Chapter 4
Vector Algebra
4.2 Angle Between Two Vectors and Scalar Product
First, we consider the angle θ between the two vectors
θ be the angle between them.
In ▵OAB, we use the cosine rule,
$$|\overrightarrow{AB}|^2 = |\overrightarrow{OA}|^2 + |\overrightarrow{OB}|^2 - 2|\overrightarrow{OA}||\overrightarrow{OB}| \, cos \, \theta,$$ $$|\vec b - \vec a|^2 = |\vec a|^2 + |\vec b|^2 - 2|\vec a||\vec b| \, cos\ \theta. \, \, \, \, (1)$$ $$\text{But}\, \, \vec b - \vec a = \begin{pmatrix}x_2\\y_2\end{pmatrix} - \begin{pmatrix}x_1\\y_1\end{pmatrix}$$ $$= \begin{pmatrix}x_2 - x_1\\y_2 - y_1\end{pmatrix},$$ equation (1) becomes
$$(x_2 - x_1)^2 + (y_2 - y_1)^2 = x_1^2 +y_1^2 + x_2^2 + y_2^2 - 2|\vec a||\vec b| \, cos \, \theta $$ which siplifies to $$x_1x_2 + y_1y_2 = |\vec a||\vec b|\ cos\ \theta $$ So, $$\cos \ \theta = \frac{x_1x_2 + y_1y_2}{|\vec a||\vec b|}. \, \, \, \, (2)$$ $$\text{The\, expression}\, x_1x_2 + y_1y_2 \, \text{can\, be}$$ $$\text{written\, by}\, \vec a \sdot \vec b. \, \text{that\, is},$$ $$\vec a \sdot \vec b = x_1x_2 + y_1y_2 $$ For two dimensions. This method can be extended into three dimensions. $$\text{If}\, \vec a = \begin{pmatrix}x_1\\y_1\\z_11\end{pmatrix}\, \text{and}\, \vec b = \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix},$$ it can be written by $$cos \, \theta = \frac{x_1x_2 + y_1y_2 + z_1z_2}{|\vec a| |\vec b|}. \, \, \, \, \, (3)$$ $$\text{Thus}\, \vec a \sdot \, \vec b \, \text{is\, also\, written\, by} $$ $$\vec a \sdot \vec b = x_1x_2 + y_1y_2 + z_1z_2. $$ $$\text{The\, expression}\, \vec a \sdot \vec b\, \text{is\, called}$$ the scalar product or dot product of the vectors $$\vec a\, \text{and}\, \vec b.$$ In both cases, we can write (2) and (3) as follows:
$$cos \, \theta = \frac{\vec a \sdot \vec b}{|\vec a| |\vec b|}. $$ This result can be written in the form, $$\vec a \sdot \vec b = |\vec a| |\vec b| \, \text{cos} \, \theta.$$ Example 8.
Find the angles between the two vectors $$\begin{pmatrix}3\\4\end{pmatrix} \, \text{and}\, \begin{pmatrix}5\\-12\end{pmatrix}.$$ Solution
$$\text{Let}\, \vec a = \begin{pmatrix}3\\4\end{pmatrix} \, \text{and}\, \vec b = \begin{pmatrix}5\\-12\end{pmatrix}.$$ We can find that $$|\vec a| = \sqrt{3^2 + 4^2} = 5 $$ and $$|\vec b| = \sqrt{5^2 + (-12)^2} = 13.$$ and $$\vec a \sdot \vec b = \begin{pmatrix}3\\4\end{pmatrix} \sdot \begin{pmatrix}5\\-12\end{pmatrix}$$ $$= 3 \times 5 + 4 \times (-12) = 15 - 48 = -33$$ $$\text{cos}\, \theta = \frac{\vec a \sdot \vec b}{|\vec a| |\vec b|}$$ $$= \text{cos}\, \theta = \frac{-33}{65}$$ $$\theta = 120.5^{\circ}.$$
Example 9.
Given points P(1, 0, -1), Q(2, 4, 1) and R(3, 5, 6), find ∠PQR.
