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Thread: Find the cartesian equation of a plane that is parallel to i + j + k - Confused!

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    Find the cartesian equation of a plane that is parallel to i + j + k - Confused!

    Hey guys! My class never learned how to do Cartesian coordinates, so I am rather lost as to how to do this question!

    Question
    Find the Cartesian equation o the plane that contains the line $\displaystyle \frac{x+3}{2} = \frac{y-2}{3} = \frac{z-1}{2} $ and is parallel to the line $\displaystyle r = (1+3t)i + (2t)j + (4-t)k $

    Solution
    I don't understand what the $\displaystyle r = (1+3t)i + (2t)j + (4-t)k $ means. Can I use this and put it into vector or parametric form? If I knew how to do that I could find the solution to the question. Especially vector form.

    Thank you!
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  2. #2
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    Quote Originally Posted by Kakariki View Post
    Find the Cartesian equation o the plane that contains the line $\displaystyle \frac{x+3}{2} = \frac{y-2}{3} = \frac{z-1}{2} $ and is parallel to the line $\displaystyle r = (1+3t)i + (2t)j + (4-t)k $
    From $\displaystyle r = (1+3t)i + (2t)j + (4-t)k $ we get $\displaystyle
    \ell (t) = \left\{ {\begin{array}{l}
    {x = 1 + 3t} \\
    {y = ~~~2t} \\
    {z = 4 - t} \\ \end{array} } \right.$
    You want the plane that contains $\displaystyle (-3,2,1)$ with normal $\displaystyle <3,2,-1>\times <2,3,2>$.
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