G12 Chapter 4: Vector Algebra

📐 Vectors & 3D Geometry
Foundations & Applications

This chapter introduces the foundational concepts of vectors, their algebraic operations, and their applications in solving geometric problems in both two-dimensional (2D) and three-dimensional (3D) coordinate spaces, culminating in the geometry of lines and planes.

1. 3D Coordinate Systems and Position Vectors

• 3D Space Representation: Any point in a 3D rectangular coordinate system is represented as an ordered triplet $(x, y, z)$.
• Position Vector: The position vector of a point $P(x, y, z)$ relative to the origin $O(0,0,0)$ is written as:
  \[ \overrightarrow{OP} = \vec{p} = x\hat{i} + y\hat{j} + z\hat{k} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \] where $\hat{i}, \hat{j}, \hat{k}$ are the standard unit vectors along the $x, y,$ and $z$ axes respectively.
• Displacement Vector: The vector connecting an initial point $P$ to a terminal point $Q$ is found by subtracting their position vectors:
  \[ \overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} \] Consequently, reversing the direction yields a negative vector: $\overrightarrow{QP} = -\overrightarrow{PQ}$.

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2. Basic Vector Operations and Properties

• Vector Addition and Subtraction: Performed component-wise. If $\vec{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$, then:
  \[ \vec{a} \pm \vec{b} = \begin{pmatrix} a_1 \pm b_1 \\ a_2 \pm b_2 \\ a_3 \pm b_3 \end{pmatrix} \]
• Scalar Multiplication: Multiplying a vector by a real number $k$ scales its components: $k\vec{a} = \begin{pmatrix} ka_1 \\ ka_2 \\ ka_3 \end{pmatrix}$.
• Magnitude (Length): The magnitude of a vector $\vec{a}$ is given by the Pythagorean distance formula in 3D:
  \[ |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
• Unit Vector: A vector with a magnitude of $1$. The unit vector $\hat{a}$ pointing in the same direction as $\vec{a}$ is calculated by:
  \[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} \]
• Parallel Vectors: Two non-zero vectors $\vec{a}$ and $\vec{b}$ are parallel ($\vec{a} \parallel \vec{b}$) if and only if one is a scalar multiple of the other: $\vec{b} = k\vec{a}$. Alternatively, their corresponding components are proportional:
  \[ \frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} \]
• Collinearity: Three points $A, B,$ and $C$ are collinear (lie on the same straight line) if the vectors formed by them (e.g., $\overrightarrow{AB}$ and $\overrightarrow{AC}$) are parallel and share a common point.

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Example 1

Illustrate the points (a) $A(0,3,0)$   (b) $B(4,0,2)$   (c) $C(-1,2,2)$

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Example 2

If $P$ is $(-3, 1, 2)$ and $Q$ is $(1, -1, 3)$, find:
(a) $\overrightarrow{OP}$   (b) $\overrightarrow{PQ}$   (c) $|\overrightarrow{PQ}|$   (d) $\overrightarrow{QP}$   (e) $|\overrightarrow{QP}|$

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(a) $\overrightarrow{OP} = \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix}$

(b) $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix} - \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$

(c) $|\overrightarrow{PQ}| = \sqrt{4^2 + (-2)^2 + 1^2} = \sqrt{21}$

(d) $\overrightarrow{QP} = -\overrightarrow{PQ} = \begin{pmatrix} -4 \\ 2 \\ -1 \end{pmatrix}$

(e) $|\overrightarrow{QP}| = \sqrt{21}$

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Example 3

If $\vec{p} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}$ and $\vec{q} = \begin{pmatrix} -2 \\ 0 \\ 2 \end{pmatrix}$, find:
(a) $\vec{p} + \vec{q}$   (b) $\vec{p} - \frac{1}{2}\vec{q}$   (c) $\frac{3}{2}\vec{q} - \vec{p}$

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(a) $\vec{p} + \vec{q} = \begin{pmatrix} -1 \\ -1 \\ 6 \end{pmatrix}$

(b) $\vec{p} - \frac{1}{2}\vec{q} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$

(c) $\frac{3}{2}\vec{q} - \vec{p} = \begin{pmatrix} -4 \\ 1 \\ -1 \end{pmatrix}$

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Example 4

Find $u$ and $v$ given that $\vec{a} = \begin{pmatrix} -1 \\ -1 \\ u \end{pmatrix}$ is parallel to $\vec{b} = \begin{pmatrix} v \\ 2 \\ -2 \end{pmatrix}$.

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Since $\vec{a} \parallel \vec{b}$, $\vec{a} = k\vec{b}$:

\[ \begin{pmatrix} -1 \\ -1 \\ u \end{pmatrix} = k \begin{pmatrix} v \\ 2 \\ -2 \end{pmatrix} \]

From $-1 = 2k$, $k = -\frac{1}{2}$. Then $v = 2$, $u = 1$.

$u = 1,\; v = 2$

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Example 5

$ABCD$ is a parallelogram. $A(-1,1,1)$, $B(2,0,-2)$, $D(3,1,4)$. Find coordinates of $C$.

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\[ \overrightarrow{DC} = \overrightarrow{AB} \implies \begin{pmatrix} x-3 \\ y-1 \\ z-4 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ -3 \end{pmatrix} \]

So $x=6,\; y=0,\; z=1$.

