# Finding the coordinates of a point on a line: Vectors

• lunds002
In summary, to find the equation of line L in parametric form, the vector AB was found to be (5,-10,25). The equation is x=1-5t, y=3+10t, z=-17-25t. To find the coordinates of point P, it was determined that OP is perpendicular to line L if the dot product of vector AB and OP equals zero. By finding the direction vector and setting the dot product to zero, the coordinates of point P can be solved for.
lunds002
Consider the points A (1,3,-17) and B (6,-7,8) which lie on the line L.

a) find an equation of line L in parametric form.

I found vector AB=(5,-10,25), and so I found the equation to be x=1-5t, y=3+10t, z=-17-25t

b) The point P is on line L such that vector OP is perpendicular to L. Find the coordinates of point P.

I know that OP is perpendicular to the line L if the dot product of vector AB and OP equals zero, but I'm not sure if that will help me find a solution to part b. Help?

lunds002 said:
I know that OP is perpendicular to the line L if the dot product of vector AB and OP equals zero, but I'm not sure if that will help me find a solution to part b.
If the vector is perpendicular to AB, what is its direction?
Using that, you can construct a dot product with the point P as a variable, then solve.

I'm unsure of how to find the direction vector..

zhermes said:
If the vector is perpendicular to AB, what is its direction?
Using that, you can construct a dot product with the point P as a variable, then solve.

lunds002 said:
I'm unsure of how to find the direction vector..
Can you find a vector OP, from the origin to an arbitrary point on your line? Since OP is perpendicular to the line, OP $\cdot$ AB = 0.

No.. I struggle with vectors so I don't really know how to do that.

Any point on your line has coordinates <1 - 5t, 3 + 10t, -17 - 25t>, so this is the same as the vector OP.

Set the dot product of this vector and AB to zero, and solve for t. That will give you the point P on your line such that OP is perpendicular to AB.

Ohh that makes sense, thanks so much! I got the answer now.

## 1. How do I find the coordinates of a point on a line using vectors?

To find the coordinates of a point on a line using vectors, you will need to have the coordinates of another point on the line and the direction vector of the line. You can then use the vector equation of the line to calculate the coordinates of the desired point.

## 2. Can I use any point on the line to find the coordinates?

No, you will need to have the coordinates of a specific point on the line in order to find the coordinates of a different point on the same line using vectors. This point is usually referred to as the initial point or anchor point.

## 3. How do I calculate the direction vector of a line?

The direction vector of a line can be calculated by finding the difference between the coordinates of two points on the line. This means subtracting the x-coordinates and the y-coordinates of the two points, respectively, to get the x-component and y-component of the direction vector.

## 4. What if the line is not in the standard form (y = mx + b)?

The vector approach can still be used to find the coordinates of a point on a line that is not in the standard form. You will just need to manipulate the equation to get it into the vector form (x = a + bt or y = a + bt) and then use the appropriate components to find the coordinates of the desired point.

## 5. Can I use vectors to find the coordinates of a point on a curved line?

No, vectors can only be used to find the coordinates of a point on a straight line. For curved lines, you will need to use other methods such as calculus or parametric equations to find the coordinates of a point.

• Precalculus Mathematics Homework Help
Replies
8
Views
2K
• Precalculus Mathematics Homework Help
Replies
18
Views
976
• Precalculus Mathematics Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
11
Views
3K
• Precalculus Mathematics Homework Help
Replies
11
Views
4K
• Precalculus Mathematics Homework Help
Replies
17
Views
1K
• Linear and Abstract Algebra
Replies
9
Views
652
• Precalculus Mathematics Homework Help
Replies
5
Views
2K
• Precalculus Mathematics Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
10
Views
1K