Is anyone able to help me with regards to this question?
Find the point on a plane $3x + 4y + z = 1$ that is closest to $(1,0,1)$
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$\begingroup$The normal vector to the plane is $\langle 3,4,1\rangle$. The point you seek would have to be some multiple of this vector added to $(1,0,1)$. $$P=(1,0,1)+c\langle 3,4,1\rangle=(1+3c,4c,1+c)$$ But this point has to satisfy the plane's equation: $$\begin{align} 3(1+3c)+4(4c)+(1+c)&=1\\ 26c+4&=1\\ c&=-\frac{3}{26} \end{align}$$
So $$P=(1,0,1)-\frac{3}{26}\langle 3,4,1\rangle=\left(\frac{17}{26},-\frac{12}{26},\frac{23}{26}\right)$$
$\endgroup$ 1 $\begingroup$With Lagrange Multipliers, we have the distance function $f(x,y,z)=(x-1)^2+y^2+(z-1)^2$ (squaring to simplify the calculations) and the constraint $3x+4y+z=1$. We get $$(x-1,y,z-1)=\lambda(3,4,1)$$ Therefore $$(x-1)/3=(y/4)=(z-1/1)=\lambda$$ and as the user above stated this gives the closest point.
$\endgroup$ $\begingroup$Yes. This is a multivariable minimization problem in which you want to minimize some function f(x,y,z) subject to the constraint g(x,y,z) - c = 0.
The first thing to understand is that the function you are minimizing is the distance from the point (1,0,1).
$$ f(x,y,z) = \sqrt{(x-1)^2+(y^2)+(z-1)^2} $$
and your constraint is
$$ 3x+4y+z-1 = 0 $$
From here you can use the standard Lagrange Multipliers method. The wikipedia intro section here covers this nicely.
One last tip: minimizing the square of the distance function will give the same result, and is much easier :)
$\endgroup$ $\begingroup$Let $(x, y, z)$ be the point in question. The distance is given by $\sqrt{(x - 1)^2 + y^2 + (z - 1)^2}$. By Cauchy Schwarz, $\left((x-1)^2 + y^2 + (z-1)^2\right)(3^2 + 4^2 + 1^2) \geq (3x + 4y + z - 4)^2$, so $\left((x-1)^2 + y^2 + (z-1)^2 \right) \geq \frac{9}{26}$
Equality is reached when $\frac{x-1}{3} = \frac{y}{4} = \frac{z-1}{1}$. Solving using $3x + 4y + z = 1$ gives $\left(\frac{17}{26}, -\frac{6}{13}, \frac{23}{26} \right)$
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