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Math Help - Find the point on the line closest to the origin?

  1. #1
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    Find the point on the line closest to the origin?

    The question asks to find the point closest to the origin of the straight line perpendicular to a surface equation previously given. I found the equation for the normal line, but I am not sure how to find the point on the line closest to the origin. Can anyone tell me how to do this? Thanks in advance.

    Here is the equation of the line...
    \frac{x-1}{6}=\frac{y-2}{3}=\frac{z-3}{2}
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  2. #2
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    I've yet to get to multi-variable calculus, but wouldn't you simply be solving an optimization problem using the distance formula?
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  3. #3
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    Quote Originally Posted by Infernorage View Post
    The question asks to find the point closest to the origin of the straight line perpendicular to a surface equation previously given. I found the equation for the normal line, but I am not sure how to find the point on the line closest to the origin. Can anyone tell me how to do this? Thanks in advance.

    Here is the equation of the line...
    \frac{x-1}{6}=\frac{y-2}{3}=\frac{z-3}{2}
    Parametrize the line as r(t) = <6t + 1, 3t + 2, 2t + 3>. Then the distance from the origin squared of r is given by (6t + 1)^2 + (3t + 2)^2 + (2t + 3)^2. Differentiate by t and set it equal to zero. Then solve for t and plug it into the distance formula.
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    Quote Originally Posted by rn443 View Post
    Parametrize the line as r(t) = <6t + 1, 3t + 2, 2t + 3>. Then the distance from the origin squared of r is given by (6t + 1)^2 + (3t + 2)^2 + (2t + 3)^2. Differentiate by t and set it equal to zero. Then solve for t and plug it into the distance formula.
    Okay, so by distance formula do you mean plug t back into the parametrized r(t) equations you have above? Do you know what the final answer should be? Thanks.
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  5. #5
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    Quote Originally Posted by Infernorage View Post
    Okay, so by distance formula do you mean plug t back into the parametrized r(t) equations you have above? Do you know what the final answer should be? Thanks.
    The distance formula of r(t) = <x(t), y(t), z(t)> is sqrt(x(t)^2 + y(t)^2 + z(t)^2).
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  6. #6
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    Quote Originally Posted by rn443 View Post
    The distance formula of r(t) = <x(t), y(t), z(t)> is sqrt(x(t)^2 + y(t)^2 + z(t)^2).
    Oh okay, thanks. Once you have the distance though, what do you do with that to get the actual point on the line?
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  7. #7
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    Quote Originally Posted by Infernorage View Post
    Oh okay, thanks. Once you have the distance though, what do you do with that to get the actual point on the line?
    Oh, whoops, I thought the question was for the distance. To find the actual point, you just plug the value of t you found back into the equation for r.
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