Vectors - Planes

Question

Consider the point $A(1, -3, 2)$ and the plane $2x-y+2z=8$. Find the coordinates of the point on the plane nearest to point $A$. I'm confused about this question because there have been no questions even similar to this in the book so far. We have a formula for the distance from a point to a plane but so far nothing like this.

Answer 1

Let $P$ be the plane and $L=\{(1,-3,2)+t(2,-1,2) : t \in \mathbb R\}$.

Then $P \cap L=\{Q\}$. Picture !!!

$Q$ is the point you are looking for !

Answer 2

Let $B$ is a needed point.

Hence, the vector normal to the plain is $(2,-1,2)$ and from here $B(1+2t,-3-t,2+2t)$.

Thus, $$2(1+2t)-(-3-t)+2(2+2t)=8,$$ which gives $t=-\frac{1}{9}$ and $B\left(\frac{7}{9},-\frac{26}{9},\frac{16}{9}\right)$.

Answer 3

Let the point closest to A be another point B, which lies on the plane.

AB must be perpendicular to the plane so that B is closest to A.

Now the direction ratios of any line perpendicular to the given plane is $(2,-1,2)$

Thus, the direction ratios of line AB are $(2,-1,2)$

The direction cosines of line AB are $(\frac{2}{3},\frac{-1}{3},\frac{2}{3})$

Any point on line AB at distance $r$ from A is given by $(1+\frac{2r}{3}, -3-\frac{r}{3},2+\frac{2r}{3})$

Now to find the coordinates of B, which is a point on line AB, we have to determine $r$ i.e the perpendicular distance of A from the plane.

$r=\vert{\frac{2(1)-(-3)+2(2)-8}{\sqrt{2^2+(-1)^2+2^2}}}\vert=\pm\frac{1}{3}$

So the coordinates of B would be $[1\pm\frac{2}{3}(\frac{1}{3}), -3\mp\frac{1}{3}(\frac{1}{3}), 2\pm\frac{2}{3}(\frac{1}{3})]= [\frac{11}{9},\frac{-28}{9},\frac{20}{9}]$ or $[\frac{7}{9},\frac{-26}{9},\frac{16}{9}]$

We see only $[\frac{7}{9},\frac{-26}{9},\frac{16}{9}]$ satisfies the equation of the plane, hence these are the required coordinates.