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Math Help - Regarding Linear Algebraaa

  1. #1
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    Regarding Linear Algebraaa

    Heres the question:
    1. The point on the line x=-1+3t, y=-2+t, z=2-4t which is closest to the point (2,-2,2) is (_,_,_)? Fill in the blanks.

    2. The point on the plane 3x+3y-2z=-2 that is closest to the point (1,-1,3) is (_,_,_)? again fill in the blanks

    appreciate if anyone can help me on this fast since this question is due for me online in like 3 hours =S
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Minimize the distance formula...
    for the 1st one...
    d=\sqrt{(-1+3t-2)^2+(-2+t+2)^2+(2-4t-2)^2}

    taking the derivative

    d'=\frac{-18+52t}{2\sqrt{9-18t+26t^2}}

    setting equal to zero and solving for t gives the solution t=\frac{9}{26}

    Plug this back into the parametric equations to get the point...
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Lagrange multipliers

    Once again we will minimize the distance formula.

    d=\sqrt{(x-1)^2+(y+1)^2+(z-3)^2}

    setting the equation of the plane equal to zero we get...

     3x+3y-2z+2=0 or..
    f(x,y,z)=3x+3y-2z+2

    This time we need lagrange multipliers...

    \nabla{d}=\lambda\nabla{f} this gives the system

    \frac{x-1}{\sqrt{(x-1)^2+(y+1)^2+(z-3)^2}}=3\lambda
    \frac{y+1}{\sqrt{(x-1)^2+(y+1)^2+(z-3)^2}}=3\lambda
    \frac{z-3}{\sqrt{(x-1)^2+(y+1)^2+(z-3)^2}}=-2\lambda

     3x+3y-2z+2=0

    Solving this system gives the solution

    x=\frac{17}{11}

    y=-\frac{5}{11}

    z=\frac{29}{11}
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