Finding the coordinates of a point on a line: Vectors

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The discussion focuses on finding the coordinates of a point P on line L defined by points A (1,3,-17) and B (6,-7,8). The parametric equation of line L is established as x=1-5t, y=3+10t, z=-17-25t, derived from vector AB=(5,-10,25). To determine point P, it is noted that vector OP must be perpendicular to line L, requiring the dot product of vector AB and OP to equal zero. Participants discuss constructing the dot product using the coordinates of point P and solving for the parameter t to find the coordinates of P. The conversation concludes with a successful resolution to the problem.
lunds002
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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?
 
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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.
 

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