Vector Algebra, Area of Parallelogram and Vector product

4.3 Area of a Parallelogram and Vector Product

The vector product is only defined when both vectors are three-dimensional. The vector product ( or cross product) of $$\vec{a} = \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} \,\, \text{and} \,\, \vec{b} = \begin{pmatrix} x_2\\y_2\\z_2 \end{pmatrix}\, , \,\, \text{is given by}$$ $$\vec{a} \times \vec{b} = \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} \times \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}$$ $$= \begin{pmatrix}y_1z_2 - z_1y_2 \\ z_1x_2 - x_1z_2 \\ x_1y_2 - y_1x_2 \end{pmatrix}$$ Instead of memorizing the formula for the vector product try using the following shortcut.
(1) Eliminate the component column that you are trying to calculate.
(2) Calculate: Down product - Up Product.

$$\text{The direction of} \,\, \vec{a} \times \vec{b} \,\, \text{is perpendicular to both} $$ $$ \vec{a} \,\, \text{and} \,\, \vec{b} \,\, \text{as shown in the diagram.}$$ $$\text{We have known that the nonzero vectors} $$ $$ \vec{a} \,\, \text{and} \,\, \vec{b} \,\, \text{are perpendicular if} \,\, \vec{a} \cdot \vec{b} = 0.$$ $$\text{We will examine:} \,\,\, (\vec{a} \times \vec{b}) \cdot \vec{a}.$$ $$\text{Let} \,\, \vec{a} = \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}\,\, \text{and} \,\, \vec{b} = \begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}.$$ fig 4.3-2

Then
$$(\vec{a} \times \vec{b})\cdot \vec{a} = \begin{pmatrix}y_1z_2 - z_1y_2 \\ z_1x_2 - x_1z_2 \\ x_1y_2 - y_1x_2 \end{pmatrix}\cdot \begin{pmatrix}x_1 \\ y_1\\ z_1 \end{pmatrix}$$ $$= (y_1z_2 - z_1y_2)x_1 + (z_1x_2 - x_1z_2)y_1 + (x_1y_2 - y_1x_2)z_1$$ $$= x_1y_1z_2 - x_1y_2z_1 + x_2y_1z_1 - x_1y_1z_2 + x_1y_2z_1 - x_2y_1z_1$$ $$=0$$ $$\text{Therefore} \,\, \vec{a} \times \vec{a} \,\, \text{is perpendicular to} \,\, \vec{a}.$$ $$\text{Similarly, we can show that} \,\, \vec{a} \times \vec{a} \,\, \text{is also perpendicular to}\,\, \vec{b}.$$ $$\text{The vector product of} \,\, \vec{a} = \begin{pmatrix}x_1\\y_1\\z_1 \end{pmatrix} \,\, \text{and} \,\, \vec{b} = \begin{pmatrix} x_2\\y_2\\z_2 \end{pmatrix}\, , $$ $$ \text{has magnitude}\,\, \left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{sin} \, \theta ,$$ $$\text{where} \,\, \theta \,\, \text{is the angle between}\,\, \vec{a} \,\, \text{and} \,\, \vec{b}.$$

Area of a Parallelogram

Consider the parallelogram OACB lying on the plane, with the vectors $$\vec{a} \,\, \text{and} \,\, \vec{b}$$ as shown in the diagram. Then the area of OACB is given by
$$OA \,\, \left|\vec{b}\right| \,\, \text{sin}\, \theta = \left|\vec{a}\right| (\left|\vec{b}\right| \,\, \text{sin}\, \theta)$$ $$= \left| \vec{a} \times \vec{b}\right|,$$ fig 4.3-3

i.e., the area of the parallelogram with sides defined by vectors
$$\overrightarrow{OA} = \vec{a} = \begin{pmatrix}x_1\\y_1\\z_1 \end{pmatrix} \,\, \text{and}$$ $$\overrightarrow{OB} = \vec{b} = \begin{pmatrix}x_2\\y_2\\z_2 \end{pmatrix}$$ is equal to the magnitude of the vector
$$\vec{a} \times \vec{b} .$$ Consequently,
$$\alpha (\triangle ACB) = \dfrac{1}{2} (\alpha (OACB)) $$ $$= \dfrac{1}{2} \left|\vec{a} \times \vec{b}\right|.$$

