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Math Help - Can anyone please help me with a question on parametric equations?

  1. #1
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    Can anyone please help me with a question on parametric equations?

    Find parametric equations of the line perpendicular to the line through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

    Can you please help me with this question, I am really stuck, don't know how to answer it ..

    Thank you
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by HorseStar2 View Post
    Find parametric equations of the line perpendicular to the line through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

    Can you please help me with this question, I am really stuck, don't know how to answer it ..

    Thank you
    use formula for line that goes through 3 points

    \begin{vmatrix}<br />
 x-x_1&y-y_1  &z-z_1 \\ <br />
 x_2-x_1&y_2-y_1  &z_2-z_1 \\ <br />
 x_3-x_1&y_3-y_1  &z_3-z_1 <br />
\end{vmatrix}

    and to intersect plane at any (let's say your point) line that you get from that up there must have point (P(3,1,-2))

    hint: draw it for your self ... it make more sense that way


    P.S. actually as i more think of it... this could be some "problem" if there is equation of the plane so you need to find where that line intersects (if intersect) plane ... but this way when you solve that determinant up there there is not much to do ... if you have 3 points there is no another way to draw line through them except like that, and there is no another condition by which you can lean it so it will go through plane at that point ...

    funny thing is that you can substitute any from P,Q,R with that P on the plane and put in determinant and you will get the same result
    Last edited by yeKciM; August 21st 2010 at 06:33 AM.
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  3. #3
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    Quote Originally Posted by HorseStar2 View Post
    Find parametric equations of the line perpendicular to the line through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

    Can you please help me with this question, I am really stuck, don't know how to answer it ..

    Thank you
    Are you sure you have the right P, Q and R. I believe that they're not colinear.
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  4. #4
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    Quote Originally Posted by Danny View Post
    Are you sure you have the right P, Q and R. I believe that they're not colinear.
    Hi, I made an error when writing the question, it should read:

    Find parametric equations of the line perpendicular to the PLANE through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

    Do I have to find the parametric equations of the two lines PQ and PR, then calculate the cross product of the two directional vectors to give a common normal to the plane? I'm not sure where to go from here if that's right....

    Thanks
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  5. #5
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    Quote Originally Posted by yeKciM View Post
    use formula for line that goes through 3 points

    \begin{vmatrix}<br />
 x-x_1&y-y_1  &z-z_1 \\ <br />
 x_2-x_1&y_2-y_1  &z_2-z_1 \\ <br />
 x_3-x_1&y_3-y_1  &z_3-z_1 <br />
\end{vmatrix}

    and to intersect plane at any (let's say your point) line that you get from that up there must have point (P(3,1,-2))

    hint: draw it for your self ... it make more sense that way


    P.S. actually as i more think of it... this could be some "problem" if there is equation of the plane so you need to find where that line intersects (if intersect) plane ... but this way when you solve that determinant up there there is not much to do ... if you have 3 points there is no another way to draw line through them except like that, and there is no another condition by which you can lean it so it will go through plane at that point ...

    funny thing is that you can substitute any from P,Q,R with that P on the plane and put in determinant and you will get the same result
    Hi, I made an error when writing the question, it should read:

    Find parametric equations of the line perpendicular to the PLANE through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

    Do I have to find the parametric equations of the two lines PQ and PR, then calculate the cross product of the two directional vectors to give a common normal to the plane? I'm not sure where to go from here if that's right....

    Thanks
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  6. #6
    Senior Member yeKciM's Avatar
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    sorry if i miss translate it to my self that "perpendicular to the PLANE" i think that means that line is normal to the plane ?
    if so your equation of that line is :
     y=1


    or am I of with this, and it's plane trough 3 points
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  7. #7
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    Quote Originally Posted by HorseStar2 View Post
    Hi, I made an error when writing the question, it should read:

    Find parametric equations of the line perpendicular to the PLANE through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

    Do I have to find the parametric equations of the two lines PQ and PR, then calculate the cross product of the two directional vectors to give a common normal to the plane? I'm not sure where to go from here if that's right....

    Thanks
    You're on the right track! Find the vector \vec{PQ} and \vec{PR}.  Then cross these (like you said). This will give the normal to the plane. The line perpendicular to the plane will follow this normal. Then use the point (3,1,-2) to finally give you your line.
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  8. #8
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    Quote Originally Posted by Danny View Post
    You're on the right track! Find the vector \vec{PQ} and \vec{PR}.  Then cross these (like you said). This will give the normal to the plane. The line perpendicular to the plane will follow this normal. Then use the point (3,1,-2) to finally give you your line.
    Hi,

    So far I have calculated.. vector PQ = (-1, 0, 2) and Vector PR = (2, 0, -1) so corresponding parametric lines are: PQ = (0, 1, 0 ) + t1(-1, 0, 2), PR = (0, 1, 0) + t2 (2, 0, -1). Completing the cross product of the vectors PQ and PR I obtained 0i + 3j + 0k. Is this right so far? so do I then have a final answer of (3, 1, -2) + t3(0, 3, 0) as my parametric equation of the line?

    Thank you
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  9. #9
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    Quote Originally Posted by HorseStar2 View Post
    Hi,

    So far I have calculated.. vector PQ = (-1, 0, 2) and Vector PR = (2, 0, -1) so corresponding parametric lines are: PQ = (0, 1, 0 ) + t1(-1, 0, 2), PR = (0, 1, 0) + t2 (2, 0, -1). Completing the cross product of the vectors PQ and PR I obtained 0i + 3j + 0k. Is this right so far? so do I then have a final answer of (3, 1, -2) + t3(0, 3, 0) as my parametric equation of the line?

    Thank you
    When I write (3, 1, -2) + t3(0, 3, 0) ... t3 is t subscript 3
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  10. #10
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    Yep - that's it!
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  11. #11
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    Quote Originally Posted by Danny View Post
    Yep - that's it!
    Really... you're not just saying that lol?
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