Vector Calculus HELP

[FaLaLollipop]

I am having a lot of trouble with vector calculus! My first question gives two parallel lines and I am suppose to find the distance between the two lines? The next question gives a point from a line through another point and I am suppose to find that distance. Next I am given three points and am suppose to find the area of the triangle. Well, that is enough to ask right now...I'll most likely be back asking some more questions. Any help will be greatly appreciated! Thanks!

 

[majik1213]

The shortest line between two parallel lines is one that is orthogonal to the two lines.
The distance formula is simply Sqrt[x^2+y^2+z^2]. Hope this helps.

 

[soroban]

Hello, FaLaLollipop!

I'll assume you know some vector formulas . . .

Find the distance from a given point to a given line.

                C
                *
              / :
           b/   :
          /     :d
        /       :
      * - - - - + - - *
      A       c       B

The given line is vector \(\displaystyle \overline{AB}\,=\,\overline c\)
The given point is \(\displaystyle C\).
Let \(\displaystyle \overline{AC}\,=\,\overline b\)

The projection of \(\displaystyle \overline{AC}\) onto \(\displaystyle \overline{AB}\) is: \(\displaystyle \:\overline v \:= \roj_cb\:=\:\left(\frac{\overline b\,\cdot\,\overline c}{|\overline c|^2}\right) \overline c\)

Then: \(\displaystyle \:\overline d\:= \:\overline b\,-\,\overline v\)

Hence: \(\displaystyle \:d \;=\:|\overline d|\)

Given three points, find the area of the triangle.

                C
                *
              /  \
           b/     \
          /        \
        / θ         \
      *--------------*
      A      c       B

From the previous problem, we know how to find \(\displaystyle d\), the height of the triangle.
\(\displaystyle \;\;\)Then: \(\displaystyle \,A\:=\:\frac{1}{2}\cdot(\text{base})\cdot(\text{height})\)


Another method:

The area of a triangle is: \(\displaystyle \:A\;= \;\frac{1}{2}\cdot b\cdot c\cdot\sin\theta\)
\(\displaystyle \;\;\)(one-half the product of two sides and the sine of the included angle)

We can find: \(\displaystyle \,b\,=\,|\overline b|,\;c\,=\,|\overline c|\)

And we can find \(\displaystyle \sin\theta\) with: \(\displaystyle \:\cos\theta \:= \:\frac{\overline b\,\cdot \overline c}{|\overline b||\overline c|}\)

 

[pka]

Here is another way to look at the problem.
Given two parallel lines: \(\displaystyle \L
l_1 (t) = P + tD\quad \& \quad l_2 (t) = Q + tD\).
Note that they are parallel because they have same direction vector, D.
Because they are two lines P is not on l<SUB>2</SUB>.
So the distance from P to l<SUB>2</SUB> is given by:
\(\displaystyle \L
d(P,l_2 ) = \frac{{\left\| {\ QP^ \to \times D} \right\|}}{{\left\| D \right\|}}\).
That will also be the distance between the lines.