Example 13.
Find the area of the triangle ABC with vertices A(1, -1, 3), B(0, 4, 1) and C(2, 7, 2).
Solution
$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$$ $$= \begin{pmatrix}0\\4\\1\end{pmatrix} - \begin{pmatrix}1\\-1\\3\end{pmatrix} = \begin{pmatrix}-1\\5\\-2\end{pmatrix}\,\, \text{and}$$ $$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}$$ $$= \begin{pmatrix}2\\7\\2\end{pmatrix} - \begin{pmatrix}1\\-1\\3\end{pmatrix} = \begin{pmatrix}1\\8\\-1\end{pmatrix}.$$ Then, we find their vector product
$$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}-1\\5\\-2\end{pmatrix} \times \begin{pmatrix}1\\8\\-1\end{pmatrix}$$ $$= \begin{pmatrix}11\\-3\\-13\end{pmatrix}.$$ $$\text{Next} \,\, \left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = \sqrt{11^2 + (-3)^2 + (-13)^2}$$ $$= \sqrt{299}.$$ Therefore, we have
$$\text{Area of triangle} \,\, ABC \,\, = \dfrac{1}{2} \left|\overrightarrow{AB} \times \overrightarrow{AC}\right|$$ $$= \dfrac{1}{2} \sqrt{299} \,\, \text{unit}^2.$$

Algebraic Properties of the Vector Product

$$\text{If} \,\, \vec{a}, \, \vec{b}\,\, \text{and} \,\, \vec{c} \,\, \text{are vectors in space and}$$ $$ k\,\, \text{is a scalar, then}$$ $$1. \,\, \vec{a} \times \vec{b} = - (\vec{b} \times \vec{a}),$$ $$2. \,\, (k\vec{a}) \times \vec{b} = k(\vec{a} \times \vec{b}),$$ $$3. \,\, \vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}),$$ $$4. \,\, \vec{a} \times \vec{0} = \vec{0}.$$

Geometric Properties of the Vector Product

$$1.\,\, \text{If two nonzero vectors}\,\, \vec{a} \,\, \text{and} \,\, \vec{b}$$ $$\text{are \textbf{parallel}, then} \,\, \vec{a} \times \vec{b} = \vec{0}.$$ $$\text{In particular,} \,\, \vec{a} \times \vec{a} = \vec{0}.$$ $$\text{2. \,\,If two nonzero vectors} \,\, \vec{a} \,\, \text{and} \,\, \vec{b}$$ are perpendicular, then
$$\left|\vec{a} \times \vec{b}\right| = \left|\vec{a}\right| \left|\vec{b}\right|.$$

Example 14.
$$\text{(a) \,\, Calculate}\,\, \vec{a} \times \vec{b} \,\, \text{when}$$ $$\vec{a} = 3\hat{i} + 2\hat{j} + 5\hat{k} \,\, \text{and}$$ $$\vec{b} = \hat{i} - 4\hat{j} + 2\hat{k}.$$ $$\text{(b) \,\, Find a unit vector}\,\, \hat{n} \,\, \text{that is perpendicular}$$ $$\text{to both} \,\, \vec{a} \,\, \text{and} \,\, \vec{b}.$$ Solution
$$\text{(a)} \quad \vec{a} \times \vec{b} = \begin{pmatrix}3\\2\\5\end{pmatrix} \times \begin{pmatrix}1\\-4\\2\end{pmatrix}$$ $$= \begin{pmatrix}2\times 2 \quad - \quad 5 \times (-4)\\ 5\times 1 \quad - \quad 3 \times 2\\ 3 \times (-4) \quad - \quad 2\times 1 \end{pmatrix}$$ $$= \begin{pmatrix}24\\-1\\-14\end{pmatrix}.$$
$$\text{(b)} \quad \vec{a} \times \vec{b} = \begin{pmatrix}24\\-1\\-14 \end{pmatrix}$$ This vector is perpendicular to
$$\vec{a} \,\, \text{and} \,\, \vec{b}.$$ $$\left|\vec{a} \times \vec{b}\right| = \sqrt{24^2 + (-1)^2 + -(14)^2}$$ $$= \sqrt{773}.$$ So, a unit vector perpendicular to both
$$\vec{a} \,\, \text{and}\,\, \vec{b} \,\, \text{is}$$ $$\hat{n} = \dfrac{1}{\sqrt{773}} \begin{pmatrix}24\\-1\\-14\end{pmatrix}.$$

