Why is the vector v=i+j+4k perpendicular to the plane given by x+y+4z=0 ?I understand that given an equation defining a plane, you can just read off the coefficients for each variable to get a normal vector.

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Clare Chung · Accepted Answer

If you have two vectors       a    1    i  +      a    2    j  +      a    3    k and       b    1    i  +      b    2    j  +      b    3    k in             R        3  , then the angle   θ between them is such that  cos  ⁡  θ  =                    a        1                    b        1            +              a        2                    b        2            +              a        3                    b        3                                      a          1                                  2                          +                  a          2                                  2                          +                  a          3                                  2                                                b          1                                  2                          +                  b          2                                  2                          +                  b          3                                  2                                .Now, take any element   x  i  +  y  j  +  z  k from that plane. Then the angle   θ between   x  i  +  y  j  +  z  k and   i  +  j  +  4  k is such that  cos  ⁡  θ  =            x      +      y      +      4      z                                x          2                +                  y          2                +                  z          2                            6              =  0.So,   θ  =      90    ∘

Trevor Rush · Accepted Answer

Let us consider the plane   P  =  {  (  x  ,  y  ,  z  )  ∈            R              3        ∣  x  +  y  +  4  z  =  0  }Clearly,   O  =  (  0  ,  0  ,  0  )  ∈  P ,   A  =  (  1  ,  −  1  ,  0  )  ∈  P and   B  =  (  0  ,  −  4  ,  1  )  ∈  PLet us denote w=A−O=(1,−1,0) and z=B−O=(0,−4,1).Thus the lines                              {                                                                      L                                      1                                                  =                {                (                x                ,                y                ,                z                )                ∈                                                      R                                                        3                                                  ∣                (                x                ,                y                ,                z                )                =                O                +                t                w                }                                                                                                                            L                                      2                                                  =                {                (                x                ,                y                ,                z                )                ∈                                                      R                                                        3                                                  ∣                (                x                ,                y                ,                z                )                =                O                +                t                z                }                                                                  are contained in P, are distinct and are not parallel.Moreover, we do also have that                              {                                                    ⟨                v                ,                w                ⟩                =                ⟨                (                1                ,                1                ,                4                )                ,                (                1                ,                −                1                ,                0                )                ⟩                =                1                −                1                +                0                =                0                                                                                                          ⟨                v                ,                z                ⟩                =                ⟨                (                1                ,                1                ,                4                )                ,                (                0                ,                −                4                ,                1                )                ⟩                =                0                −                4                +                4                +                0                =                0                                                                  Consequently, since   v  ⊥      L          1       and   v  ⊥      L          2      , we conclude that   v  ⊥  P, and we are done.

Why is the vector v=i+j+4k perpendicular to the plane given by x+y+4z=0 ?

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