1. ## 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

2. 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...

3. ## 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

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

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

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