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?

 

[Dick]

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?

 

[Dick]

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.

 

[Dick]

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!

 

[HallsofIvy]

Here's the way I would do that with no "dot product": If he started at (-2, 3) and walks in the direction of the vector 3i+ 2j, then he walked along the line given by the parametric equations x= -2+ 3t, y= 3+ 2t. A vector perpendicular to 3i+ 2j is 2i- 3j so if he turned at right angles and walked directly to (0, 6), he walked along the line x= 2s, y= 6- 3s. Find where those two lines intersect.

Perhaps you can use that as a check.