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    Points A,B,C,D in a point have position vectors a=6i+8j, b=(3/2)a, c=6i+3j, d=(5/3)c respectively. Write down the vector equations of the lines AD and BC and find the position vector of their point of intersection?

    Can I have some help answering this question please? Thanks in advance!!
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    (Original post by Lavanyaa)
    Points A,B,C,D in a point have position vectors a=6i+8j, b=(3/2)a, c=6i+3j, d=(5/3)c respectively. Write down the vector equations of the lines AD and BC and find the position vector of their point of intersection?

    Can I have some help answering this question please? Thanks in advance!!
    AD= d-a = (10i+5j)-(6i+8j)= 4i-3j

    BC= c-b = (6i+3j)-(9i+12j)= -6i-9j

    Equation for the line passing through AD is (6i+8j)+t(4i-3j)
    equation for the line passing through BC is (9i+12j)+s(-6i-12j)/(-1i-2j)

    To find point of intersection you must find the value of either S or T and sub that value into the equation of any of the equations. This will give the point of intersection after you use simultaneous equations to find s and T

    Note Im sure I went wrong somewhere with my calculations of the Vector So I recommend you do on your own but follow the steps I showed. I was not sure what the question was actually asking because you never mentioned the line passing through AD or BC so depending on that you choose what equation to use

    6+4t=9-6s (written as 12+8t=18-12s)
    8-3t=12-12j

    20+5t=30
    5t=10
    t=2

    Thus sub in 2 into your first equation to get the point of intersection 6i+2j+8i-6j= 14i-4j (once again Im sure this is wrong because I have no checked it once but I hope you get the idea)
 
 
 
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