Vector Calculus - Having trouble finding a Line

In summary, the problem is trying to get perpendicular lines by finding the line that passes through a point and a line and is perpendicular to the given line.
  • #1
DougD720
47
0
I know this has been posted before, and I've read the post concerning the same problem and I've googled this a million times, but i can't seem to get it. So here's the problem:

Given a Point and a Line, find the Line that passes through the point (3,1,-2) and is perpendicular to the line x=-1+t, y=-2+t, z=-1+t.

Okay, so i found the direction vector of the given line (which i'll call l) to be (1,1,1).
I know that a vector lying on l and a vector lying on the line i need to find (which i'll call l'), when you apply the dot product to those 2 vectors it must equal 0 in order for them to be perpendicular.

I know i have to use orthagonal projection, well i think i need to. So far i took the point given (which i called Q) and a point on the line given (which i called P) to find the vector QP, which (in my case) is (4,3,-1).

From here I'm lost with what to do, I've been sitting here racking my brain trying different things but i just can't seem to figure out how to find the line. I'm not asking for a solution but a *simple* set of steps or something like that to follow would be greatly appreciated.

Thanks!
 
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  • #2
Try the following system:
given line L=(-1,-2,-1)+(1,1,1)t
other line K=(3,1,-2)+(a,b,c)s

Known equations:
a+b+c=0 (perpendicular)
a^2+b^2+c^2=1 (normalization, since an arbitrary multiple is allowed)
To complete the problem definition, we need to assume K and L must meet at some point, otherwise, K could be any line in a plane passing through (3,1,-2) and perpendicular to L. This then gives us 3 more equations K(s)=L(t).

Solve the system (5 equations - 4 linear and 1 quadratic) for 5 unknowns (a,b,c,s,t).

Good luck!
 
  • #3
Thanks! I think i get it now and i'll try it in a minute i just have one question. What do you mean by the second step of your "Known Equations" section? I don't see where this one equation is coming from, but i get the rest. Thanks so much!
 
  • #4
Also, i don't know where to begin solving this system of equations :(
 
  • #5
a^2+b^2+c^2=1 is thrown in because (a,b,c) can be replaced by (fa,fb,fc), where f is arbitrary. Otherwise the whole system floats.

As far as solving the system, I suggest solving the four linear equations to get rid of s and t and either a,b, or c. Since the relationships among a,b, and c will be linear, these can be inserted into the normalization equation to get a quadratic in one remaining variable.
 
  • #6
Thanks so much for the help! I figured it out, our TA did it differently but i prefer the approach suggested - it's more difficult but more concrete. Our TA suggested minimizing the distance equation between the point and the line, which is also effective, i thought of doing that but i liked the more difficult approach suggested. Thanks again!
 

1. What is a vector in vector calculus?

A vector in vector calculus is a mathematical entity that has both magnitude and direction. It is represented by an arrow pointing in the direction of the vector with a length that represents its magnitude.

2. How is a line defined in vector calculus?

In vector calculus, a line is defined as a set of points that satisfy a linear equation. This equation can be written in the form of ax + by + cz = d, where a, b, and c represent the direction of the line and d is a constant term.

3. What is the difference between a point and a vector in vector calculus?

In vector calculus, a point is a specific location in space, while a vector is a mathematical representation of magnitude and direction. A point can be identified by its coordinates, while a vector is represented by an arrow pointing in a specific direction.

4. How do I find a line in vector calculus?

Finding a line in vector calculus involves determining the direction and location of the line. This can be achieved by finding two points on the line and using them to create a vector equation or by using the slope and a point on the line to create a parametric equation.

5. What are some applications of vector calculus in real life?

Vector calculus has many applications in real life, including physics, engineering, computer graphics, and economics. It is used to calculate forces, motion, and energy in physical systems, design structures and analyze fluid flow, create realistic computer animations, and model financial markets, among others.

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