# Further maths vector question

#1
I don't understand the first line of working. Why does the vector equation of AB have the same direction vector (5,4,3) as the two lines, when AB should technically be perpendicular to those lines and have a different direction vector? What am I missing here?
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1 year ago
#2
(Original post by alulaaustralis)
I don't understand the first line of working. Why does the vector equation of AB have the same direction vector (5,4,3) as the two lines, when AB should technically be perpendicular to those lines and have a different direction vector? What am I missing here?
Note that they aren't giving a vector equation for the line AB, they are giving the particular vector that takes you from A to B. The t in this expression is not a variable t that can take any value, it is one particular value that gives you the vector from A to B.

What you say about AB being perpendicular to the lines is of course true, and this is what they do in the next stage (taking the scalar product of the vector from A to B with the direction vector of the lines) to work out the particular value of t that makes this work.
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#3
(Original post by Pangol)
Note that they aren't giving a vector equation for the line AB, they are giving the particular vector that takes you from A to B. The t in this expression is not a variable t that can take any value, it is one particular value that gives you the vector from A to B.

What you say about AB being perpendicular to the lines is of course true, and this is what they do in the next stage (taking the scalar product of the vector from A to B with the direction vector of the lines) to work out the particular value of t that makes this work.
I see, so they have simply taken A away from B hence the t meaning μ - λ ?
Thank you so much!
0
1 year ago
#4
(Original post by alulaaustralis)
I don't understand the first line of working. Why does the vector equation of AB have the same direction vector (5,4,3) as the two lines, when AB should technically be perpendicular to those lines and have a different direction vector? What am I missing here?
In the first line you set up a vector for the distance AB by determining the position and direction vector. So far, so good. For a direction vector you need a parameter in form of t. The general parameter t can be got by the difference of the Greek letters given in the first line (can't see them in attachment) or read out in the equation for r. To get the concrete value of parameter t, you calculate the scalar product for perpendicular direction. The rest is a piece of cake: put the value t in your vector AB, get the coordinates and as last step you calculate the length of the vector with these coordinates.
0
1 year ago
#5
(Original post by alulaaustralis)
I see, so they have simply taken A away from B hence the t meaning μ - λ ?
Thank you so much!
Yeah, that's exactly it. I can see why it looks confusing, as λ and μ are used as variable parameters in the definition of the lines themselves, but at the start of their working they mean to take the particular values of λ and μ that get you to A and B. So when they then say that t = μ - λ, that is one particular value of t that gives you the vector from A to B, which they then go on to determine.
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