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 a⃗=(x1y1z1)  and  b⃗=(x2y2z2) ,  is given by\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} a⃗×b⃗=(x1y1z1)×(x2y2z2)\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} =(y1z2−z1y2z1x2−x1z2x1y2−y1x2)= \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.

fig 4.3-1

The direction of  a⃗×b⃗  is perpendicular to both\text{The direction of} \,\, \vec{a} \times \vec{b} \,\, \text{is perpendicular to both} a⃗  and  b⃗  as shown in the diagram. \vec{a} \,\, \text{and} \,\, \vec{b} \,\, \text{as shown in the diagram.} We have known that the nonzero vectors\text{We have known that the nonzero vectors} a⃗  and  b⃗  are perpendicular if  a⃗⋅b⃗=0. \vec{a} \,\, \text{and} \,\, \vec{b} \,\, \text{are perpendicular if} \,\, \vec{a} \cdot \vec{b} = 0. We will examine:   (a⃗×b⃗)⋅a⃗.\text{We will examine:} \,\,\, (\vec{a} \times \vec{b}) \cdot \vec{a}. Let  a⃗=(x1y1z1)  and  b⃗=(x2y2z2).\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

(a⃗×b⃗)⋅a⃗=(y1z2−z1y2z1x2−x1z2x1y2−y1x2)⋅(x1y1z1)(\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} =(y1z2−z1y2)x1+(z1x2−x1z2)y1+(x1y2−y1x2)z1= (y_1z_2 - z_1y_2)x_1 + (z_1x_2 - x_1z_2)y_1 + (x_1y_2 - y_1x_2)z_1 =x1y1z2−x1y2z1+x2y1z1−x1y1z2+x1y2z1−x2y1z1= 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=0 Therefore  a⃗×a⃗  is perpendicular to  a⃗.\text{Therefore} \,\, \vec{a} \times \vec{a} \,\, \text{is perpendicular to} \,\, \vec{a}. Similarly, we can show that  a⃗×a⃗  is also perpendicular to  b⃗.\text{Similarly, we can show that} \,\, \vec{a} \times \vec{a} \,\, \text{is also perpendicular to}\,\, \vec{b}. The vector product of  a⃗=(x1y1z1)  and  b⃗=(x2y2z2) ,\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}\, , has magnitude  ∣a⃗∣∣b⃗∣  sin θ, \text{has magnitude}\,\, \left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{sin} \, \theta , where  θ  is the angle between  a⃗  and  b⃗.\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
a⃗  and  b⃗\vec{a} \,\, \text{and} \,\, \vec{b}
as shown in the diagram. Then the area of OACB is given by
OA  ∣b⃗∣  sin θ=∣a⃗∣(∣b⃗∣  sin θ)OA \,\, \left|\vec{b}\right| \,\, \text{sin}\, \theta = \left|\vec{a}\right| (\left|\vec{b}\right| \,\, \text{sin}\, \theta) =∣a⃗×b⃗∣,= \left| \vec{a} \times \vec{b}\right|,
fig 4.3-3

i.e., the area of the parallelogram with sides defined by vectors
OA→=a⃗=(x1y1z1)  and\overrightarrow{OA} = \vec{a} = \begin{pmatrix}x_1\\y_1\\z_1 \end{pmatrix} \,\, \text{and} OB→=b⃗=(x2y2z2)\overrightarrow{OB} = \vec{b} = \begin{pmatrix}x_2\\y_2\\z_2 \end{pmatrix} is equal to the magnitude of the vector
a⃗×b⃗.\vec{a} \times \vec{b} . Consequently,
α(△ACB)=12(α(OACB))\alpha (\triangle ACB) = \dfrac{1}{2} (\alpha (OACB)) =12∣a⃗×b⃗∣.= \dfrac{1}{2} \left|\vec{a} \times \vec{b}\right|.
Example 12.
Find the area of the parallelogram determined by the vectors
a⃗=(123)  and  b⃗=(14−1)\vec{a} = \begin{pmatrix}1\\2\\3\end{pmatrix} \,\, \text{and} \,\, \vec{b} = \begin{pmatrix}1\\4\\-1\end{pmatrix}
Solution
We can find that
a⃗×b⃗=(123)×(14−1)\vec{a} \times \vec{b} = \begin{pmatrix}1\\2\\3\end{pmatrix} \times \begin{pmatrix}1\\4\\-1\end{pmatrix} =(2×(−1)−3×43×1−1×(−1)1×4−2×1)= \begin{pmatrix}2\times(-1) \quad - \quad 3\times 4\\ 3\times 1 \quad - \quad 1\times(-1)\\ 1\times 4 \quad - \quad 2\times 1 \end{pmatrix} =(−1442).= \begin{pmatrix}-14\\4\\2\end{pmatrix} . Now,∣a⃗×b⃗∣=196+16+4=216.\text{Now,}\quad \left|\vec{a} \times \vec{b}\right| = \sqrt{196 + 16 + 4} = \sqrt{216}. Therefore the area of the parallelogram is
216  unit2.\sqrt{216} \,\, \text{unit}^2.

