 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

Apply Lagrange Multipliers:
$D={\left({\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}+{\left(z-1\right)}^{2}\right)}^{\frac{1}{2}}$
Now, we can use Constraint Surface:
$x+2\cdot y+3\cdot z=13$to $g\left(x,y,z\right)=x+2\cdot y+3\cdot z-13$
Formulate Condition (for constant scalarmultiplier "$\lambda$"):
$\mathrm{△}downD=\lambda \cdot \mathrm{△}downg$

$=\lambda \cdot \left[\left(1\right)\cdot i+\left(2\right)\cdot j+\left(3\right)\cdot k\right]$
$\left(x-1\right)=\left(1\right)\cdot \lambda \cdot {D}^{\frac{1}{2}}$
$\left(y-1\right)=\left(2\right)\cdot \lambda \cdot {D}^{\frac{1}{2}}$
$\left(z-1\right)=\left(3\right)\cdot \lambda \cdot {D}^{\frac{1}{2}}$
$\to \left(x-1\right)=\frac{y-1}{2}=\frac{z-1}{3}$
$\to \left[y=2\cdot x-1\right]$ and $\left[z=3\cdot x-2\right]$
$x+2\cdot y+3\cdot z=13$
$\to x+2\cdot \left[2\cdot x-1\right]+3\cdot \left[3\cdot x-2\right]=13$
$\to x+4\cdot x-2+9\cdot x-6=13$
$\to 14\cdot x=21$
$\to x=\left(\frac{3}{2}\right)$
$\left(x=\frac{3}{2}\right)$
$y=2\cdot \left(\frac{3}{2}\right)-1=\left(2\right)$
$z=3\cdot \left(\frac{3}{2}\right)-2=\frac{5}{2}$ Jeffrey Jordon

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)

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

=(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 $\left(1+\frac{1}{2},1+2×\frac{1}{2},1+3×\frac{1}{2}\right)$

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

Do you have a similar question?