Example 15.
$$\text{Given that} \,\, \left|\vec{a}\right| = 4, \,\, \left|\vec{b}\right| = 5 \,\, \text{and that}$$ $$\vec{a} \,\, \text{and}\,\, \vec{b} \,\, \text{are perpendicular, evaluate}$$ $$\left|(2\vec{a} - \vec{b}) \times (\vec{a} + 3\vec{b})\right|.$$ Solution $$(2\vec{a} - \vec{b}) \times (\vec{a} + 3\vec{b})$$ $$= 2\vec{a}\times\vec{a} \,\, + \,\, 2\vec{a}\times 3\vec{b} \,\, - \vec{b}\times \vec{a} \,\, - \,\, \vec{b}\times 3\vec{b}$$ $$= 2(\vec{a}\times\vec{a}) + 6(\vec{a}\times \vec{b}) - (\vec{b}\times \vec{a}) - 3(\vec{b}\times\vec{b})$$ $$= 6(\vec{a}\times\vec{b}) - (\vec{b}\times\vec{a})$$ $$= 6(\vec{a}\times\vec{b}) + (\vec{a}\times\vec{b})$$ $$= 7(\vec{a}\times\vec{b}).$$ $$\left|(2\vec{a} - \vec{b}) \times (\vec{a} + 3\vec{b})\right| = 7\left|\vec{a}\times\vec{b}\right|$$ $$= 7\left|\vec{a}\right| \left|\vec{b}\right|$$ $$=140 .$$

Exercise 4.3

1. Find a vector perpendicular to the following pair of vectors:
$$\text{(a)} \,\, \begin{pmatrix}3\\1\\1\end{pmatrix} \,\, \text{and} \,\, \begin{pmatrix}1\\2\\3 \end{pmatrix}$$ $$\text{(b)} \,\, \begin{pmatrix}3\\-1\\4\end{pmatrix} \,\, \text{and} \,\, \begin{pmatrix}-1\\1\\5\end{pmatrix}$$ Solution
$$\vec{a} \times \vec{b} = \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}\times\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix} = \begin{pmatrix}y_1z_2 - z_1y_2 \\ z_1x_2 - x_1z_2 \\ x_1y_2 - y_1x_2 \end{pmatrix}$$ $$(a) \,\,\text{Let} \,\, \vec{a} = \begin{pmatrix}3\\1\\1\end{pmatrix}, \,\, \vec{b} = \begin{pmatrix}1\\2\\3\end{pmatrix}$$ $$\text{The required vector} = \vec{a}\times\vec{b}$$ $$= \begin{pmatrix}3\\1\\1\end{pmatrix} \times \begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}1\times3 \,\, - \,\, 1\times 2\\ 1\times 1 \,\, - \,\, 3\times 3\\ 3\times 2 \,\, - \,\, 1\times 1 \end{pmatrix}$$ $$= \begin{pmatrix}3-2\\1-9\\6-1\end{pmatrix} = \begin{pmatrix}1\\-8\\5\end{pmatrix}$$

$$\text{(b)\,\, Let} \,\, \vec{a} = \begin{pmatrix}3\\-1\\4\end{pmatrix} \,\, \text{and} \,\, \vec{b} = \begin{pmatrix}-1\\1\\5\end{pmatrix}$$ $$= \begin{pmatrix}3\\-1\\4\end{pmatrix}\times \begin{pmatrix}-1\\1\\5\end{pmatrix}$$ fig 4.3-1

$$= \begin{pmatrix}-1\times 5 \quad - \quad 4\times 1\\ 4\times (-1) \quad - \quad 3\times 5\\ 3\times 1 \quad - \quad (-1)\times (-1)\end{pmatrix}$$ $$= \begin{pmatrix}-5 - 4\\-4 - 15\\ 3 - 1\end{pmatrix} = \begin{pmatrix}-9\\-19\\2\end{pmatrix}$$

