Isa Trevino

2021-04-06

Can u find the point on the plane $x+2y+3z=13$ that is closest to the point (1,1,1). You need to minimize the function

rogreenhoxa8

Skilled2021-04-08Added 109 answers

Apply Lagrange Multipliers:

$D={({(x-1)}^{2}+{(y-1)}^{2}+{(z-1)}^{2})}^{\frac{1}{2}}$

Now, we can use Constraint Surface:

$x+2\cdot y+3\cdot z=13$to $g(x,y,z)=x+2\cdot y+3\cdot z-13$

Formulate Condition (for constant scalarmultiplier "$\lambda$"):

$\mathrm{\u25b3}downD=\lambda \cdot \mathrm{\u25b3}downg$

$\to (\frac{1}{2}\cdot {D}^{\frac{-1}{2}}\cdot [2\cdot (x-1)\cdot i+2\cdot (y-1)\cdot j+2\cdot (z-1)\cdot k]$

$=\lambda \cdot [\left(1\right)\cdot i+\left(2\right)\cdot j+\left(3\right)\cdot k]$

$(x-1)=\left(1\right)\cdot \lambda \cdot {D}^{\frac{1}{2}}$

$(y-1)=\left(2\right)\cdot \lambda \cdot {D}^{\frac{1}{2}}$

$(z-1)=\left(3\right)\cdot \lambda \cdot {D}^{\frac{1}{2}}$

$\to (x-1)=\frac{y-1}{2}=\frac{z-1}{3}$

$\to [y=2\cdot x-1]$ and $[z=3\cdot x-2]$

$x+2\cdot y+3\cdot z=13$

$\to x+2\cdot [2\cdot x-1]+3\cdot [3\cdot x-2]=13$

$\to x+4\cdot x-2+9\cdot x-6=13$

$\to 14\cdot x=21$

$\to x=\left(\frac{3}{2}\right)$

$(x=\frac{3}{2})$

$y=2\cdot \left(\frac{3}{2}\right)-1=\left(2\right)$

$z=3\cdot \left(\frac{3}{2}\right)-2=\frac{5}{2}$

Jeffrey Jordon

Expert2021-10-14Added 2605 answers

Given x+2y+3z=13 Point (1,1,1)

The normal vector to the plane is <1,2,3>

The point you seek would have to be multiple of this vector added to (1,1,1)

$\therefore P=(1,1,1)+c<1,2,3>$

=(1+c,1+2c,1+3c)

But this point has to satisfy the plane`s equation.

(1+c)+2(1+2c)+3(1+3c)=13

1+c+2+4c+3+9c=13

14c=13-6

$c=\frac{7}{14}=\frac{1}{2}$

$\therefore $ the point P is $(1+\frac{1}{2},1+2\times \frac{1}{2},1+3\times \frac{1}{2})$

$=(\frac{3}{2},2,\frac{5}{2})$

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