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Thread: 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...
    $\displaystyle d=\sqrt{(-1+3t-2)^2+(-2+t+2)^2+(2-4t-2)^2}$

    taking the derivative

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

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

    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.

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

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

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

    This time we need lagrange multipliers...

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

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

    $\displaystyle 3x+3y-2z+2=0$

    Solving this system gives the solution

    $\displaystyle x=\frac{17}{11}$

    $\displaystyle y=-\frac{5}{11}$

    $\displaystyle z=\frac{29}{11}$
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