$$\text{2. \,\, Consider} \,\, \vec{a} = \begin{pmatrix}2\\-1\\3\end{pmatrix} \,\, \text{and} \,\, \vec{b} = \begin{pmatrix}1\\0\\-1\end{pmatrix}.$$ $$\text{(a)\,\, Find} \,\, \vec{a} \times \vec{b} .$$ $$\text{(b)\,\, Find sin}\,\, \theta \,\, \text{using} \,\, \left|\vec{a} \times \vec{b}\right| = \left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{sin} \,\, \theta.$$ Solution
$$\text{(a)}\,\, \vec{a}\times \vec{b} = \begin{pmatrix}2\\-1\\3\end{pmatrix}\times \begin{pmatrix}1\\0\\-1\end{pmatrix}$$ $$= \begin{pmatrix}-1\times(-1) \quad - \quad 3\times 0\\ 3\times 1 \quad - \quad 2 \times (-1)\\ 2\times 0 \quad - \quad (-1) \times 1 \end{pmatrix}$$ $$= \begin{pmatrix}1-0 \\ 3+2\\ 0+1\end{pmatrix} = \begin{pmatrix}1\\5\\1 \end{pmatrix}$$
$$\text{(b)}\,\, \left|\vec{a}\times\vec{b}\right| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{27}$$ $$\left|\vec{a}\right| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}$$ $$\left|\vec{b}\right| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$$ $$\left|\vec{a}\times\vec{b}\right| = \left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{sin}\,\, \theta$$ $$\dfrac{\left|\vec{a}\times\vec{b}\right|}{\left|\vec{a}\right| \left|\vec{b}\right|} = \text{sin}\,\, \theta$$ $$\text{sin}\,\, \theta = \dfrac{\sqrt{27}}{\sqrt{14} \, \sqrt{2}}$$

$$\text{3. \,\, Prove that for any two vectors}\,\, \vec{a}\,\, \text{and} \,\, \vec{b},$$ $$\left|\vec{a} \times \vec{b}\right|^2 \,\, + \,\, \left(\vec{a}\cdot\vec{b}\right)^2 \,\, =\,\, \left|\vec{a}\right|^2 \left|\vec{b}\right|^2.$$ Solution
$$\scriptsize{\text{Let} \,\, \vec{a} = \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} \,\, \text{and} \,\, \vec{b} = \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}}$$ $$\scriptsize{\vec{a}\times\vec{b} = \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} \times \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}}$$ $$\scriptsize{= \begin{pmatrix}y_1z_2 - y_2z_1\\ z_1x_2 - z_2x_1\\ x_1y_2 - x_2y_1\end{pmatrix}}$$ $$\scriptsize{\left|\vec{a}\times\vec{b}\right|^2 = (y_1z_2 - y_2z_1)^2 + (z_1x_2 - z_2x_1)^2 + (x_1y_2 - x_2y_1)^2}$$ $$\scriptsize{= (y_1z_2)^2 - 2y_1z_2y_2z_1 + (y_2z_1)^2 +(z_1x_2)^2 - 2z_1x_2z_2x_1 \,+}$$ $$\scriptsize{(z_2x_1)^2 + (x_1y_2)^2 - 2x_1y_2x_2y_1 + (x_2y_1)^2}$$ $$\scriptsize{= (y_1z_2)^2 - 2y_1y_2z_1z_2 + (y_2z_1)^2 + (z_1x_2)^2 - 2x_1x_2z_1z_2 \, +}$$ $$\scriptsize{(z_2x_1)^2 + (x_1y_2)^2 - 2x_1x_2y_1y_2 + (x_2y_1)^2}$$
$$\scriptsize{\vec{a}\cdot\vec{b} = x_1x_2 + y_1y_2 + z_1z_2}$$ $$\scriptsize{(\vec{a}\cdot\vec{b})^2 = (x_1x_2 + y_1y_2 + z_1z_2) (x_1x_2 + y_1y_2 + z_1z_2)}$$ $$\scriptsize{= x_1x_2\cdot x_1x_2 + x_1x_2\cdot y_1y_2 + x_1x_2\cdot z_1z_2 \,\, + \,\, y_1y_2\cdot x_1x_2 \,+ }$$ $$\scriptsize{y_1y_2\cdot y_1y_2 + y_1y_2\cdot z_1z_2 \,\, + \,\, z_1z_2\cdot x_1x_2 + z_1z_2\cdot y_1y_2 + z_1z_2\cdot z_1z_2}$$ $$\scriptsize{= (x_1x_2)^2 + x_1x_2y_1y_2 + z_1z_2x_1x_2 + x_1x_2y_1y_2 + (y_1y_2)^2 \, +}$$ $$\scriptsize{y_1y_2z_1z_2 + z_1z_2x_1x_2 + y_1y_2z_1z_2 + (z_1z_2)^2}$$ $$\scriptsize{= (x_1x_2)^2 + 2x_1x_2y_1y_2 + (y_1y_2)^2 + 2y_1y_2z_1z_2 + (z_1z_2)^2 + 2z_1z_2x_1x_2}$$
$$\scriptsize{\left|\vec{a}\times\vec{b}\right|^2 + (\vec{a}\cdot\vec{b})^2 = (y_1z_2)^2 + (y_2z_1)^2 + (z_1x_2)^2 + (z_2x_1)^2 \,+ }$$ $$\scriptsize{(x_1y_2)^2 + (x_2y_1)^2 + (x_1x_2)^2 + (y_1y_2)^2 + (z_1z_2)^2}$$
$$\scriptsize{\left|\vec{a}\right|^2 \left|\vec{b}\right|^2 = (x_1^2 + y_1^2 + z_1^2) \, (x_2^2 + y_2^2 + z_2^2)}$$ $$\scriptsize{= x_1^2x_2^2 + x_1^2y_2^2 + x_1^2z_2^2 + y_1^2x_2^2 + y_1^2y_2^2 \, +}$$ $$\scriptsize{y_1^2z_2^2 + z_1^2x_2^2 + z_1^2y_2^2 + z_1^2z_2^2}$$ $$\scriptsize{= (x_1x_2)^2 + (x_1y_2)^2 + (x_1z_2)^2 + (x_2y_1)^2 + (y_1y_2)^2 \, +}$$ $$\scriptsize{(y_1z_2)^2 + (z_1x_2)^2 + (z_1y_2)^2 + (z_1z_2)^2}$$ $$\scriptsize{\therefore \left|\vec{a}\times\vec{b}\right|^2 + (\vec{a}\cdot \vec{b})^2 = \left|\vec{a}\right|^2 \left|\vec{b}\right|^2}$$
or
$$\left|\vec{a}\times\vec{b}\right| = \left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{sin}\,\, \theta$$ $$\left|\vec{a}\times\vec{b}\right|^2 = (\left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{sin}\,\, \theta)^2$$ $$= \left|\vec{a}\right|^2 \left|\vec{b}\right|^2\,\, \text{sin}^2\,\, \theta \qquad \_\_\_\_\_\_(1)$$ $$\vec{a}\cdot\vec{b} = \left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{cos}\,\, \theta$$ $$(\vec{a}\cdot\vec{b})^2 = (\left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{cos}\,\, \theta)^2$$ $$ = \left|\vec{a}\right|^2 \left|\vec{b}\right|^2 \,\, \text{cos}^2\,\, \theta \qquad \_\_\_\_\_\_(2)$$ $$eq \_\_\_\_\_\_(1) + (2)$$ $$\left|\vec{a}\times\vec{b}\right|^2 + (\vec{a}\cdot\vec{b})^2 = \left|\vec{a}\right|^2 \left|\vec{b}\right|^2 \,\, \text{sin}^2\,\, \theta \,\, + \left|\vec{a}\right|^2 \left|\vec{b}\right|^2 \,\, \text{cos}^2\,\, \theta$$ $$= \left|\vec{a}\right|^2 \left|\vec{b}\right|^2 \,\, (\text{sin}^2 \,\, \theta + \text{cos}^2\,\, \theta)$$ $$= \left|\vec{a}\right|^2 \left|\vec{b}\right|^2 \cdot 1$$ $$= \left|\vec{a}\right|^2 \left|\vec{b}\right|^2 $$

