Question

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

Answer 1

Hint: We first use the formula of finding the vector normal to a given plane. We put the values for $6x-2y+3z+8=0$ in the equation to find the normal. We convert it into a vector form and then find the unit form by dividing it with the modulus value.

Complete answer:
The general formula for the plane equation of $ax+by+cz=d$, the vector normal to the plane is given by \[\left( a,b,c \right)\].
Converting $6x-2y+3z+8=0$ to its general form we get $6x-2y+3z=-8$.
The value of the coefficients will be $a=6,b=-2,c=3,d=-8$.
Therefore, the vector normal to the plane $6x-2y+3z+8=0$ is $\left( 6,-2,3 \right)$.
The vector form with coordinates will be $6\widehat{i}-2\widehat{j}+3\widehat{k}$.
Now we find the modulus value of the vector.
For $\overrightarrow{A}=a\widehat{i}+b\widehat{j}+c\widehat{k}$, the modulus value will be $\left| \overrightarrow{A} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$.
So, for $6\widehat{i}-2\widehat{j}+3\widehat{k}$, the modulus value will be $\left| \overrightarrow{A} \right|=\sqrt{{{6}^{2}}+{{\left( -2 \right)}^{2}}+{{3}^{2}}}=\sqrt{49}=7$.
Now the unit vector along $6\widehat{i}-2\widehat{j}+3\widehat{k}$ will be $\dfrac{\overrightarrow{A}}{\left| \overrightarrow{A} \right|}=\dfrac{6\widehat{i}-2\widehat{j}+3\widehat{k}}{7}=\dfrac{6}{7}\widehat{i}-\dfrac{2}{7}\widehat{j}+\dfrac{3}{7}\widehat{k}$.
Therefore, unit vector perpendicular to plane $6x-2y+3z+8=0$ will be $\dfrac{6}{7}\widehat{i}-\dfrac{2}{7}\widehat{j}+\dfrac{3}{7}\widehat{k}$.

Note:
The normal vector, often simply called the "normal," to a surface is always perpendicular at a fixed point. Therefore, instead of taking \[\left( a,b,c \right)\] for $ax+by+cz=d$ we can define the normal as partial derivatives and taking the determinant form of $N=\left| \begin{matrix}
   {{f}_{x}}\left( {{x}_{0}},{{y}_{0}} \right) \\
   {{f}_{y}}\left( {{x}_{0}},{{y}_{0}} \right) \\
   1 \\
\end{matrix} \right|$ where the ${{f}_{x}}\left( {{x}_{0}},{{y}_{0}} \right)$ and ${{f}_{y}}\left( {{x}_{0}},{{y}_{0}} \right)$ are the partial derivatives for main function $f\left( x,y \right)$ at point $\left( {{x}_{0}},{{y}_{0}} \right)$.