Intersection of Line Segments (1 Viewer)
- Thread starter zacn
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firstly, these are vector equations of lines, not line segments, unless the domain of lambda is restricted between some finite values.
Also, not that the lambda in these two lines aren't really the same variable- they can take on different values. It might be easier to notate one of them as mu or something else. (though in questions they might give them both to you as lambda)
When the lines intersect, 2i+3j +lambda(-2i+3j) = 6i +2j + mu(i-4j).
Equating i and j components will give you a set of simultaneous equations to solve for lambda and mu. Then subbing the value of lambda (or mu) into the appropriate line equation will give you the position vector of the point of intersection
Also, not that the lambda in these two lines aren't really the same variable- they can take on different values. It might be easier to notate one of them as mu or something else. (though in questions they might give them both to you as lambda)
When the lines intersect, 2i+3j +lambda(-2i+3j) = 6i +2j + mu(i-4j).
Equating i and j components will give you a set of simultaneous equations to solve for lambda and mu. Then subbing the value of lambda (or mu) into the appropriate line equation will give you the position vector of the point of intersection
As pointed out by ultra908, question asks for the point of intersection of 2 lines:
Rewriting these vector equations in their parametric form:
Equating x and y for the 2 lines, we get after simplifying:
Using this, we get: x = 2 -2(-3) = 8 and y = 3 + 3(-3) = -6
So the point of intersection is (8, -6)
The line L1 passes thru the point (2,3) with gradient 3/(-2) = -3/2 with eqn: 3x + 2y = 12
and L2 passes thru the point (6,2) with gradient (-4)/1 = -4 with eqn: 4x + y = 26
The point (8 ,-6) satisfies both these equations; therefore (8, -6) lies on both lines and is therefore the point of intersection of L1 and L2. This also indirectly gives me the satisfaction that I had got the value of the parameter
Remark
One way to specify a line is to provide one point on the line and its direction. In typical co-ord geometry this direction is the gradient.
In vector method, the direction is provided by the direction vector (not unique), e.g. for L1 it is
In L!: the vector eqn is based on the one point on L1, (2,3) and the said direction vector.
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yes using vectors