Ive been stuck on this problem for a while now,
Consider the two lines in the xy-plane, given the equations:
L_{1}:ax+by=k
L_{2}:cx+dy=l
where all four a,b,c and d are nonzero
i) Give the parametric Vector equations for both L_{1 }and L_{2}
ii) Show that L_{1} and L_{2} intersect in a single point if and only if ad-bc≠0
iii) Suppose that ad-bc≠0, and let P be the point where L_{1} and L_{2} intersect. Find the parametric scalar equations of the line passing through P perpendicular to both L_{1} and L_{2}.
Now if this didn't have a,b,c and d as variables i thought i would have subbed in points on L_{1} to give me points to use as r_{0} in the equation r=r_{0}+tv but im not sure how to tackle it.
any help would be much appreciated.
we can pick a point on L_{1} to use as r_{0}, it really doesn't matter which point we use.
the obvious one is taking x = 0, in which case y = k/b, so we can use r_{0} = (0,k/b) (note we need b non-zero for this to work).
to find the vector v, we need to know the slope of the line. we can find this various ways, but let's just find out what y is when x = 1.
this gives: a + by = k, so y = (k-a)/b. now we have two points: (0,k/b), and (1,(k-a)/b). so the slope is rise/run = [(k-a)/b - k/b]/1 = (k - a - k)/b = -a/b.
so the line parallel to L_{1}, that passes through the origin is:
y = -(a/b)x, or vectors of the form (x,-ax/b) = x(1,-a/b).
so we can let v = (1,-a/b), so that:
L_{1} = (0,k/b) + t(1,-a/b)
if t = x, this means a point on our line is (0,k/b) + (x,-a/b) = (x,(k-ax)/b), so that:
y = (k-ax)/b
by = k - ax
ax + by = k.
L_{2} can be derived the same way.
note that L_{2} has slope -c/d, and since ad - bc ≠ 0, -bc ≠ -ad, so -a/b ≠ -c/d (they are not PARALLEL lines). therefore, they DO intersect.
where?
suppose we have (0,k/b) + t(1,-a/b) = (0,l/d) + t(1,-c/d), that is:
(t,(k-at)/b) = (t,(l-ct)/d). clearly:
d(k-at) = b(l-ct) so:
dk -adt = bl - bct, so (ad - bc)t = dk - bl, so t = (dk - bl)/(ad - bc) (again, you see why we MUST have ad - bc ≠ 0). thus they intersect at the point:
P = ((dk - bl)/(ad - bc), (ck - al)/(ad - bc))
however, if ad - bc = 0, then ad = bc, so a/b = c/d, so -a/b = -c/d, and the two lines are parallel (have the same slope) and cannot intersect, unless they coincide (in which case they do not intersect in a single point, but everywhere along them).
for iii) you should be able to take the cross-product of the two v's you found, and you already have the point P. or: you can write P = (x,y), in which case in three dimensions, the desired line is:
L_{3} = (x,y,0) + t(0,0,1). you cannot find a line perpendicular to BOTH lines that lies in the xy-plane.
if, however, you are supposed to find a line perpendicular to *each* line (this will be two new lines), you can proceed as follows:
L_{3} perpendicular to L_{1} must have slope b/a. so it is of the form P + t(1,b/a) (you could also write this as P + t(a,b)). L_{4} should be easy.