Griffiths problem 1.4 Use the cross product to find the components of the unit vector n perpendicular to the plane shown in Fig. 1.11.

# Griffiths problem 1.4 Use the cross product to find the components of the unit vector n perpendicular to the plane shown in Fig. 1.11.

P
945 points

Use the cross product to find the components of the unit vector n perpendicular to the plane shown in Fig. 1.11.

Griffiths problem
Pairoaj

P
945 points

To find a unit vector $\hat{n}$ that is perpendicular to the plane just take the cross product of two vectors in the plane and then divide that result by its length to get a unit vector.

So choose two vectors (you can find a vector in the plane by picking any two points in the plane and taking their difference),

$A = (0-1)\hat{x}+(0-0)\hat{y}+(3-0)\hat{z}=-1\hat{x}+0\hat{y}+3\hat{z}$

$B = (0-0)\hat{x}+(2-0)\hat{y}+(0-3)\hat{z}=0\hat{x}+2\hat{y}-3\hat{z}$

now take the cross product $A \times B$

$$A \times B=\begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ -1 & 0 & 3 \\ 0 & 2 & -3 \\ \end{vmatrix} = -6\hat{x}+3\hat{y}-2\hat{z}$$

Which is a vector that is perpendicular to the plane, but you need a unit vector so take

$$\frac{A \times B}{\left\lVert A\times B \right\rVert}=\frac{-6\hat{x}+3\hat{y}-2\hat{z}}{\sqrt{36+9+4}}=-\frac{6}{7}\hat{x}+\frac{3}{7}\hat{y}-\frac{2}{7}\hat{z}$$

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