# Dot product - two points and a projection

Mugen112

## Homework Statement

A person starts at coordinates (-2, 3) and arrived at coordinates (0, 6). If he began walking in the direction of the vector v=3i+2j and changes direction only once, when he turns at a right angle, what are the coordinates of the point where he makes the turn.

## Homework Equations

projection of u onto v = (u dot v/abs(v)^2)(v)
scalar projection = abs(u)cos($$\theta$$)

## The Attempt at a Solution

I'm not even sure what equations to use! I'm looking at the problem trying to think about ways I can handle it, but nothing works. I've tried moving the first point to the origin and moving the second point to its corresponding point from the first, then I can get the vector of those two points. Then take the vector of where he first started walking to then project it onto the first, then that gives me the projection of the second onto the first. I can get the scalar projection as well, but how am I supposed to get the points at which he stops and turns?

Homework Helper
The points lying along his initial walking direction can be expressed as (-2+3t,3+2t) with t a real number. At some value of t the vector difference between (-2+3t,3+2t) and (0,6) will be orthogonal to the walking direction 3i+2j. Can you find that value of t?

Mugen112
Still not sure how to do the problem. Getting kind of frustrated. So the vector difference.. meaning ((-2+3t, 3+2t) - (0,6)) (dot) (3i+2j) = 0? Then solve for t?

Last edited:
Homework Helper
Still not sure how to do the problem. Getting kind of frustrated. So the vector difference.. meaning ((-2+3t, 3+2t) - (0,6)) (dot) (3i+2j) = 0? Then solve for t?

Yes. Do you see why?

Mugen112
I can see the concept, but I don't understand how it would work. After I take the difference, I will be solving for t. There would be two equations with t correct? One for i and one for j? I tried this and got the wrong answer.

Homework Helper
I can see the concept, but I don't understand how it would work. After I take the difference, I will be solving for t. There would be two equations with t correct? One for i and one for j? I tried this and got the wrong answer.

No. One equation for t. After you take the dot product i and j disappear, don't they? The dot product is just a number.

Mugen112
Oh right, dot product is just one number. So I did the calculations, and I get..

(-2+3t, 3+2t) - (0,6) (dot) (3i+2j) = 0

(-2+3t, -3+2t) (dot) (3i + 2j) = 0

(-6+9t) + (-6+4t) = 0

(-12+13t)=0

t=12/13 ?

So after I get t, I plug it back into (-2 + 3t, 3 +2t) ? I tried this and still get the incorrect answer.

Mugen112
Ah, caught my own mistake. I used the vector difference instead of the first vector. Thanks for all the help!!!!