How to find unit vector perpendicular to plane: 6x-2y+3z+8=0?

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Kieran Orozco · Accepted Answer

If we have a plane       a    x    +    b    y    +    c    z    =    d  , the vector normal to the plane is given by             (      a      ,      b      ,      c      )      Thus as given plane is       6    x    -    2    y    +    3    z    +    8    =    0   or       6    x    -    2    y    +    3    z    =    -    8  and vector normal to this plane is             (      6      ,      -      2      ,      3      )      Now the magnitude of vector             (      6      ,      -      2      ,      3      )       is                               6          2                +                              (            -            2            )                    2                +                  3          2                      =                  36        +        4        +        9              =          49        =    7  Therefore, unit vector perpendicular to plane       6    x    -    2    y    +    3    z    +    8    =    0   is             (              6        7            ,      -              2        7            ,              3        7            )

Chestonky1a · Accepted Answer

Using the outcome, we can get a perpendicular vector.
that

ϕ (x, y, z) = 0 \Rightarrow \nabla ϕ (x, y, z)

is a vector perpendicular to the plane
where

\nabla

is the del operator and is defined as:

\nabla \equiv \vec{i} \frac{\partial}{d x} + \vec{j} \frac{\partial}{\partial y} + \vec{k} \frac{\partial}{\partial z}

keeping in mind that we treat the other variables as constants when we partially discriminate with respect to a variable

ϕ (x, y, z) = 6 x - 2 y + 3 z + 8 = 0

∴ \nabla ϕ (x, y, z) =

\vec{i} \frac{\partial}{\partial x} (6 x - 2 y + 3 z + 8)

+ \vec{j} \frac{\partial}{\partial y} (6 x - 2 y + 3 z + 8)

+ \vec{k} \frac{\partial}{\partial z} (6 x - 2 y + 3 z + 8)

= 6 \vec{i} - 2 \vec{j} + 3 \vec{k}

call this

\vec{n}

a unit vector is then

\hat{\vec{n}} = \frac{\vec{n}}{| \vec{n} |}

∴ \hat{\vec{n}} = \frac{6 \vec{i} - 2 \vec{j} + 3 \vec{k}}{\sqrt{6^{2} + 2^{2} + 3^{2}}}

= \frac{6 \vec{i} - 2 \vec{j} + 3 \vec{k}}{7}

= \frac{6}{7} \vec{i} - \frac{2}{7} \vec{j} + \frac{3}{7} \vec{k}

How to find unit vector perpendicular to plane: 6x-2y+3z+8=0?

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