The perpendicular from a point to a line minimises the distance from the point to that line. Use quadratic theory to find the coordinates of the foot of the perpendicular from (1,1,2) to the line which has the equations x=1+µ y=2-µ z=3+µ
I can't seem to understand this question because the vector µ(1,-1,1) seems to have 2 different perpendicular lines. I also don't know how to use the quadratic theory to the question.
On
the vector from point $(1,2,3)$ on the line to the point $(1,1,2)$ not on the line.
$v = (0,-1,-1)$
This vector is perpendicular to the line... that is $(1,-1,1)\cdot(1,-1,-1) = 0$
Normally you would have to find the projection of the vector, and then the orthogonal vector. But that is not necessary here.
Someone is being very nice to you....
$\|v\| = \sqrt 2$
In fact, there is an infinite number of vectors perpendicular to $(1,-1,1)$.
It appears that whoever posed this problem wants you to use the definition of the distance between two points to solve it. So, form the expression for the distance between an arbitrary point on the line and the point $(1,1,2)$: $$\sqrt{(1+\mu-1)^2+(2-\mu-1)^2+(3+\mu-2)^2}=\sqrt{3\mu^2+2}.$$ Now use what you know of quadratic equations to find the value of $\mu$ that minimizes this.
Alternatively, you can use the fact that the vector from the point to the foot must be perpendicular to the line. This gives you the linear equation $(1+\mu-1,2-\mu-1,3+\mu-2)\cdot(1,-1,1)=0$.