4. Given points A, B and C with coordinates (3, -5, 1), (7, 7, 2) and (-1, -1- 3).
$$\text{(a) \,\, Calculate} \,\, \vec{p} = \overrightarrow{AB} \times \overrightarrow{AC} \,\, \text{and} $$ $$ \vec{q} = \overrightarrow{BA} \times \overrightarrow{BC}.$$ $$\text{(b) \,\, What can you say about vectors} \,\, \vec{p}\,\, \text{and}\,\, \vec{q}\,?$$ Solution
A(3, -5, 1), B(7, 7, 2), C(-1, 1, 3).
$$\text{(a)} \quad \overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{OB} - \overrightarrow{OA}$$ $$\begin{pmatrix}7\\7\\2\end{pmatrix} - \begin{pmatrix}3\\-5\\1\end{pmatrix} = \begin{pmatrix}4\\12\\1\end{pmatrix}$$ $$\quad \overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = \overrightarrow{OC} - \overrightarrow{OA}$$ $$\begin{pmatrix}-1\\1\\3\end{pmatrix} - \begin{pmatrix}3\\-5\\1\end{pmatrix} = \begin{pmatrix}-4\\6\\2\end{pmatrix}$$ $$\vec{p} = \overrightarrow{AB}\times\overrightarrow{AC}$$ $$= \begin{pmatrix}4\\12\\1\end{pmatrix} \times \begin{pmatrix}-4\\6\\2\end{pmatrix}$$
$$= \begin{pmatrix}12\times 2 \quad - \quad 6\times 1 \\ 1\times (-4) \quad - \quad 2\times 4 \\ 4\times 6 \quad - \quad (-4)\times 12 \end{pmatrix}$$ $$= \begin{pmatrix}24 - 6 \\ -4 - 8 \\ 24 + 48 \end{pmatrix} = \begin{pmatrix}18 \\ -12\\ 72\end{pmatrix}$$ $$\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \overrightarrow{OC} - \overrightarrow{OB}$$ $$= \begin{pmatrix}-1\\1\\3 \end{pmatrix} - \begin{pmatrix}7\\7\\2\end{pmatrix} = \begin{pmatrix}-8\\-6\\1 \end{pmatrix}$$ $$\overrightarrow{BA} = - \overrightarrow{AB} = - \begin{pmatrix}4 \\ 12 \\ 1 \end{pmatrix} = \begin{pmatrix}-4 \\ -12 \\ -1 \end{pmatrix}$$
$$\vec{q} = \overrightarrow{BA} \times \overrightarrow{BC}$$ $$= \begin{pmatrix}-4\\-12\\-1 \end{pmatrix} \times \begin{pmatrix}-8 \\ -6 \\1 \end{pmatrix}$$
$$= \begin{pmatrix}-12 \times 1 \quad - \quad (-6)\times (-1) \\ -1\times (-8) \quad - \quad 1\times (-4) \\ -4\times (-6) \quad - \quad (-8)\times (-12) \end{pmatrix}$$ $$= \begin{pmatrix}-12 - 6\\ 8 + 4 \\ 24 - 96 \end{pmatrix} = \begin{pmatrix}-18 \\ 12 \\ -72 \end{pmatrix} = - \vec{p}$$
(b) p = q
$$\vec{p} \,\, \text{and} \,\, \vec{q} \,\, \text{are parallel and opposite direction.}$$

