Vector dot and cross product problem

Given

$\mathbf A = (2, 3, 1)$, $\mathbf B = (4, -1, 2)$.

(a) Dot product

$$\mathbf A\cdot\mathbf B = 2(4)+3(-1)+1(2) = 8-3+2 = 7$$

(b) Cross product

$$\mathbf A\times\mathbf B= \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&3&1\\4&-1&2\end{vmatrix} = (6+1)\mathbf i-(4-4)\mathbf j+(-2-12)\mathbf k = 7\mathbf i+0\mathbf j-14\mathbf k$$

(c) Angle

$|\mathbf A|=\sqrt{14}=3.742$, $|\mathbf B|=\sqrt{21}=4.583$.

$$\cos\theta=\frac{7}{3.742\times4.583}=0.4082,\qquad \theta=65.9^\circ$$

(d) Unit normal

$|\mathbf A\times\mathbf B|=\sqrt{49+0+196}=15.652$

$$\hat{\mathbf n}=\frac{(7,0,-14)}{15.652}=(0.447,\ 0,\ -0.894)$$

(e) Area of parallelogram

Area $=|\mathbf A\times\mathbf B| = 15.65$ square units.

Uses in modelling

• Dot product: angle between edges, projection of a point on a line, checking perpendicularity (dot product = 0).
• Cross product: surface normal of a face (used in shading, offsetting, STL facet normals), area of a face, and the sign of a turn.

Answer: A.B = 7; AxB = (7, 0, -14); angle = 65.9 deg; unit normal = (0.447, 0, -0.894); area = 15.65 square units.