1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

3-D Vector question

  1. Jul 3, 2008 #1
    I'm doing some multi-variable calculus review, and I had a question (my understanding of the class was not as good as I would have liked it to be).

    <b> 1. The problem statement, all variables and given/known data </b>.
    Find a plane containing the line r(t) = <6,-6,4> + t<-2,7,-4> and orthogonal to the plane -7x+8y+5z=1.

    <b> 2. Relevant equations </b>.
    I think I need to use a cross product. I cross <-2,7,-4> and <-7,8,5> to get a vector orthogonal to to the plane (and the line). Then, I use n (dot) (r-r_0), but I keep getting the wrong answer. I fear that my approach is wrong though.
    answer is: 635/2

    <b> 3. The attempt at a solution </b>.
    So when I cross <-2,7,-4> and <-7,8,5> I get <67, 38, 33>, and my plane is 67x+38y+33z=46.

    Help please.
    Last edited: Jul 3, 2008
  2. jcsd
  3. Jul 3, 2008 #2


    User Avatar
    Homework Helper

    That numerical answer is for which part of the question? You're asked to find the plane containing the line right? I got the same normal vector as you did <67,68,33>. But then you're told to find the equation of the plane, which you know has the general form [tex]\vec{n} \cdot (\vec{r}-\vec{r_0})[/tex]

    So what you're missing is a point which resides on the plane. Look at the question again, how can you get that missing point? How did you get 67x + 38y + 33z = 46?
  4. Jul 3, 2008 #3
    Yeah that fraction was part of what I copied and pasted, so ignore that. Ok, so all I need is a point, meaning I can use t=1 to get the point (4,1,0)? Then it would be:


    I got 67x + 38y + 33z = 46 by doing <67,38,33> \cdot <x-6, y+6, z-4>. I used t=0 for my point. I used the wrong normal vector (just a writing mistake).
  5. Jul 3, 2008 #4


    User Avatar
    Homework Helper

    Typo error on my part. The normal vector should be <67,38,33>, not <67,68,33>. Use that vector and just apply the formula for any value of t. After all, the line lies on the plane, does it not?
  6. Jul 4, 2008 #5
    This is true. But I let t=0 and still got the wrong answer.
  7. Jul 4, 2008 #6


    User Avatar
    Homework Helper

    What's the answer supposed to be?
  8. Jul 5, 2008 #7
    I got the correct answer. My arithmetic was wrong, as it usually is when doing cross-products. Thank you for your help!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: 3-D Vector question
  1. Graphing in 3-d (Replies: 5)