5. The points A(3, 1, 2), B(-1, 1, 5) and C(7, 2, 3) are vectors of parallelogram ABCD.
(a) Find the coordinates of D.
(b) Calculate the area of parallelogram.
Solution
ABCD is a parallelogram.
$$\therefore \overrightarrow{AB} = \overrightarrow{DC}$$ $$\overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{DO} + \overrightarrow{OC}$$ $$\overrightarrow{OB} - \overrightarrow{OA} = \overrightarrow{OC} - \overrightarrow{OD}$$ $$\begin{pmatrix}-1 \\1 \\5 \end{pmatrix} - \begin{pmatrix}3\\1\\2\end{pmatrix} = \begin{pmatrix}7\\2\\3\end{pmatrix} - \begin{pmatrix}x\\y\\z\end{pmatrix}$$ $$\begin{pmatrix}-4\\0\\3\end{pmatrix} = \begin{pmatrix}7 - x\\ 2 - y\\ 3 - z\end{pmatrix}$$

-4 = 7 - x	,	0 = 2 - y	,	3 = 3 - z
x - 4 = 7	,	-2 = -y	,	z + 3 = 3
x = 11	,	y = 2	,	z = 0

∴ D(11, 2, 0).
$$\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \overrightarrow{OC} - \overrightarrow{OB}$$ $$= \begin{pmatrix}7\\2\\3 \end{pmatrix} - \begin{pmatrix}-1\\1\\5\end{pmatrix} = \begin{pmatrix}8\\1\\-2\end{pmatrix}$$ $$\overrightarrow{AB} = \begin{pmatrix}-4\\0\\3\end{pmatrix}$$ $$\overrightarrow{AB}\times \overrightarrow{BC} = \begin{pmatrix}-4\\0\\3 \end{pmatrix} \times \begin{pmatrix}8\\1\\-2 \end{pmatrix}$$
$$= \begin{pmatrix}0\times (-2) \quad -\quad 1\times 3 \\ 3\times 8 \quad - \quad (-2)\times(-4) \\ -4 \times 1 \quad - \quad 8\times 0\end{pmatrix}$$ $$= \begin{pmatrix}0 - 3\\ 24 - 8\\ -4 - 0 \end{pmatrix} = \begin{pmatrix}-3\\16\\-4 \end{pmatrix}$$ $$\left|\overrightarrow{AB}\times \overrightarrow{BC}\right| = \sqrt{(-3)^2 + 16^2 + (-4)^2}$$ $$=\sqrt{9 + 256 + 16} = \sqrt{281} \, \text{unit}^2$$