Solution
$$\text{The\, angle\, between}\, \overrightarrow{PQ}\, \text{and}\, \overrightarrow{PR}$$ $$\texttt{is\, given\, by}\, \theta\, \text{in} $$ $$\text{cos}\, \theta = \frac{\overrightarrow{PQ} \sdot \overrightarrow{PR}}{|\overrightarrow{PQ}| |\overrightarrow{PR}|}.$$ $$\text{Since}\, \overrightarrow{PQ} = \begin{pmatrix}2\\4\\1\end{pmatrix} - \begin{pmatrix}1\\0\\-1\end{pmatrix}$$ $$= \begin{pmatrix}1\\4\\2\end{pmatrix}, \, \text{we\, get}$$ $$|\overrightarrow{PR}| = \sqrt{1^2 + 4^2 + 2^2} = \sqrt{21}.$$ $$\overrightarrow{PQ} \sdot \overrightarrow{PR} = \begin{pmatrix}1\\4\\2\end{pmatrix} \sdot \begin{pmatrix}2\\5\\7\end{pmatrix} = 36,$$ $$\text{cos}\, \theta\, = \frac{36}{\sqrt{21} \times \sqrt{78}},$$ $$\theta = 27.2^\circ .$$
Algebraic Properties of the Scalar Product
The scalar product of two vectors has the following algebraic properties.$$\text{If}\, \vec a,\, \vec b\, \text{and}\, \vec c \, \, \text{are}$$ $$\text{vectors\, in\, space\, and}\, k\, \text{is\, a\, scalar,\, then}$$ $$1. \, \, \, \vec a \sdot \vec b = \vec b \sdot \vec a, $$ $$2.\, \, \, (-\vec a) \sdot vec b = \vec a \sdot (-\vec b) = -(\vec a \sdot \vec b),$$ $$3. \, \, \, \vec a \sdot(\vec b + \vec c) = \vec a |sdot \vec b + \vec a \sdot \vec c,$$ $$4. \, \, \, (k\vec a) \sdot \vec b = k(\vec a \sdot \vec b) = \vec a \sdot(k \vec b).$$ $$\vec 0 \sdot \vec a = 0.$$
Geometric Properties of the Scalar Product
$$1.\, \, \, \text{If\, nonzero\, vectors}\, \vec a\, \text{and}\, \vec b\, \text{are} $$ $$\text{\textbf{perpendicular}\, then}\, \vec a \sdot \, \vec b = 0.$$ $$2. \, \, \, \text{If\, nonzero \, vectors}\, \vec a\, \text{and}\, \vec b\, \text{are} $$ $$\text{\textbf{parallel}\, in\, same\, direction,\, then}$$ $$\vec a \sdot \vec b = |\vec a| |\vec b|,\, \text{in\, particular,}$$ $$\vec a \sdot \vec a = |\vec a|^2.$$ $$\text{If\, nonzero\, vectors}\, \vec a \, \text{and}\, \vec b\, \text{are}$$ $$\text{\textbf{parallel}\, in\, opposite\, directions,\, then}$$ $$\vec a \sdot \vec b = -|\vec a| |\vec b|.$$Example 10.
$$\text{Given\, that}\, \vec a\, \text{and}\, \vec b\, \text{are\, perpendicular\, vectors}$$ $$\text{such\, that}\, |\vec a| = 3\, \text{and}\, \, |\vec b| = 1,$$ $$\text{evaluate}\, (\vec a - \vec b)\sdot(\vec a + 5\vec b). $$ Solution
$$\text{Since}\, \vec a \,\text{and}\, \vec b\, \text{are\, perpendicular},$$ $$\text{so}\, \vec a \sdot \vec b = \vec b \sdot \vec a = 0.$$ $$(\vec a - \vec b) \sdot (\vec a + 5\vec b) = \vec a \sdot \vec a + 5(\vec a \sdot \vec b) - \vec b\sdot \vec a - 5(\vec b\sdot \vec b)$$ $$= \vec a\sdot \vec a - 5(\vec b\sdot \vec b)$$ $$= |\vec a|^2 - 5|\vec b|^2 = 3^2 - 5 \times 1^2 = 4.$$
Example 11.
