Find the point on a plane $3x + 4y + z = 1$ that is closest to $(1,0,1)$ 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 $(1,0,1)$
http://i.imgur.com/ywdsJi7.png 

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 $\left(\frac{17}{26}, -\frac{6}{13}, \frac{23}{26} \right)$
