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Math Help - Finding Parametric and Symmetric Equations for a line

  1. #1
    Junior Member Shadow236's Avatar
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    Finding Parametric and Symmetric Equations for a line

    I got stuck on this question because it says that the vector value must be perpendicular to two other vectors. What am I supposed to do?

    Find the Parametric and Symmetric Equations for the line through (2,1,0) and perpendicular to both i + j and j + k.

    Thank you
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    Re: Finding Parametric and Symmetric Equations for a line

    Hey Shadow236.

    Firstly you need to find the direction vector of the line and this is the vector perpendicular to i + j and j + k (hint: cross product).

    The final hint is to note the equation of a line: r(t) = a + d*t where a is a point on the line, d is a direction vector (usually normalized) and t is a parameter (scalar number).
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    Junior Member Shadow236's Avatar
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    Re: Finding Parametric and Symmetric Equations for a line

    Oh okay, so I use the cross product to find the d value. After doing that I ended up with d = < 1, -1, 1 >.

    After finding that I ended up with the Parametric equations:
    x = 2 + t
    y = 1 - t
    z = t

    With that being said, the Symmetric equations are:
    x - 2 = -y + 1 = z

    Thanks for your help, please let me know if I am wrong.
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    Re: Finding Parametric and Symmetric Equations for a line

    Quote Originally Posted by Shadow236 View Post
    Find the Parametric and Symmetric Equations for the line through (2,1,0) and perpendicular to both i + j and j + k.
    Notice that \left( {i + j} \right) \times \left( {j + k} \right) = \left\langle {1, - 1,1} \right\rangle .

    Thus we have \ell (t):\left\langle {2 + t,1 - t,t} \right\rangle in parametric form.

    In symmetric form: \frac{{x - 2}}{1} = \frac{{y - 1}}{{ - 1}} - \frac{z}{1}
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    Re: Finding Parametric and Symmetric Equations for a line

    Another way to do this (for those of us who just don't like the cross product!):
    Any line through (2, 1, 0) can be written in the form x= at+ 2, y= bt+ 1, z= ct. A vector in the direction of that line is ai+ bj + ck. For that to be perpendicular to i+ j, their dot products must be 0: (ai+ bj + ck).(i+ j)= a+ b= 0 so a= -b. For that to be perpendicular to j+ k, their dot products must be 0: (ai+ bj + ck).(j+ k)= b+ c= 0 so c= -b. That is, the vector must be of the form -bi+ bj- bk. Since the length does not matter, we can take b= 1 to get -i+ j- k. The line has parametric equations x= -t+ 2, y= t+ 1, z= -t. (My parameter "t" is the negative of Plato's.)

    We can solve each of those for t: t= -x+ 2, t= y- 1, and t= -z. Since those are all equal to t, they are equal to each other: -x+ 2= y- 1= -z. Or course, we can multiply through by -1 to get x- 2= 1- y= z, Plato's form for the "symmetric form".
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