91Ó°ÊÓ

To find the vector perpendicular to a plane, the boundary vector must be comuted and then cross product of the boundary vector of the concerned plane must be evaluated. The given plane is drawn as follows:

The plane in y-z plane has unit vector $\hat{n}$, perpendicular to it as shown in the figure.

The end points of the base A of given plane are $\left(0,0,0\right)$and $(1,0,0)$and end points of the base B, which is left of given plane are $(0,0,3)$and $(1,0,0)$,

Find the position vector of base A and B.

$$\begin{gathered} A=(0-1)i+(2-0)j+(0-0)k \\ =-i+2j+0k \end{gathered}$$

Solve for vector B.

$$\begin{gathered} B=(0-1)i+(2-0)k+(0-0)k \\ =-i+0j+3k \end{gathered}$$

The formula of the cross product of the vectors $A$and $B$is

$A×B=\left|A\right|\left|B\right|\sin \mathrm{θ}\hat{n}$, $\mathrm{θ}$is the angle between the vectos $A$and $B$.

Find the cross product of the vectors $A$and $B$.

$$\begin{gathered} A×B=\left|\begin{array}{ccc}i& j& k\\ -1& 2& 0\\ -1& 0& 3\end{array}\right| \\ =i(6-0)-j(-3-0)+k(0+2) \\ =6i+3y+2k \end{gathered}$$

The unit vector $\hat{n}$is obtained as

$$\begin{gathered} \hat{n}=\frac{A×\mathbf{ }B}{\left|A×B\right|} \\ =\frac{6i+3j+2k}{\sqrt{6^2+3^2+2^2}} \\ =\frac{6i+3j+2k}{\sqrt{49}} \\ =\frac{6i+3j+2k}{7} \\ =\frac{6}{7}i+\frac{3}{7}j+\frac{2}{7}k \end{gathered}$$

Thus, the unit vector is $\hat{n}=\frac{6}{7}i+\frac{3}{7}j+\frac{2}{7}k$.