Points A, B and C have position vectors $$\vec a = k\begin{pmatrix}2\\-1\\1\end{pmatrix}, \, \vec b = \begin{pmatrix}3\\2\\-2\end{pmatrix}$$ $$\text{and}\, \vec c = \begin{pmatrix}1\\1\\4\end{pmatrix}.$$ $$\text{(a) \, \, Find}\, \overrightarrow{BC}.$$ $$\text{(b)\, \, Find}\, \overrightarrow{AB}\, \text{in\, terms\, of}\, k.$$ $$\text{(c)\, \, Find\, the\, value\, of\,} k\, \text{for\, which}$$ $$\overrightarrow{AB}\, \text{is\, perpendicular\, to}\, \overrightarrow{BC}.$$ Solution
$$\text{(a)}\, \, \overrightarrow{BC} = \vec c - \vec b$$ $$= \begin{pmatrix}1\\1\\4\end{pmatrix} - \begin{pmatrix}3\\2\\-2\end{pmatrix}$$ $$= \begin{pmatrix}-2\\-1\\6\end{pmatrix}.$$ $$\text{(b)}\, \, \overrightarrow{AB} = \vec b - \vec a$$ $$= \begin{pmatrix}3\\2\\-2\end{pmatrix} - k\begin{pmatrix}2\\-1\\1\end{pmatrix}$$ $$= \begin{pmatrix}3-2k\\2+k\\-2-k\end{pmatrix}.$$ $$\text{(c)\, \, Since}\, \overrightarrow{AB} \, \text{is\, perpendicular to}\, \overrightarrow{BC},$$ $$\text{it\, follows\, that}$$ $$\overrightarrow{AB} \sdot \overrightarrow{BC} = 0$$ $$\begin{pmatrix}3-2k\\2+k\\-2-k\end{pmatrix} \sdot \begin{pmatrix}-2\\-1\\6\end{pmatrix} = 0$$ $$-6 + 4k -2 - k -12 - 6k = 0$$ $$-3k = 20$$ $$k = - \frac{20}{3}.$$
Exercise 4.2
$$1. \hspace{0.2cm} For \, \, \vec{p} = \begin{pmatrix}3\\2\end{pmatrix}, \hspace{0.3cm} \vec{q} = \begin{pmatrix}-1\\5\end{pmatrix}\,
and \, \, \vec{r} = \begin{pmatrix}-2\\4\end{pmatrix}, \, find:$$
$$(a) \quad \vec{q} \cdot \vec{p} \hspace{1cm} (b) \quad \vec{q} \cdot \vec{r} \hspace{1cm}(c) \quad \vec{q} \cdot (\vec{p} + \vec{r}) \newline
(d) \quad \hat{i} \cdot \vec{p} \hspace{1cm} (e) \quad \vec{q} \cdot \hat{j} \hspace{1cm} (f) \quad \hat{i} \cdot \hat{i}.$$
$$\textbf{Solution}$$
$$(a) \quad \vec{q} \cdot \vec{p} = \begin{pmatrix}-1\\5\end{pmatrix} \cdot \begin{pmatrix}3\\2\end{pmatrix}$$
$$= (-1) \times 3 \quad + \quad 5 \times 2$$
$$= -3 + 10$$
$$= 7\newline$$
$$(b) \quad \vec{q} \cdot \vec{r} = \begin{pmatrix} -1\\5\end{pmatrix} \cdot \begin{pmatrix}-2\\4\end{pmatrix}$$
$$= (-1) \times (-2) \quad + \quad 5 \times 4$$
$$= 2 + 20$$
$$= 22 \newline$$
$$(c) \quad \vec{q} \cdot (\vec{p} + \vec{r}) = \begin{pmatrix} -1\\5\end{pmatrix} \cdot \left[\begin{pmatrix}3\\2\end{pmatrix} + \begin{pmatrix}-2\\4\end{pmatrix}
\right]$$
$$= \begin{pmatrix}-1\\5\end{pmatrix} \cdot \begin{pmatrix}1\\6\end{pmatrix}$$
$$= (-1) \times 1 \quad + \quad 5 \times 6$$
$$= -1 + 30$$
$$= 29$$
$$\newline$$
$$(d) \quad \hat{i} \cdot \vec{p} = \begin{pmatrix} 1\\0\end{pmatrix} \cdot \begin{pmatrix}3\\2\end{pmatrix}$$
