# Thread: Major trouble on vector calculus

1. ## Major trouble on vector calculus

We barely even touched projection vectors (?) and I am completely lost on how to even do this problem because I have no idea what to do.

The distance d of a point P to the line through points A and B is the length of the component of AP¯¯¯¯¯¯¯¯ that is orthogonal to AB¯¯¯¯¯¯¯¯, as indicated in the diagram. So the distance from P=(2,2,3) to the line through the points A=(1,4,2) and B=(5,4,3) is ...

What do I even do here? Pythagorean Theorem at some point?

2. ## Re: Major trouble on vector calculus

What diagram?

3. ## Re: Major trouble on vector calculus

what this is saying is that the vector $\vec{AB}$ is orthogonal to the vector $\vec{PP^\prime}$ where $P^\prime$ is the point on the line $\overline{AB}$ closest to the point $P$

$\vec{AB} = B-A = (−5,4,−3) -(1,4,−2) = (-6,0,-1)$

let $P^\prime = A + \lambda \vec{AB} = (1-6 \lambda ,4,2-5 \lambda)$ (do you understand why we can write it this way?)

then $\vec{PP^\prime} = P^\prime - P =(3-6 \lambda ,6,-5 \lambda -1)$

and we want

$\vec{PP^\prime}\cdot \vec{AB} = 0$

$\vec{PP^\prime}\cdot \vec{AB} = 61 \lambda -13$

$\lambda = \dfrac{13}{61}$

$P^\prime = (1-6 \lambda ,4,2-5 \lambda) =\left( -\dfrac{17}{61},4,\dfrac{57}{61}\right)$

$d = | \overline{PP^\prime} | =\left |\left( -\dfrac{17}{61},4,\dfrac{57}{61}\right)\right| = 3 \sqrt{\dfrac{293}{61}}$

4. ## Re: Major trouble on vector calculus

Originally Posted by TheTideCaller
We barely even touched projection vectors (?) and I am completely lost on how to even do this problem because I have no idea what to do.

The distance d of a point P to the line through points A and B is the length of the component of AP¯¯¯¯¯¯¯¯ that is orthogonal to AB¯¯¯¯¯¯¯¯, as indicated in the diagram. So the distance from P=(−2,−2,3) to the line through the points A=(1,4,−2) and B=(−5,4,−3) is ... ?
You learn about vectors. That is way one does these questions.
If $\ell(t)=A+tD$ is a line and $P\notin\ell$ then the distance from $P$ to $\ell$, $\mathscr{D}(P;\ell)=\dfrac{\|\overrightarrow {AP}\times D\|}{\|D\|}$
SEE HERE

In this case $\overrightarrow {AP} = \left\langle { - 3, - 6,5} \right\rangle~\&~D=\left\langle { - 1,0,-6} \right\rangle$

Now you have at it.