$C(6,0,1)$

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Example 6

(a) Unit vector in same direction as $\vec{a} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$.
(b) Vector of magnitude 5 parallel to $\vec{a}$.

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(a) $|\vec{a}| = 3$, so $\hat{a} = \frac{1}{3}\begin{pmatrix}2\\-2\\1\end{pmatrix} = \begin{pmatrix}\frac{2}{3}\\-\frac{2}{3}\\\frac{1}{3}\end{pmatrix}$

(b) $\vec{v} = \pm 5\hat{a} = \pm \begin{pmatrix}\frac{10}{3}\\-\frac{10}{3}\\\frac{5}{3}\end{pmatrix}$

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Example 7

Prove that $A(8,2,2)$, $C(20,5,5)$, $B(12,3,3)$ are collinear.

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\[ \overrightarrow{AB} = \begin{pmatrix}4\\1\\1\end{pmatrix}, \quad \overrightarrow{BC} = \begin{pmatrix}8\\2\\2\end{pmatrix} = 2\overrightarrow{AB} \]

Thus $A,B,C$ are collinear.

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Exercise 4.1

Question 1.

Let $\overrightarrow{a}=\begin{pmatrix}5\\-2\\-4\end{pmatrix},\overrightarrow{b}=\begin{pmatrix}3\\-6\\1\end{pmatrix}$ and $\overrightarrow{c}=\begin{pmatrix}0\\7\\-1\end{pmatrix}.$ Find the following vectors:
(a) 3$\overrightarrow{a}$

Short Answer\[ \begin{pmatrix}15\\-6\\-12\end{pmatrix}\]

(b) $4\overrightarrow{b}$

Short Answer\[\begin{pmatrix}12\\-24\\4\end{pmatrix}\]

(c) $\begin{array}{c}\overrightarrow{a}-\overrightarrow{b}\end{array}$

Short Answer$\begin{pmatrix}2\\4\\-5\end{pmatrix}$

(d) $\begin{array}{c}\overrightarrow{b}+\overrightarrow{c}\end{array}$

Short Answer$\begin{pmatrix}3\\1\\0\end{pmatrix}$

(e) 2$\overrightarrow{b} + \overrightarrow{c}$

Short Answer$\begin{pmatrix}6\\-5\\1\end{pmatrix}$

(f) $\overrightarrow{a} - 2\overrightarrow{b}$

Short Answer$ \begin{pmatrix}-1\\10\\-6\end{pmatrix}$

(g) $\overrightarrow{a} + \overrightarrow{b} - 2\overrightarrow{c}$

Short Answer$\begin{pmatrix}8\\-22\\-1\end{pmatrix}$

(h) 3$\overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c}$

Short Answer$\begin{pmatrix}12\\7\\-14\end{pmatrix}$

Question 2.

Given vectors $\overrightarrow{a}=\begin{pmatrix}1\\2\\7\end{pmatrix},\overrightarrow{b}=\begin{pmatrix}-3\\4\\2\end{pmatrix}$ and $\overrightarrow{c}=\begin{pmatrix}-2\\p\\q\end{pmatrix}.$
(a) Find the values of $p$ and $q$ such that $\overrightarrow{c}$ is parallel to $\overrightarrow{a}.$

Short Answer $p = 2m = 2(-2) = -4$, and $q = 7m = 7(-2) = -14$

(b) Find the value of scalar $k$ such that $\overrightarrow{a}+k\overrightarrow{b}$ is parallel to vector $\begin{pmatrix}0\\10\\23\end{pmatrix}.$

Short Answer$ k = \frac{1}{3}$

Question 3.

Points $A,B,C$ and $D$ have position vectors $\overrightarrow{a}=\left(\begin{array}{c}3\\-1\\1\\\end{array}\right),\overrightarrow{b}=\left(\begin{array}{c}5\\0\\3\\\end{array}\right)$ $\overrightarrow{c}=\begin{pmatrix}7\\8\\-3\end{pmatrix}$ and $\overrightarrow{d}=\begin{pmatrix}4\\3\\-2\end{pmatrix}$, respectively. Point $E$ is the midpoint of $BC.$
(a) Find the position vector of $E.$

Short Answer$$\begin{pmatrix}6\\4\\0\end{pmatrix}$$

(b) Show that $ABED$ is a parallelogram.

Short AnswerProof

Question 4.

Points $A,B$ and $C$ have position vectors $\overrightarrow{a}=\begin{pmatrix}2\\-1\\4\end{pmatrix},\overrightarrow{b}=\begin{pmatrix}5\\1\\2\end{pmatrix}$ and $\overrightarrow{c}=\left(\begin{array}{c}{3}\\{1}\\{4}\\\end{array}\right)$,respectively. Find the position vector of point $D$ such that $ABCD$ is a paralílelogram.

Short Answer$$\begin{pmatrix}0\\-1\\6\end{pmatrix}$$

Question 5.

$K( 1, - 1, 0) , L( 4, - 3, 7)$ and $M(a,2;b)$ are collinear. Find $a$ and $b.$

Short Answer$$a=-\frac{7}{2},b= = -\frac{21}{2}$$,