine where the two lines intersect, by solving ax+ by= c and $bx- ay= b\alpha- a\beta$ simultaneously and then determine the distance between that point of intersection and $(\alpha, \beta)$.

That should get the formula you want.

thank you.

mfb said:
What are H and h? What is l?
Do you know the cross-product?
Alternatively, can you transform your line to get ##\sqrt{a^2+b^2}=1##? This is always possible (as a!=0 or b!=0 for a meaningful line).

thank you but i we didn't study that "theory" so i don't think I'm allowed to apply it anyway the hallsofivy provided a rather simple solution .

Andrax said:
thank you but i we didn't study that "theory" so i don't think I'm allowed to apply it anyway the hallsofivy provided a rather simple solution .

Strictly speaking, the distance between a line and a point is defined to be the *shortest* distance between the given point and points on the line. It follows (basically, as a theorem) that the shortest line is perpendicular to the given line, as HallsOfIvy has stated.

Ray Vickson said:
Strictly speaking, the distance between a line and a point is defined to be the *shortest* distance between the given point and points on the line. It follows (basically, as a theorem) that the shortest line is perpendicular to the given line, as HallsOfIvy has stated.
In fact, it follows from the Pythagorean theorem. If we measure from point P to line l along a NON perpendicular line, we can drop a perpendicular from P to l giving us a right triangle in which the original, non-perpendicular, distance is the hypotenuse. By $c^2= a^2+ b^2$, c, the length of the hypotenuse is longer than the length of either leg.

Conversely, you can define the "distance between point P and line l" to be measured along the perpendicular and use that theorem to show that it is the shortest distance.

## 1. What is a dot product analysis proof?

A dot product analysis proof is a mathematical method used to prove the relationship between two vectors and their dot product. It involves multiplying the corresponding components of the vectors and adding them together to determine the dot product.

## 2. How is a dot product analysis proof used in science?

Dot product analysis proofs are commonly used in science, particularly in physics and engineering, to determine the work done by a force or the angle between two vectors. They are also used in computer graphics to calculate lighting and shading effects.

## 3. What are the steps involved in a dot product analysis proof?

The first step is to write out the two vectors in their component form. Then, multiply the corresponding components of the vectors and add them together. Finally, simplify the resulting expression to obtain the dot product value.

## 4. Can a dot product analysis proof be used for any number of dimensions?

Yes, a dot product analysis proof can be used for any number of dimensions. The number of components in each vector will increase with the number of dimensions, but the basic steps of multiplying and adding the components remain the same.

## 5. Are there any applications of dot product analysis proof in everyday life?

Yes, there are many applications of dot product analysis proof in everyday life. For example, it is used in determining the work done by a force in physics, calculating the angles between roads and buildings in urban planning, and determining the direction of a moving object in sports and video games.

• Calculus and Beyond Homework Help
Replies
13
Views
2K
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
901
• Calculus and Beyond Homework Help
Replies
1
Views
900
• Calculus and Beyond Homework Help
Replies
3
Views
840
• Quantum Physics
Replies
14
Views
1K