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
AB→=OB→−OA→\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} =(041)−(1−13)=(−15−2)  and= \begin{pmatrix}0\\4\\1\end{pmatrix} - \begin{pmatrix}1\\-1\\3\end{pmatrix} = \begin{pmatrix}-1\\5\\-2\end{pmatrix}\,\, \text{and} AC→=OC→−OA→\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} =(272)−(1−13)=(18−1).= \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
AB→×AC→=(−15−2)×(18−1)\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}-1\\5\\-2\end{pmatrix} \times \begin{pmatrix}1\\8\\-1\end{pmatrix} =(11−3−13).= \begin{pmatrix}11\\-3\\-13\end{pmatrix}. Next  ∣AB→×AC→∣=112+(−3)2+(−13)2\text{Next} \,\, \left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = \sqrt{11^2 + (-3)^2 + (-13)^2} =299.= \sqrt{299}. Therefore, we have
Area of triangle  ABC  =12∣AB→×AC→∣\text{Area of triangle} \,\, ABC \,\, = \dfrac{1}{2} \left|\overrightarrow{AB} \times \overrightarrow{AC}\right| =12299  unit2.= \dfrac{1}{2} \sqrt{299} \,\, \text{unit}^2.

Algebraic Properties of the Vector Product

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



Geometric Properties of the Vector Product

1.  If two nonzero vectors  a⃗  and  b⃗1.\,\, \text{If two nonzero vectors}\,\, \vec{a} \,\, \text{and} \,\, \vec{b} are parallel, then  a⃗×b⃗=0⃗.\text{are \textbf{parallel}, then} \,\, \vec{a} \times \vec{b} = \vec{0}. In particular,  a⃗×a⃗=0⃗.\text{In particular,} \,\, \vec{a} \times \vec{a} = \vec{0}. 2.   If two nonzero vectors  a⃗  and  b⃗\text{2. \,\,If two nonzero vectors} \,\, \vec{a} \,\, \text{and} \,\, \vec{b} are perpendicular, then
∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣.\left|\vec{a} \times \vec{b}\right| = \left|\vec{a}\right| \left|\vec{b}\right|.

Example 14.
(a)    Calculate  a⃗×b⃗  when\text{(a) \,\, Calculate}\,\, \vec{a} \times \vec{b} \,\, \text{when} a⃗=3i^+2j^+5k^  and\vec{a} = 3\hat{i} + 2\hat{j} + 5\hat{k} \,\, \text{and} b⃗=i^−4j^+2k^.\vec{b} = \hat{i} - 4\hat{j} + 2\hat{k}. (b)    Find a unit vector  n^  that is perpendicular\text{(b) \,\, Find a unit vector}\,\, \hat{n} \,\, \text{that is perpendicular} to both  a⃗  and  b⃗.\text{to both} \,\, \vec{a} \,\, \text{and} \,\, \vec{b}. Solution
(a)a⃗×b⃗=(325)×(1−42)\text{(a)} \quad \vec{a} \times \vec{b} = \begin{pmatrix}3\\2\\5\end{pmatrix} \times \begin{pmatrix}1\\-4\\2\end{pmatrix} =(2×2−5×(−4)5×1−3×23×(−4)−2×1)= \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} =(24−1−14).= \begin{pmatrix}24\\-1\\-14\end{pmatrix}.
(b)a⃗×b⃗=(24−1−14)\text{(b)} \quad \vec{a} \times \vec{b} = \begin{pmatrix}24\\-1\\-14 \end{pmatrix} This vector is perpendicular to
a⃗  and  b⃗.\vec{a} \,\, \text{and} \,\, \vec{b}. ∣a⃗×b⃗∣=242+(−1)2+−(14)2\left|\vec{a} \times \vec{b}\right| = \sqrt{24^2 + (-1)^2 + -(14)^2} =773.= \sqrt{773}. So, a unit vector perpendicular to both
a⃗  and  b⃗  is\vec{a} \,\, \text{and}\,\, \vec{b} \,\, \text{is} n^=1773(24−1−14).\hat{n} = \dfrac{1}{\sqrt{773}} \begin{pmatrix}24\\-1\\-14\end{pmatrix}.

