Let vector v=<2,3,4> and point P=(1,2,3).
All vectors parallel to v can be represented by tv=t<2,3,4>=<2t,3t,4t> where t is an arbitrary scalar.
We can convert point P to a vector i+2j+3k represented by <1,2,3>. If O is the origin the vector OP=<1,2,3>. If we introduce an arbitrary point Q(x,y,z) we can create the vector OQ=<x,y,z>.
OP+PQ=OQ, so PQ=OQ-OP=<x,y,z>-<1,2,3>=<x-1,y-2,z-3>.
If v is the vector associated with PQ, then <x-1,y-2,z-3>=<2t,3t,4t>. Therefore:
x-1=2t, y-2=3t, z-3=4t, from which t=(x-1)/2=(y-2)/3=(z-3)/4. (Note that (1-x)/(-2)=(x-1)/2.) So the line vector PQ is the given line. Therefore v=<2,3,4> is the vector passing through the point P(1,2,3).
The normal n to the given plane 3x+4y+5z=2 is <3,4,5> (the coefficients). The line vector is neither parallel to the normal (a multiple of the normal vector) nor parallel to the plane (perpendicular to the normal, because the dot product of the line and normal vectors is non-zero).
We can use the parameterised equation of the line to write x, y and z in terms of the parameter t:
x=2t+1, y=3t+2, z=4t+3. We can find the value of t so that we find where the line meets the plane:
3(2t+1)+4(3t+2)+5(4t+3)=2,
6t+3+12t+8+20t+15=2,
38t=2-26=-24, t=-24/38=-12/19.
x=2t+1=-5/19, y=3t+2=2/19, z=4t+3=9/19.
So we have one endpoint A(-5/19,2/19,9/19) of the line on the plane. Another point on the line is the point P(1,2,3), identified at the beginning of this solution. We need to find N, where the perpendicular from P meets the plane. This perpendicular is parallel to the normal of the plane n=<3,4,5>. As we've seen earlier all parallels to the normal are given by <3t,4t,5t> where t is a scalar. We will have a right triangle PAN, where PN is the perpendicular, AP the hypotenuse and AP the projection of the line on to the plane. First we need to find the equation of the line through P parallel to the normal of the plane.
Again, we can utilise an arbitrary point Q(x,y,z) and create vector OQ=<x,y,z> and the vector PQ=<x-1,y-2,z-3>, but the vector we use this time is tn=<3t,4t,5t>, which is parallel to the normal of the plane.
<x-1,y-2,z-3>=<3t,4t,5t>, and x-1=3t, y-2=4t, z-3=5t, t=(x-1)/3=(y-2)/4=(z-3)/5, so x=3t+1, y=4t+2, z=5t+3, to substitute into the equation of the plane.
We can now find the point N where the perpendicular line meets the plane:
3(3t+1)+4(4t+2)+5(5t+3)=2, 50t+26=2, 50t=-24, t=-24/50=-12/25.
So N is the point (-36/25+1,-48/25+2,-12/5+3)=(-11/25,2/25,3/5). We now have both end points of the line AN. We need the equation of this line. As before we convert the points into vectors OA and ON by joining them to O.
Whereas at the beginning of this exercise we started with a point and a vector to establish the equation of a line, now we need to derive the vector from the vectors OA and ON.
OA+AN=ON, AN=ON-OA=<(-11/25,2/25,3/5)>-<-5/19,2/19,9/19>=
<-84/475,-12/475,12/95>.
We can use point A or N as our reference point and, again, we can use an arbitrary point Q(x,y,z) and vector parallel to AN=<-84/475t,-12/475t,12/95t>.
Using point N, we end up with x+11/25=-84/475t, y-2/25=-12/475t, z-3/5=12/95t, giving us:
(x+11/25)/(-84/475)=(y-2/25)/(-12/475)=(z-3/5)/(12/95), that is:
-(475x+209)/84=-(475y-38)/12=(95z-57)/12 as the equation of the projection line.