Writing a vector a a sum of two vectorsWrite u = [ 9 − 1 2 ] as a sum of two vectors where one is orthogonal to the plane x+2y−2z=0 and the other is parallel to the same plane.I've attempted to use the fact that the normal vector to the plane, [ 1 2 − 2 ] , can be seen from the given plane equation above. We should be able to write u as u = [ 9 − 1 2 ] = s [ 1 2 − 2 ] − t [ a b c ] .How do I find a,b,c in the vector that is parallel to the plane?

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Writing a vector a a sum of two vectorsWrite       u    =      [                            9                                      −          1                                      2                      ]    as a sum of two vectors where one is orthogonal to the plane x+2y−2z=0 and the other is parallel to the same plane.I&#039;ve attempted to use the fact that the normal vector to the plane,       [                            1                                      2                                      −          2                      ]  , can be seen from the given plane equation above. We should be able to write u as       u    =      [                            9                                      −          1                                      2                      ]    =  s      [                            1                                      2                                      −          2                      ]    −  t      [                            a                                      b                                      c                      ]    .How do I find a,b,c in the vector that is parallel to the plane?

Andres Barrett · Accepted Answer

Vector projection is your best friend. By projecting             u      →       onto             n      →        :=  ⟨  1  ,  2  ,  −  2  ⟩, you capture ALL of the movement of             u      →       that is in the direction of             n      →       ; therefore, what is left over should be orthogonal.Vector projection is most easily expressed when the target vector is a unit vector; so, let&#039;s compute            n      ^        :=                    n        →                    ‖                        n          →                    ‖        =      [                                        1            3                                                            2            3                                                −                      2            3                                ]    .Then we can compute      proj                            n          →                      (            u      →        )  =      proj                            n          ^                      (            u      →        )  :=  (            u      →        ⋅            n      ^        )              n      ^        =  1  ⋅      [                                        1            3                                                            2            3                                                −                      2            3                                ]    =      [                                        1            3                                                            2            3                                                −                      2            3                                ]    .Then, consider what happens if we &#039;remove&#039; this component from             u      →       :            u      →        −      proj                            n          ^                      (            u      →        )  =      [                                        26            3                                                −                      5            3                                                            8            3                                ]    .You can verify that this is orthogonal to             n      →       (or             n      ^      ) via the dot product.

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