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

Exercise 4.3

1.   Find a vector perpendicular to the following pair of vectors:
(a)  (311)  and  (123)\text{(a)} \,\, \begin{pmatrix}3\\1\\1\end{pmatrix} \,\, \text{and} \,\, \begin{pmatrix}1\\2\\3 \end{pmatrix} (b)  (3−14)  and  (−115)\text{(b)} \,\, \begin{pmatrix}3\\-1\\4\end{pmatrix} \,\, \text{and} \,\, \begin{pmatrix}-1\\1\\5\end{pmatrix} Solution
a⃗×b⃗=(x1y1z1)×(x2y2z2)=(y1z2−z1y2z1x2−x1z2x1y2−y1x2)\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)  Let  a⃗=(311),  b⃗=(123)(a) \,\,\text{Let} \,\, \vec{a} = \begin{pmatrix}3\\1\\1\end{pmatrix}, \,\, \vec{b} = \begin{pmatrix}1\\2\\3\end{pmatrix} The required vector=a⃗×b⃗\text{The required vector} = \vec{a}\times\vec{b} =(311)×(123)=(1×3  −  1×21×1  −  3×33×2  −  1×1)= \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} =(3−21−96−1)=(1−85)= \begin{pmatrix}3-2\\1-9\\6-1\end{pmatrix} = \begin{pmatrix}1\\-8\\5\end{pmatrix}

fig 4.3-1

(b)   Let  a⃗=(3−14)  and  b⃗=(−115)\text{(b)\,\, Let} \,\, \vec{a} = \begin{pmatrix}3\\-1\\4\end{pmatrix} \,\, \text{and} \,\, \vec{b} = \begin{pmatrix}-1\\1\\5\end{pmatrix} =(3−14)×(−115)= \begin{pmatrix}3\\-1\\4\end{pmatrix}\times \begin{pmatrix}-1\\1\\5\end{pmatrix}
fig 4.3-1

=(−1×5−4×14×(−1)−3×53×1−(−1)×(−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} =(−5−4−4−153−1)=(−9−192)= \begin{pmatrix}-5 - 4\\-4 - 15\\ 3 - 1\end{pmatrix} = \begin{pmatrix}-9\\-19\\2\end{pmatrix}
2.    Consider  a⃗=(2−13)  and  b⃗=(10−1).\text{2. \,\, Consider} \,\, \vec{a} = \begin{pmatrix}2\\-1\\3\end{pmatrix} \,\, \text{and} \,\, \vec{b} = \begin{pmatrix}1\\0\\-1\end{pmatrix}. (a)   Find  a⃗×b⃗.\text{(a)\,\, Find} \,\, \vec{a} \times \vec{b} . (b)   Find sin  θ  using  ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣  sin  θ.\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
(a)  a⃗×b⃗=(2−13)×(10−1)\text{(a)}\,\, \vec{a}\times \vec{b} = \begin{pmatrix}2\\-1\\3\end{pmatrix}\times \begin{pmatrix}1\\0\\-1\end{pmatrix} =(−1×(−1)−3×03×1−2×(−1)2×0−(−1)×1)= \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} =(1−03+20+1)=(151)= \begin{pmatrix}1-0 \\ 3+2\\ 0+1\end{pmatrix} = \begin{pmatrix}1\\5\\1 \end{pmatrix}
(b)  ∣a⃗×b⃗∣=12+52+12=27\text{(b)}\,\, \left|\vec{a}\times\vec{b}\right| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{27} ∣a⃗∣=22+(−1)2+32=14\left|\vec{a}\right| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14} ∣b⃗∣=12+02+(−1)2=2\left|\vec{b}\right| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2} ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣  sin  θ\left|\vec{a}\times\vec{b}\right| = \left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{sin}\,\, \theta ∣a⃗×b⃗∣∣a⃗∣∣b⃗∣=sin  θ\dfrac{\left|\vec{a}\times\vec{b}\right|}{\left|\vec{a}\right| \left|\vec{b}\right|} = \text{sin}\,\, \theta sin  θ=2714 2\text{sin}\,\, \theta = \dfrac{\sqrt{27}}{\sqrt{14} \, \sqrt{2}}

