Parametric Vector and Scalar equations

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.

Re: Parametric Vector and Scalar equations

Quote:

Originally Posted by

**olliethechum** 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.

Hi there,

just a heads up, I think it's suppose to be L_{2}: **d**cx+dy=l

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Somebody please help with this problem!

Re: Parametric Vector and Scalar equations

Quote:

Originally Posted by

**olliethechum** 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}.

..........

Re: Parametric Vector and Scalar equations

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.