How do you find the parametric equation through the point (6, 0, 10) and parallel to the plane − 3 x + 2 y − 4 z = 0 ?

Question

How do you find the parametric equation through the point (6, 0, 10) and parallel to the plane   −  3  x  +  2  y  −  4  z  =  0 ?

Alexis Fields · Accepted Answer

Step 1Any plane parallel to the plane       -    3    x    +    2    y    -    4    z    =    0   will have an equation of the same form but equal to a different constant:      -    3    x    +    2    y    -    4    z    =    C                 [1]      To force the point             (      6      ,      0      ,      10      )       to be in the plane, substitute the point into equation [1] and solve for C:      -    3          (      6      )        +    2          (      0      )        -    4          (      10      )        =    C        C    =    -    58  The scalar equation of the plane parallel to the given plane is:      -    3    x    +    2    y    -    4    z    =    -    58                 [2]      The vector equation of the plane is            (      x      ,      y      ,      z      )        =          (      6      ,      0      ,      10      )        +    s          (              u        ¯            )        +    t          (              v        ¯            )      Where             u      ¯                    and                    v      ¯       are any two vectors in the plane.Let&#039;s use equation [2] to find two other points in the plane:Let       x    =    6                and              y    =    2        -    3          (      6      )        +    2          (      2      )        -    4    z    =    -    58        -    4    z    =    -    44        z    =    11  The vector from the point             (      6      ,      0      ,      10      )       to the point             (      6      ,      2      ,      11      )       is:            u      ¯        =    2          j      ^        +          k      ^      Let       x    =    10                and              y    =    0        -    3          (      10      )        +    2          (      0      )        -    4    z    =    -    58        -    4    z    =    -    28        z    =    7  The vector from point             (      6      ,      0      ,      10      )       to the point             (      6      ,      0      ,      7      )       is:            v      ¯        =    4          i      ^        -    3          k      ^      A vector equation of the plane is:            (      x      ,      y      ,      z      )        =          (      6      ,      0      ,      10      )        +    s          (      2              j        ^            +              k        ^            )        +    t          (      4              i        ^            -      3              k        ^            )      From the above equation, we can write a set of parametric equations:      x    =    4    t    +    6        y    =    2    s        z    =    s    -    3    t    +    10  NOTE: These are not unique equations, because we could have chosen other points and obtained different vectors.

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