FP3 vectors
I don't understand this,
Why does the line AB have the same direction as the two parallel lines? Surely the line is AB is perpendicular to both of the lines?
Also,
Why do we need a unit vector for the common perpendicular? (in the question, they go on to find PQ, where P is a point on one line, and Q is a point on another, and project that onto the unit vector perpendicular to both the lines).
Please some explanations, talk as if you were talking to a 3 year old as I just don't get any of this.
Why does the line AB have the same direction as the two parallel lines? Surely the line is AB is perpendicular to both of the lines?
Also,
Why do we need a unit vector for the common perpendicular? (in the question, they go on to find PQ, where P is a point on one line, and Q is a point on another, and project that onto the unit vector perpendicular to both the lines).
Please some explanations, talk as if you were talking to a 3 year old as I just don't get any of this.
Original post by DH.
I don't understand this,
Why does the line AB have the same direction as the two parallel lines? Surely the line is AB is perpendicular to both of the lines?
Also,
Why do we need a unit vector for the common perpendicular? (in the question, they go on to find PQ, where P is a point on one line, and Q is a point on another, and project that onto the unit vector perpendicular to both the lines).
Please some explanations, talk as if you were talking to a 3 year old as I just don't get any of this.
Why does the line AB have the same direction as the two parallel lines? Surely the line is AB is perpendicular to both of the lines?
Also,
Why do we need a unit vector for the common perpendicular? (in the question, they go on to find PQ, where P is a point on one line, and Q is a point on another, and project that onto the unit vector perpendicular to both the lines).
Please some explanations, talk as if you were talking to a 3 year old as I just don't get any of this.
For example 34:
Assuming A lies on the first line given in the question, it has general position vector (1+5lambda, 2+4lambda, -1+3lambda). Assuming B lies on the second line, it has position vector (2+5mu, 4mu, 1+3mu). Then the vector A to B = (2+5mu, 4mu, 1+3mu) - (1+5lambda, 2+4lambda, -1+3lambda). This equals (1+5mu-5lambda, -2+4mu-4lambda, 2+3mu-3lambda). Simplifying gets you (1,-2,2) + (mu-lambda)(5,4,3). For this step I factored out 5, 4 and 3 respectively from each component for the A to B vector. Writing (mu-lambda) as a single paramter t, gets you the equation of AB.
For the second question:
You need to find a vector which is perpendicular to both lines and project a know vector across this perpendicular vector to find the distance between the skew lines. Therefore, you do the cross product of the direction vectors of each line. Call this vector n. A point on the first line is a and a point on the second line is c. You find the vector (c-a) and project it across the vector n using the dot product. Basically, modulus of ((c-a) dot n) / modulus of n). The reason this gives you the length between the two skew lines is because the dot product itself multiplies a vector by another vector which is in the same direction as itself. i.e. a dot b = mod a * mod b* costheta. This: mod b* costheta is the component of b in the direction of a. Therefore, ((c-a) dot n) / modulus of n) = mod (c-a)*modn*costheta/modn, and the mod n's cancel giving the absolute value of (c-a)cos theta which is the component of (c-a) in the direction of n, thus giving the shortest distance between the skew lines.
I may have complicated it a bit, hopefully you understand a bit better. Also, there should be a video on Khan Academy regarding vector projection, which is the underlying feature of distance between skew lines. I can't remember the name of the video, something like vector dot product I think.
(edited 10 years ago)
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