3.    Prove that for any two vectors  a⃗  and  b⃗,\text{3. \,\, Prove that for any two vectors}\,\, \vec{a}\,\, \text{and} \,\, \vec{b}, ∣a⃗×b⃗∣2  +  (a⃗⋅b⃗)2  =  ∣a⃗∣2∣b⃗∣2.\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
Leta⃗=(x1y1z1)andb⃗=(x2y2z2)\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}} a⃗×b⃗=(x1y1z1)×(x2y2z2)\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}} =(y1z2−y2z1z1x2−z2x1x1y2−x2y1)\scriptsize{= \begin{pmatrix}y_1z_2 - y_2z_1\\ z_1x_2 - z_2x_1\\ x_1y_2 - x_2y_1\end{pmatrix}} ∣a⃗×b⃗∣2=(y1z2−y2z1)2+(z1x2−z2x1)2+(x1y2−x2y1)2\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} =(y1z2)2−2y1z2y2z1+(y2z1)2+(z1x2)2−2z1x2z2x1+\scriptsize{= (y_1z_2)^2 - 2y_1z_2y_2z_1 + (y_2z_1)^2 +(z_1x_2)^2 - 2z_1x_2z_2x_1 \,+} (z2x1)2+(x1y2)2−2x1y2x2y1+(x2y1)2\scriptsize{(z_2x_1)^2 + (x_1y_2)^2 - 2x_1y_2x_2y_1 + (x_2y_1)^2} =(y1z2)2−2y1y2z1z2+(y2z1)2+(z1x2)2−2x1x2z1z2+\scriptsize{= (y_1z_2)^2 - 2y_1y_2z_1z_2 + (y_2z_1)^2 + (z_1x_2)^2 - 2x_1x_2z_1z_2 \, +} (z2x1)2+(x1y2)2−2x1x2y1y2+(x2y1)2\scriptsize{(z_2x_1)^2 + (x_1y_2)^2 - 2x_1x_2y_1y_2 + (x_2y_1)^2}
a⃗⋅b⃗=x1x2+y1y2+z1z2\scriptsize{\vec{a}\cdot\vec{b} = x_1x_2 + y_1y_2 + z_1z_2} (a⃗⋅b⃗)2=(x1x2+y1y2+z1z2)(x1x2+y1y2+z1z2)\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)} =x1x2⋅x1x2+x1x2⋅y1y2+x1x2⋅z1z2+y1y2⋅x1x2+\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 \,+ } y1y2⋅y1y2+y1y2⋅z1z2+z1z2⋅x1x2+z1z2⋅y1y2+z1z2⋅z1z2\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} =(x1x2)2+x1x2y1y2+z1z2x1x2+x1x2y1y2+(y1y2)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 \, +} y1y2z1z2+z1z2x1x2+y1y2z1z2+(z1z2)2\scriptsize{y_1y_2z_1z_2 + z_1z_2x_1x_2 + y_1y_2z_1z_2 + (z_1z_2)^2} =(x1x2)2+2x1x2y1y2+(y1y2)2+2y1y2z1z2+(z1z2)2+2z1z2x1x2\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}
∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=(y1z2)2+(y2z1)2+(z1x2)2+(z2x1)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 \,+ } (x1y2)2+(x2y1)2+(x1x2)2+(y1y2)2+(z1z2)2\scriptsize{(x_1y_2)^2 + (x_2y_1)^2 + (x_1x_2)^2 + (y_1y_2)^2 + (z_1z_2)^2}
∣a⃗∣2∣b⃗∣2=(x12+y12+z12)(x22+y22+z22)\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)} =x12x22+x12y22+x12z22+y12x22+y12y22+\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 \, +} y12z22+z12x22+z12y22+z12z22\scriptsize{y_1^2z_2^2 + z_1^2x_2^2 + z_1^2y_2^2 + z_1^2z_2^2} =(x1x2)2+(x1y2)2+(x1z2)2+(x2y1)2+(y1y2)2+\scriptsize{= (x_1x_2)^2 + (x_1y_2)^2 + (x_1z_2)^2 + (x_2y_1)^2 + (y_1y_2)^2 \, +} (y1z2)2+(z1x2)2+(z1y2)2+(z1z2)2\scriptsize{(y_1z_2)^2 + (z_1x_2)^2 + (z_1y_2)^2 + (z_1z_2)^2} ∴∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣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

∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣  sin  θ\left|\vec{a}\times\vec{b}\right| = \left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{sin}\,\, \theta ∣a⃗×b⃗∣2=(∣a⃗∣∣b⃗∣  sin  θ)2\left|\vec{a}\times\vec{b}\right|^2 = (\left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{sin}\,\, \theta)^2 =∣a⃗∣2∣b⃗∣2  sin2  θ______(1)= \left|\vec{a}\right|^2 \left|\vec{b}\right|^2\,\, \text{sin}^2\,\, \theta \qquad \_\_\_\_\_\_(1) a⃗⋅b⃗=∣a⃗∣∣b⃗∣  cos  θ\vec{a}\cdot\vec{b} = \left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{cos}\,\, \theta (a⃗⋅b⃗)2=(∣a⃗∣∣b⃗∣  cos  θ)2(\vec{a}\cdot\vec{b})^2 = (\left|\vec{a}\right| \left|\vec{b}\right| \,\, \text{cos}\,\, \theta)^2 =∣a⃗∣2∣b⃗∣2  cos2  θ______(2) = \left|\vec{a}\right|^2 \left|\vec{b}\right|^2 \,\, \text{cos}^2\,\, \theta \qquad \_\_\_\_\_\_(2) eq______(1)+(2)eq \_\_\_\_\_\_(1) + (2) ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2  sin2  θ  +∣a⃗∣2∣b⃗∣2  cos2  θ\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 =∣a⃗∣2∣b⃗∣2  (sin2  θ+cos2  θ)= \left|\vec{a}\right|^2 \left|\vec{b}\right|^2 \,\, (\text{sin}^2 \,\, \theta + \text{cos}^2\,\, \theta) =∣a⃗∣2∣b⃗∣2⋅1= \left|\vec{a}\right|^2 \left|\vec{b}\right|^2 \cdot 1 =∣a⃗∣2∣b⃗∣2= \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).
(a)    Calculate  p⃗=AB→×AC→  and\text{(a) \,\, Calculate} \,\, \vec{p} = \overrightarrow{AB} \times \overrightarrow{AC} \,\, \text{and} q⃗=BA→×BC→. \vec{q} = \overrightarrow{BA} \times \overrightarrow{BC}. (b)    What can you say about vectors  p⃗  and  q⃗ ?\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).
(a)AB→=AO→+OB→=OB→−OA→\text{(a)} \quad \overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{OB} - \overrightarrow{OA} (772)−(3−51)=(4121)\begin{pmatrix}7\\7\\2\end{pmatrix} - \begin{pmatrix}3\\-5\\1\end{pmatrix} = \begin{pmatrix}4\\12\\1\end{pmatrix} AC→=AO→+OC→=OC→−OA→\quad \overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = \overrightarrow{OC} - \overrightarrow{OA} (−113)−(3−51)=(−462)\begin{pmatrix}-1\\1\\3\end{pmatrix} - \begin{pmatrix}3\\-5\\1\end{pmatrix} = \begin{pmatrix}-4\\6\\2\end{pmatrix} p⃗=AB→×AC→\vec{p} = \overrightarrow{AB}\times\overrightarrow{AC} =(4121)×(−462)= \begin{pmatrix}4\\12\\1\end{pmatrix} \times \begin{pmatrix}-4\\6\\2\end{pmatrix} 4           12           1           4           12 -4           6           2           -4           6
=(12×2−6×11×(−4)−2×44×6−(−4)×12)= \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} =(24−6−4−824+48)=(18−1272)= \begin{pmatrix}24 - 6 \\ -4 - 8 \\ 24 + 48 \end{pmatrix} = \begin{pmatrix}18 \\ -12\\ 72\end{pmatrix} BC→=BO→+OC→=OC→−OB→\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \overrightarrow{OC} - \overrightarrow{OB} =(−113)−(772)=(−8−61)= \begin{pmatrix}-1\\1\\3 \end{pmatrix} - \begin{pmatrix}7\\7\\2\end{pmatrix} = \begin{pmatrix}-8\\-6\\1 \end{pmatrix} BA→=−AB→=−(4121)=(−4−12−1)\overrightarrow{BA} = - \overrightarrow{AB} = - \begin{pmatrix}4 \\ 12 \\ 1 \end{pmatrix} = \begin{pmatrix}-4 \\ -12 \\ -1 \end{pmatrix}
q⃗=BA→×BC→\vec{q} = \overrightarrow{BA} \times \overrightarrow{BC} =(−4−12−1)×(−8−61)= \begin{pmatrix}-4\\-12\\-1 \end{pmatrix} \times \begin{pmatrix}-8 \\ -6 \\1 \end{pmatrix} -4       -12           -1           -4           -12 -8         -6           1           -8           -6
=(−12×1−(−6)×(−1)−1×(−8)−1×(−4)−4×(−6)−(−8)×(−12))= \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} =(−12−68+424−96)=(−1812−72)=−p⃗= \begin{pmatrix}-12 - 6\\ 8 + 4 \\ 24 - 96 \end{pmatrix} = \begin{pmatrix}-18 \\ 12 \\ -72 \end{pmatrix} = - \vec{p}
(b)   p = q
p⃗  and  q⃗  are parallel and opposite direction.\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.
∴AB→=DC→\therefore \overrightarrow{AB} = \overrightarrow{DC} AO→+OB→=DO→+OC→\overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{DO} + \overrightarrow{OC} OB→−OA→=OC→−OD→\overrightarrow{OB} - \overrightarrow{OA} = \overrightarrow{OC} - \overrightarrow{OD} (−115)−(312)=(723)−(xyz)\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} (−403)=(7−x2−y3−z)\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).

