Exercise: r = {(1,1,1)} + Sp{(1,2,3)}
a)Find the distance between the point(5,6,7) and r
My Resolution Method:
Finding the orthogonal complenet of r:
$$(x,y,z).(1,2,3) = 0 <=> Sp((-2,1,0),(-3,0,1))$$
What do I have to do next? I know I have to use the orthogonal projection but I dont know how
Thank you so much for your help
One way to get the distance is to get the orthogonal projection of the point over the line. To do that you need to get the plane perpendicular to the line through the point and then the intersection.
Since the direction of the line is given by $(1,2,3)$ this vector is the normal vector of the plane. If we want it to contain the point $(5,6,7)$ then it has the equation $$x-5+2(y-6)+3(z-7)=0.$$ That is $$x+2y+3z=38.$$ Now we write the parametric equations of the line $$(x,y,z)=(1+t,1+2t,1+3t).$$ We get the intersection point by solving the equation
$$1+t+2(1+2t)+3(1+3t)=38.$$ We get $t=32/14=16/7.$ That is, the intersection point is $$(1+16/7,1+2\cdot 16/7,1+3\cdot 16/7).$$ This point is the orthogonal projection. Now, to finish, you get the distance between both points which is the distance between the given point and the line.