Question

1. find the shortest distance from the point (0, 0, 8) to the plane x + 2y+3z = 6 in three different ways: (a) by projection like you would have done for the first test (b) by minimizing the square of the distance function (written as a function of two variables) and then taking a square root (c) by using lagrange multipliers

Explanation:

Step1: Recall distance formula

The distance $d$ from a point $(x_0,y_0,z_0)$ to a plane $Ax + By+ Cz+D = 0$ is given by $d=\frac{\vert Ax_0 + By_0+ Cz_0+D\vert}{\sqrt{A^{2}+B^{2}+C^{2}}}$. First, rewrite the plane equation $x + 2y+3z=6$ as $x + 2y+3z - 6=0$, and the point is $(0,0,8)$.

Step2: Substitute values into formula

Substitute $x_0 = 0,y_0 = 0,z_0 = 8,A = 1,B = 2,C = 3,D=-6$ into the distance formula. We get $d=\frac{\vert1\times0 + 2\times0+3\times8 - 6\vert}{\sqrt{1^{2}+2^{2}+3^{2}}}$.

Step3: Simplify the expression

First, simplify the numerator: $1\times0 + 2\times0+3\times8 - 6=24 - 6=18$. Then, simplify the denominator: $\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1 + 4+9}=\sqrt{14}$. So $d=\frac{18}{\sqrt{14}}=\frac{9\sqrt{14}}{7}$.

For part (b):
Let the point on the plane $x + 2y+3z=6$ be $(x,y,z)$. The distance $D$ from $(0,0,8)$ to $(x,y,z)$ is $D=\sqrt{(x - 0)^{2}+(y - 0)^{2}+(z - 8)^{2}}$. Since $z = 2-\frac{1}{3}x-\frac{2}{3}y$, we can write the square - of the distance function $f(x,y)=(x - 0)^{2}+(y - 0)^{2}+(2-\frac{1}{3}x-\frac{2}{3}y - 8)^{2}=x^{2}+y^{2}+(-6-\frac{1}{3}x-\frac{2}{3}y)^{2}$.
Take partial derivatives:
$\frac{\partial f}{\partial x}=2x+2(-6-\frac{1}{3}x-\frac{2}{3}y)(-\frac{1}{3})=2x+\frac{2}{3}(6+\frac{1}{3}x+\frac{2}{3}y)$
$\frac{\partial f}{\partial y}=2y+2(-6-\frac{1}{3}x-\frac{2}{3}y)(-\frac{2}{3})=2y+\frac{4}{3}(6+\frac{1}{3}x+\frac{2}{3}y)$
Set $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$ and solve the system of equations to find the critical points. Then use the second - derivative test to confirm it's a minimum. After finding the point $(x,y)$ on the plane, calculate the distance.

For part (c):
Let $f(x,y,z)=(x - 0)^{2}+(y - 0)^{2}+(z - 8)^{2}$ and $g(x,y,z)=x + 2y+3z-6$.
The Lagrange function is $L(x,y,z,\lambda)=f(x,y,z)-\lambda g(x,y,z)=x^{2}+y^{2}+(z - 8)^{2}-\lambda(x + 2y+3z - 6)$.
Take partial derivatives:
$\frac{\partial L}{\partial x}=2x-\lambda = 0\Rightarrow x=\frac{\lambda}{2}$
$\frac{\partial L}{\partial y}=2y-2\lambda = 0\Rightarrow y=\lambda$
$\frac{\partial L}{\partial z}=2(z - 8)-3\lambda = 0\Rightarrow z = 8+\frac{3\lambda}{2}$
Substitute $x,y,z$ into $g(x,y,z)=0$: $\frac{\lambda}{2}+2\lambda+3(8+\frac{3\lambda}{2})-6 = 0$.
Solve for $\lambda$:
$\frac{\lambda}{2}+2\lambda + 24+\frac{9\lambda}{2}-6=0$
$(\frac{1 + 4+9}{2})\lambda+18 = 0$
$7\lambda=-18\Rightarrow\lambda=-\frac{18}{7}$
Then $x=-\frac{9}{7},y =-\frac{18}{7},z=\frac{38}{7}$.
The distance $d=\sqrt{(-\frac{9}{7}-0)^{2}+(-\frac{18}{7}-0)^{2}+(\frac{38}{7}-8)^{2}}=\sqrt{\frac{81 + 324+( - 18)^{2}}{49}}=\sqrt{\frac{81+324 + 324}{49}}=\frac{9\sqrt{14}}{7}$

Answer:

The shortest distance is $\frac{9\sqrt{14}}{7}$ for all three methods.