$$= 1 \times 3 \quad + \quad 0 \times 2$$
$$= 3 + 0$$
$$ = 3$$
$$(e) \quad \vec{q} \cdot \vec{j} = \begin{pmatrix} -1\\ 5\end{pmatrix} \cdot \begin{pmatrix}0\\1\end{pmatrix}$$ $$= (-1) \times 0 \quad + \quad 5 \times 1$$ $$= 0 + 5$$ $$= 5$$
$$(f) \quad \hat{i} \cdot \hat{i} = \begin{pmatrix}1\\0\end{pmatrix} \cdot \begin{pmatrix}1\\0\end{pmatrix}$$ $$ = 1 \times 1 \quad 0 \times 0 $$ $$ = 1 $$
$$\maroonE{\hat{i} = (1, 0) \,\,\text{and} \,\, \hat{j} = (0, 1)\,\, \text{are called standard unit vectors.}}$$
$$(e) \quad \vec{q} \cdot \vec{j} = \begin{pmatrix} -1\\ 5\end{pmatrix} \cdot \begin{pmatrix}0\\1\end{pmatrix}$$ $$= (-1) \times 0 \quad + \quad 5 \times 1$$ $$= 0 + 5$$ $$= 5$$
$$(f) \quad \hat{i} \cdot \hat{i} = \begin{pmatrix}1\\0\end{pmatrix} \cdot \begin{pmatrix}1\\0\end{pmatrix}$$ $$ = 1 \times 1 \quad 0 \times 0 $$ $$ = 1 $$
$$\maroonE{\hat{i} = (1, 0) \,\,\text{and} \,\, \hat{j} = (0, 1)\,\, \text{are called standard unit vectors.}}$$
$$2. \quad For \,\, \vec{a} = \begin{pmatrix}2\\1\\3 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \quad and \, \,
\vec{c} = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} , \quad find:$$
$$(a) \quad \vec{a} \cdot \vec{b} \qquad (b) \,\, \vec{b} \cdot \vec{a} \qquad (c) \,\, \left|\vec{a}\right|^2 $$
$$ (d) \,\, \vec{a} \cdot \vec{a} \qquad (e) \,\, \vec{a} \cdot \left( \vec{b} + \vec{c} \right) \qquad (f) \,\, \vec{a} \cdot \vec{b} \quad + \quad \vec{a} \cdot
\vec{c}.$$
Solution
$$(a) \quad \vec{a} \cdot \vec{b} = \begin{pmatrix} 2 \\ 1\\ 3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$$
$$= 2 \times (-1) \quad + \quad 1 \times 1 \quad + \quad 3 \times 1$$
$$= -2 + 1 + 3$$
$$= 2$$
$$(b) \quad \vec{b} \cdot \vec{a} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix}2 \\ 1 \\ 3 \end{pmatrix}$$ $$= (-1) \times 2 \quad + \quad 1 \times 1 \quad + \quad 1 \times 3$$ $$= -2 + 1 + 3 $$ $$= 2$$
$$(c) \quad \left|\vec{a}\right| = \sqrt{2^2 + 1^2 + 3^2} $$ $$= \sqrt{14} $$ $$ \left|\vec{a}\right|^2 = \left( \sqrt{14} \right)^2$$ $$= 14$$
$$(d) \quad \vec{a} \cdot \vec{a} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$$ $$= 2 \times 2 \quad + \quad 1 \times 1 \quad + \quad 3 \times 3 $$ $$= 4 + 1 + 9 $$ $$= 14$$
$$(e) \quad \vec{a} \cdot (\vec{b} + \vec{c}) = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \cdot \left[\begin{pmatrix}-1 \\ 1 \\ 1\end{pmatrix} + \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\right] $$ $$= \begin{pmatrix}2 \\ 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} $$ $$ = 2 \times (-1) \quad + \quad 1 \times 0 \quad + \quad 3 \times 2 $$ $$= -2 + 0 + 6 $$ $$= 4$$
