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yrelhan4

  • one year ago

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.

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  1. Spacelimbus
    • one year ago
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    Seems like a work for the dot product to me.

  2. yrelhan4
    • one year ago
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    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.

  3. Spacelimbus
    • one year ago
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    |dw:1360345934670:dw|

  4. Spacelimbus
    • one year ago
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    careful, a line in 3D space shouldn't be written in cartesian form,

  5. Spacelimbus
    • one year ago
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    \[\Large \vec{v} \cdot \vec{SP}= 0 \]

  6. Spacelimbus
    • one year ago
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    \[\Large 0S \in g \]

  7. Spacelimbus
    • one year ago
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    |dw:1360346169283:dw|

  8. yrelhan4
    • one year ago
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    well it can be written in cartesian form with direction ratios.

  9. Spacelimbus
    • one year ago
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    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.

  10. TuringTest
    • one year ago
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    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?

  11. yrelhan4
    • one year ago
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    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?

  12. Spacelimbus
    • one year ago
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    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.

  13. yrelhan4
    • one year ago
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    @TuringTest well its the position vector as they call it. its basically a point.

  14. TuringTest
    • one year ago
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    ok, I think I understand the question better now, thanks

  15. yrelhan4
    • one year ago
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    @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.

  16. TuringTest
    • one year ago
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    the line is parallel to the vector in my interpretation

  17. sirm3d
    • one year ago
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    |dw:1360346673680:dw|

  18. Spacelimbus
    • one year ago
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    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: \[\Large r_x=0P + \lambda \vec{v} \]

  19. Spacelimbus
    • one year ago
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    I guess @sirm3d and I agree then, 6i+4j-4k is a direction vector of the parametric line

  20. sirm3d
    • one year ago
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    as @Spacelimbus said, it's the dot product at work here.

  21. TuringTest
    • one year ago
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    I agree as well, for what that's worth :p

  22. yrelhan4
    • one year ago
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    hmm. and whats wrong with the method i'm using?

  23. Spacelimbus
    • one year ago
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    |dw:1360346938724:dw|

  24. Spacelimbus
    • one year ago
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    stealing @sirm3d superior model

  25. sirm3d
    • one year ago
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    |dw:1360347140886:dw| \[\cos \theta = \frac{\vec{PR} \cdot \vec{SR}}{|\vec{PR}||\vec{SR}|}\\|\vec{PS}|=|\vec{PR}|\sin\theta\]

  26. yrelhan4
    • one year ago
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    i think i can do it now. thank you guys.

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