Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

yrelhan4

the distance from the point -i + 2j + 6k to the straight line through the point (2,3,-4) and parallel to the vector 6 i+ 3j - 4k.

  • one year ago
  • one year ago

  • This Question is Closed
  1. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    Seems like a work for the dot product to me.

    • one year ago
  2. yrelhan4
    Best Response
    You've already chosen the best response.
    Medals 0

    so i converted all of it into cartesian form. wrote the line as (x-2)/6=(y-3)/3=(z+4)/-4 then i put all of it =r. found out random points. then found out direction ratios of the perpendicular. through condition of perpendicularity found r. and found those random points. and the found the distance. the answer i am getting is wrong.

    • one year ago
  3. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1360345934670:dw|

    • one year ago
  4. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    careful, a line in 3D space shouldn't be written in cartesian form,

    • one year ago
  5. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\Large \vec{v} \cdot \vec{SP}= 0 \]

    • one year ago
  6. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\Large 0S \in g \]

    • one year ago
  7. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1360346169283:dw|

    • one year ago
  8. yrelhan4
    Best Response
    You've already chosen the best response.
    Medals 0

    well it can be written in cartesian form with direction ratios.

    • one year ago
  9. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    well if you want to do it like that, then unfortunately I can't help you. I haven't seen such a thing, what I would agree with, is saying that you can always set up an equation of a plane that includes the line and from there figure out the distance. But as soon as I see something of the form x+y+z=c, I think of a plane, not of a line.

    • one year ago
  10. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 0

    I am confused for many reasons: "the distance from the point -i + 2j + 6k..." but -i +2j + 6k is not a point. and the distance to which point on the other line? the minimum distance I presume?

    • one year ago
  11. yrelhan4
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm. i just want to confirm if i am interpreting the question right. its saying the line is parallel to the vector 6 i+ 3j - 4k and not that it wants the distance parallel to this vector?

    • one year ago
  12. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    Mathematical definition is always considering it's normal, orthogonal point. That is a distance. So if you have a point given, as I tried to sketch it in the above picture, then it's distance to the line is the one where you start from the line, draw a right angle through that point and figure out the distance.

    • one year ago
  13. yrelhan4
    Best Response
    You've already chosen the best response.
    Medals 0

    @TuringTest well its the position vector as they call it. its basically a point.

    • one year ago
  14. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, I think I understand the question better now, thanks

    • one year ago
  15. yrelhan4
    Best Response
    You've already chosen the best response.
    Medals 0

    @Spacelimbus well i still have the doubt. does it ask for the distance parallel to that vector? or the line is parallel to that vector? 6 i+ 3j - 4k.

    • one year ago
  16. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 0

    the line is parallel to the vector in my interpretation

    • one year ago
  17. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1360346673680:dw|

    • one year ago
  18. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    To set up an equation in the 3D-Space, you need a direction vector, an a point, that's how I interpret it, because that satisfies the equation: \[\Large r_x=0P + \lambda \vec{v} \]

    • one year ago
  19. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    I guess @sirm3d and I agree then, 6i+4j-4k is a direction vector of the parametric line

    • one year ago
  20. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    as @Spacelimbus said, it's the dot product at work here.

    • one year ago
  21. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 0

    I agree as well, for what that's worth :p

    • one year ago
  22. yrelhan4
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm. and whats wrong with the method i'm using?

    • one year ago
  23. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1360346938724:dw|

    • one year ago
  24. Spacelimbus
    Best Response
    You've already chosen the best response.
    Medals 2

    stealing @sirm3d superior model

    • one year ago
  25. sirm3d
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1360347140886:dw| \[\cos \theta = \frac{\vec{PR} \cdot \vec{SR}}{|\vec{PR}||\vec{SR}|}\\|\vec{PS}|=|\vec{PR}|\sin\theta\]

    • one year ago
  26. yrelhan4
    Best Response
    You've already chosen the best response.
    Medals 0

    i think i can do it now. thank you guys.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.