$$(f) \quad \vec{a} \cdot \vec{b} \quad + \quad \vec{a} \cdot \vec{c} = \begin{pmatrix}2 \\ 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \quad + \quad \begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix} \cdot \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$$ $$= [2 \times (-1) \,\, + \,\, 1 \times 1 \,\, + \,\, 3 \times 1] \quad + \quad [2 \times 0 \,\, + \,\, 1 \times (-1) \,\, + \,\, 3 \times 1]$$ $$= (-2 + 1 + 3) \quad + \quad (0 - 1 + 3)$$ $$= 2 + 2$$ $$= 4$$
$$(b) \quad \vec{b} \cdot \vec{a} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix}2 \\ 1 \\ 3 \end{pmatrix}$$ $$= (-1) \times 2 \quad + \quad 1 \times 1 \quad + \quad 1 \times 3$$ $$= -2 + 1 + 3 $$ $$= 2$$
$$(c) \quad \left|\vec{a}\right| = \sqrt{2^2 + 1^2 + 3^2} $$ $$= \sqrt{14} $$ $$ \left|\vec{a}\right|^2 = \left( \sqrt{14} \right)^2$$ $$= 14$$
$$(d) \quad \vec{a} \cdot \vec{a} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$$ $$= 2 \times 2 \quad + \quad 1 \times 1 \quad + \quad 3 \times 3 $$ $$= 4 + 1 + 9 $$ $$= 14$$
$$(e) \quad \vec{a} \cdot (\vec{b} + \vec{c}) = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \cdot \left[\begin{pmatrix}-1 \\ 1 \\ 1\end{pmatrix} + \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\right] $$ $$= \begin{pmatrix}2 \\ 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} $$ $$ = 2 \times (-1) \quad + \quad 1 \times 0 \quad + \quad 3 \times 2 $$ $$= -2 + 0 + 6 $$ $$= 4$$
$$(f) \quad \vec{a} \cdot \vec{b} \quad + \quad \vec{a} \cdot \vec{c} = \begin{pmatrix}2 \\ 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \quad + \quad \begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix} \cdot \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$$ $$= [2 \times (-1) \,\, + \,\, 1 \times 1 \,\, + \,\, 3 \times 1] \quad + \quad [2 \times 0 \,\, + \,\, 1 \times (-1) \,\, + \,\, 3 \times 1]$$ $$= (-2 + 1 + 3) \quad + \quad (0 - 1 + 3)$$ $$= 2 + 2$$ $$= 4$$
$$3. \quad \text{Find the angle between} \,\, \vec{m} \,\, \text{and} \,\, \vec{n} \,\, \text{if:}$$
$$\text{(a)} \quad \vec{m} = \begin{pmatrix}2 \\ -1 \\ -1 \end{pmatrix} \quad \text{and} \quad \vec{n} = \begin{pmatrix}-1 \\ 3 \\ 2\end{pmatrix}$$
$$\text{(b)} \quad \vec{m} = 2\hat{j} - \hat{k} \quad and \quad \vec{n} = \hat{i} + 2\hat{k}.$$
Solution
$$\text{(a)} \quad \left|\vec{m}\right| = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{6} $$
