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Thread: position vector of the point to the intersection of the line

  1. #1
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    position vector of the point to the intersection of the line

    Find the position vector of the point of intersection of the line

    x= 28110 +t -120-1

    and the plane

    x= -253-18 +λ-10-2-1 +μ-130-9


    The syntax for entering the column vector abc is < a,b,c > .

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  2. #2
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    Re: position vector of the point to the intersection of the line

    The problem with writing a vector as "abc" is that if one or more of "a", "b", and "c" has more than one digit, it is impossible to tell which! That is, is "28110" supposed to be <28, 1, 10> or <2, 81, 10> or <2, 8, 110>? Is "-253-18" supposed to be <-25, 3, -18> or <-2, 53, -18> or <-253, -1, 8>?

    If, for example, if the line is <2, 81, 10>+ t<-1, 20, -1> and the plane is $\displaystyle <-25, -3, -18>+ \lambda<-10,-2, -1>+ \mu<-13, 0, 9>$, then eliminating $\displaystyle \lambda$ and $\displaystyle \mu$ from the equations for the plane, -18x- 77y+ 26z= 213. The equation of the line gives x= 2- t, y= 81+ 20t, and z= 10- t. Replace x, y, and z in the equation for the plane with those so you have a single linear equation in "t". Solve that equation for t, then put that t into the equations for x, y, and z.
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  3. #3
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    Re: position vector of the point to the intersection of the line

    Quote Originally Posted by ckhu199 View Post
    Find the position vector of the point of intersection of the line

    x= 28,1,10 +t -12,0,-1

    and the plane

    x= -25,3,-18 +λ-10,-2,-1 +μ-13,0,-9


    The syntax for entering the column vector abc is < a,b,c > .

    Quoting the text, it was possible to see what the breakdown was, so I added commas for clarity.
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