BC→=BO→+OC→=OC→−OB→\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \overrightarrow{OC} - \overrightarrow{OB} =(723)−(−115)=(81−2)= \begin{pmatrix}7\\2\\3 \end{pmatrix} - \begin{pmatrix}-1\\1\\5\end{pmatrix} = \begin{pmatrix}8\\1\\-2\end{pmatrix} AB→=(−403)\overrightarrow{AB} = \begin{pmatrix}-4\\0\\3\end{pmatrix} AB→×BC→=(−403)×(81−2)\overrightarrow{AB}\times \overrightarrow{BC} = \begin{pmatrix}-4\\0\\3 \end{pmatrix} \times \begin{pmatrix}8\\1\\-2 \end{pmatrix} -4           0           3           -4           0 8           1           -2           8          1
=(0×(−2)−1×33×8−(−2)×(−4)−4×1−8×0)= \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} =(0−324−8−4−0)=(−316−4)= \begin{pmatrix}0 - 3\\ 24 - 8\\ -4 - 0 \end{pmatrix} = \begin{pmatrix}-3\\16\\-4 \end{pmatrix} ∣AB→×BC→∣=(−3)2+162+(−4)2\left|\overrightarrow{AB}\times \overrightarrow{BC}\right| = \sqrt{(-3)^2 + 16^2 + (-4)^2} =9+256+16=281 unit2=\sqrt{9 + 256 + 16} = \sqrt{281} \, \text{unit}^2