$$\left|\vec{n}\right| = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{14}$$
$$\vec{m} \cdot \vec{n} = \begin{pmatrix}2 \\ -1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}$$
$$= 2 \times (-1) \quad + \quad (-1) \times 3 \quad + \quad (-1) \times 2$$
$$= -2 - 3 - 2$$
$$= -7$$
$$\text{cos}\, \theta = \dfrac{\vec{m} \cdot \vec{n}}{\left|\vec{m}\right| \left|\vec{n}\right|}$$ $$= \dfrac{-7}{\sqrt{6} \sqrt{14} }$$ $$= \dfrac{-7} {\sqrt{84}}$$ $$\text{cos} \, \theta = - 0.7638$$ $$\theta = cos^{-1} (- 0.7638)$$ $$ = - cos^{-1} 0.7638)$$ $$= - 40.1996^{\circ}$$The solution of cos θ = - 0.7638 is
θ = 180 - 40.1996
= 139.8004°
$$\text{(b)} \quad \vec{m}= 2\hat{j} - \hat{k}$$ $$= 2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} - \begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}$$ $$= \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} - \begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}$$ $$= \begin{pmatrix}0 \\ 2 \\ -1 \end{pmatrix}$$
$$\vec{n} = \hat{i} + 2\hat{k}$$ $$= \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} + 2\begin{pmatrix}0\\ 0 \\ 1 \end{pmatrix}$$ $$= \begin{pmatrix}1\\0\\0\end{pmatrix} + \begin{pmatrix}0\\0\\2 \end{pmatrix}$$ $$= \begin{pmatrix}1\\0\\2\end{pmatrix}$$ $$\left|\vec{m}\right| = \sqrt{0^2 + 2^2 + (-1)^2} = \sqrt{5}$$ $$\left|\vec{n}\right| = \sqrt{1^2 + 0^2 + 2^2} = \sqrt{5}$$ $$\vec{m} \cdot \vec{n} = \begin{pmatrix}0\\2\\-1 \end{pmatrix} \cdot \begin{pmatrix}1\\0\\2 \end{pmatrix}$$ $$= 0 \times 1 \quad + \quad 2 \times 0 \quad + \quad (-1) \times 2$$ $$= 0 + 0 - 2$$ $$= -2$$
$$\text{cos} \,\, \theta = \dfrac{\vec{m} \cdot \vec{n}} {\left|\vec{m}\right| \left|\vec{n}\right|}$$ $$= \dfrac{-2} {\sqrt{5} \sqrt{5}} = \dfrac{-2}{5} = -0.4$$
θ = cos-1 (-0.4)
= - cos-1 0.4
= - 66.4218
The solution of cos θ = -0.4 is
θ = 180 - 66.4218
= 113.5782°
$$\text{cos}\, \theta = \dfrac{\vec{m} \cdot \vec{n}}{\left|\vec{m}\right| \left|\vec{n}\right|}$$ $$= \dfrac{-7}{\sqrt{6} \sqrt{14} }$$ $$= \dfrac{-7} {\sqrt{84}}$$ $$\text{cos} \, \theta = - 0.7638$$ $$\theta = cos^{-1} (- 0.7638)$$ $$ = - cos^{-1} 0.7638)$$ $$= - 40.1996^{\circ}$$
θ = 180 - 40.1996
= 139.8004°
$$\text{(b)} \quad \vec{m}= 2\hat{j} - \hat{k}$$ $$= 2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} - \begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}$$ $$= \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} - \begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}$$ $$= \begin{pmatrix}0 \\ 2 \\ -1 \end{pmatrix}$$
