Question 1010756
consider the following scalar equation of a plane. x+3y+z=0

Find the orthogonal complement of the vector <1,2,1>. 
<pre>
We want to find the set of vectors parallel to the plane that are
perpendicular to the vector <1,2,1>.

Vectors in or parallel to the plane are vectors < x,y,z > such that 

x+2y+z = 0

Vectors perpendicular to <1,3,1> are < x,y,z > such that

<1,3,1>&#8729;< x,y,z > = x+3y+z = 0

So we have this under-determined system

{{{system(x+2y+z=0,x+3y+z=0)}}}

Subtracting them we get y = 0
Substituting we get x+2(0)+z = 0, or x+z=0, or z=-x

So the orthogonal complement is the set of vectors of the form

< k,0,-k >
</pre>
Also find the point on the plane which is closest to (3,4,1).
<pre>
That is the perpendicular distance from the point. We use the
formula for the perpendicular distance from the point
 
(x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>) to the plane

ax + by + cz + d = 0, which is:

is {{{D}}}{{{""=""}}}{{{abs(ax[1]+bx[1]+cx[1]+d)/sqrt(a^2+b^2+c^2)}}}

So in our case (x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>) = (3,4,1),
a=1, b=3, c=1, d=0

{{{D}}}{{{""=""}}}{{{abs((1)(3)+(3)(4)+(1)(1)+0)/sqrt((1)^2+(3)^2+(1)^2)}}}

{{{D}}}{{{""=""}}}{{{abs(3+12+1)/sqrt(1+3+1)}}}

{{{D}}}{{{""=""}}}{{{16/sqrt(5)}}}

Edwin</pre>