$$\vec{n} = \hat{i} + 2\hat{k}$$ $$= \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} + 2\begin{pmatrix}0\\ 0 \\ 1 \end{pmatrix}$$ $$= \begin{pmatrix}1\\0\\0\end{pmatrix} + \begin{pmatrix}0\\0\\2 \end{pmatrix}$$ $$= \begin{pmatrix}1\\0\\2\end{pmatrix}$$ $$\left|\vec{m}\right| = \sqrt{0^2 + 2^2 + (-1)^2} = \sqrt{5}$$ $$\left|\vec{n}\right| = \sqrt{1^2 + 0^2 + 2^2} = \sqrt{5}$$ $$\vec{m} \cdot \vec{n} = \begin{pmatrix}0\\2\\-1 \end{pmatrix} \cdot \begin{pmatrix}1\\0\\2 \end{pmatrix}$$ $$= 0 \times 1 \quad + \quad 2 \times 0 \quad + \quad (-1) \times 2$$ $$= 0 + 0 - 2$$ $$= -2$$
$$\text{cos} \,\, \theta = \dfrac{\vec{m} \cdot \vec{n}} {\left|\vec{m}\right| \left|\vec{n}\right|}$$ $$= \dfrac{-2} {\sqrt{5} \sqrt{5}} = \dfrac{-2}{5} = -0.4$$
= - cos-1 0.4
= - 66.4218
The solution of cos θ = -0.4 is
θ = 180 - 66.4218
= 113.5782°
(i) Perpendicular (ii) parallel
$$\text{(a)} \quad \vec{p} = \begin{pmatrix}3\\t \end{pmatrix} \quad and \quad \vec{q} = \begin{pmatrix}-2\\1 \end{pmatrix},$$ $$\text{(b)} \quad \vec{r} = \begin{pmatrix}t \\ t + 2 \end{pmatrix} \quad and \quad \vec{s} = \begin{pmatrix}t \\ -4 \end{pmatrix},$$ $$\text{(c)} \quad \vec{a} = \begin{pmatrix}0 \\ t + 2 \end{pmatrix} \quad and \quad \vec{b} = \begin{pmatrix}2 - 3t \\ t \end{pmatrix}.$$
$$\text{(a)} \, (i) \quad \text{If} \,\, \vec{p} \,\, \text{and}\,\, \vec{q} \,\, \text{are perpendicular, then}$$ $$\vec{p} \cdot \vec{q} = 0$$ $$\begin{pmatrix}3\\t\end{pmatrix} \cdot \begin{pmatrix}-2 \\ 1 \end{pmatrix} = 0 $$ $$3 \times (-2) \quad + \quad t \times 1 = 0$$ $$-6 + t = 0$$ $$t = 6$$
$$\text{(ii) \quad If} \,\, \vec{p} \,\, \text{and} \,\, \vec{q} \,\, \text{are parallel, then}$$ $$\vec{p} \cdot \vec{q} = \left|\vec{p}\right| \left|\vec{q}\right|$$ $$\begin{pmatrix}3\\t\end{pmatrix} \cdot \begin{pmatrix}-2\\1 \end{pmatrix} = \sqrt{3^2 + t^2} \sqrt{(-2)^2 + 1^2}$$ $$3 \times (-2) \quad + \quad t \times 1 = \sqrt{9 + t^2} \sqrt{5}$$
(-6 + t)2 = ( √ 9 + t2 - √ 5 )2
36 - 12t + t2 = 45 + 5t2
0 = 4t2 + 12t + 9
(2t + 3) (2t + 3) = 0
(2t + 3)2 = 0
[ (2t + 3)2 = 0 ,] 1 &frasl 2
2t + 3 = 0
2t = -3
t = - 3 ⁄2
$$\text{(b)\, (i) \quad If} \,\, \vec{r} \,\, \text{and} \,\, \vec{s} \,\, \text{are perpendicular, then} $$
$$\vec{r} \cdot \vec{s} = 0$$
$$\begin{pmatrix}t \\ t + 2 \end{pmatrix} \cdot \begin{pmatrix}t \\ -4 \end{pmatrix} = 0$$
t2 - 4t - 8 = 0
t2 - 4t = 8
(t - 2)2 - 4 = 8
(t - 2)2 = 12
√ (t - 2)2 = ± √ 12
t - 2 = ± √ 12
t = 2 ± √ 12
= 2 ± √ 4 × 3
= 2 ± 2√ 3
t2 - 4t - 8 = √ 2t2 + 4t + 4 √ t2 + 16
Squaring both sides
(t2 - 4t - 8)2 = (2t2 + 4t + 4) (t2 + 16)
$$t^4 - 8t^3 + 64t + 64 = 2t^4 + 4t^3 + 36t^2 + 64t + 64$$ $$t^4 - 8t^3 + \cancel{64t} + \cancel{\purpleD{64}} = 2t^4 + 4t^3 + 36t^2 + \cancel{64t} + \cancel{\purpleD{64}}$$ $$0 = t^4 + 12t^3 + 36t^2$$ $$t^2 (t^2 + 12t + 36) = 0$$ $$t^2 = 0 \quad \text{or} \quad t^2 + 12t + 36 = 0$$ $$t \cdot t = 0 \quad \text{or} \quad (t + 6)(t + 6) = 0$$ $$t = \dfrac{0}{t} \quad \text{or} \quad t + 6 = \dfrac{0}{t + 6}$$ $$t = 0 \quad \text{or} \quad t + 6 = 0$$ $$ t = 0 \quad \text{or} \quad t = -6$$
$$\text{(c) \,\, (i) \quad If} \,\, \vec{a} \,\, \text{and} \,\, \vec{b} \,\, \text{are perpendicular, then} $$ $$\vec{a} \cdot \vec{b} = 0$$ $$\begin{pmatrix}0\\t + 2 \end{pmatrix} \cdot \begin{pmatrix}2 - 3t\\ t \end{pmatrix} = 0$$ $$0 \times (2 - 3t) \quad + \quad (t + 2) \times t = 0$$ $$0 + t^2 + 2t = 0$$ $$t (t + 2) = 0$$ $$t = 0 \quad \text{or} \quad t + 2 = 0$$ $$t = 0 \quad \text{or} \quad t = -2$$
$$\text{(ii)} \quad \vec{a} \,\, \text{and} \,\, \vec{b} \text{are parallel.}$$ $$\vec{b} = k\vec{a}$$ $$\begin{pmatrix}2 - 3t\\ t \end{pmatrix} = k \begin{pmatrix}0 \\ t + 2 \end{pmatrix}$$ $$\begin{pmatrix}2 - 3t \\ t \end{pmatrix} = \begin{pmatrix}0 \\ k (t + 2) \end{pmatrix}$$ $$2 - 3t = 0$$ $$-3t = -2$$ $$t = \dfrac{\cancel{-}2} {\cancel{-} 3}$$ $$t = \dfrac{2}{3}$$
$$\text{(a)} \quad \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$$ $$= \vec{a} + \vec{b}$$ $$\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD}$$ $$= - \overrightarrow{AB} + \overrightarrow{AD}$$ $$= - \vec{a} + \vec{b}$$
$$\text{(b)} \quad (\vec{a} + \vec{b}) \cdot (\vec{b} - \vec{a}) = \vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} - \vec{b}\cdot\vec{a}$$ $$= \vec{a}\cdot\vec{b} - \left|\vec{a}\right|^2 + \left|\vec{b}\right|^2 - \vec{a}\cdot\vec{b}$$ $$= \left|\vec{b}\right|^2 - \left|\vec{a}\right|^2$$
$$\text{(c) \quad If} \,\, ABCD \,\, \text{is a rhombus, then}$$ $$\left|\overrightarrow{AB}\right| = \left|\overrightarrow{AD}\right|$$ $$\therefore \left|\vec{a}\right| = \left|\vec{b}\right| $$ $$ \left|\vec{a}\right|^2 = \left|\vec{b}\right|^2 $$ $$\overrightarrow{AC} \cdot \overrightarrow{BD} = (\vec{a} + \vec{b}) \cdot (\vec{b} - \vec{a}) $$ $$= \vec{a}\cdot\vec{b} - \left|\vec{a}\right|^2 + \left|\vec{b}\right|^2 - \vec{a}\cdot\vec{b}$$ $$= \left|\vec{b}\right|^2 - \left|\vec{a}\right|^2$$ $$= \left|\vec{b}\right|^2 - \left|\vec{b}\right|^2 \qquad (\because \left|\vec{a}\right|^2 = \left|\vec{b}\right|^2)$$ $$= 0$$ $$\therefore \overrightarrow{AC}\,\, \perp \,\, \overrightarrow{BD}$$