📐 Vectors & 3D Geometry
Foundations & Applications
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: $\overrightarrow{QP} = -\overrightarrow{PQ}$.
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
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Example 2
If $P$ is $(-3, 1, 2)$ and $Q$ is $(1, -1, 3)$, find:
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Solution
(a) $\overrightarrow{OP} = \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix}$
(b) $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix} - \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 + 3 \\ -1 - 1 \\ 3 - 2 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}.$
(c) $|\overrightarrow{PQ}| = \sqrt{4^2 + (-2)^2 + 1^2} = \sqrt{21}.$
(d) $\overrightarrow{QP} = \overrightarrow{OP} - \overrightarrow{OQ} = \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 - 1 \\ 1 + 1 \\ 2 - 3 \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \\ -1 \end{pmatrix}.$
(e) $|\overrightarrow{QP}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{21}.$
Example 3
Example 3
If $\vec{p} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}$ and $\vec{q} = \begin{pmatrix} -2 \\ 0 \\ 2 \end{pmatrix}$, find:
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Solution
(a) $\vec{p} + \vec{q} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix} + \begin{pmatrix} -2 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 - 2 \\ -1 + 0 \\ 4 + 2 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \\ 6 \end{pmatrix}.$
(b) $\vec{p} - \frac{1}{2}\vec{q} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix} - \frac{1}{2}\begin{pmatrix} -2 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 + 1 \\ -1 - 0 \\ 4 - 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}.$
(c) $\frac{3}{2}\vec{q} - \vec{p} = \frac{3}{2}\begin{pmatrix} -2 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 - 1 \\ 0 + 1 \\ 3 - 4 \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \\ -1 \end{pmatrix}.$
Example 4
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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Solution
Since $\vec{a}$ and $\vec{b}$ are parallel, we have $\vec{a} = k \vec{b}$ for some scalar $k$.
We have $k = -\frac{1}{2}.$
Thus $u = 1 \quad \text{and} \quad v = 2.$
Example 5
Example 5
$ABCD$ is a parallelogram. $\quad$ $A$ is $(-1, 1, 1)$, $B$ is $(2, 0, -2)$, and $D$ is $(3, 1, 4)$.
Find the coordinates of $C$.
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Solution
Let $C$ be $(x, y, z)$.
Since $ABCD$ is a parallelogram, $AB \parallel DC$, and they have the same length, so
[Schematic diagram of parallelogram $ABCD$ with vertices labelled sequentially counterclockwise from top-left, showing directed vector pairs $\overrightarrow{AB}$ and $\overrightarrow{DC}$.]
Therefore $C$ is $(6, 0, 1)$.
Example 6
Example 6
(a) Find the unit vector in the same direction as $\vec{a} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$.
(b) Find a vector of magnitude 5 that is parallel to $\vec{a}$.
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Solution
(a) $|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = 3.$
The unit vector of $\vec{a}$ is
(b) Let $\vec{b}$ be parallel to $\vec{a}$ and $|\vec{b}| = 5$.
Then $\vec{b} = 5\hat{a} = \begin{pmatrix} \frac{10}{3} \\ -\frac{10}{3} \\ \frac{5}{3} \end{pmatrix} \qquad \text{or} \qquad \vec{b} = -5\hat{a} = \begin{pmatrix} -\frac{10}{3} \\ \frac{10}{3} \\ -\frac{5}{3} \end{pmatrix}.$
Example 7
Example 7
Prove that $A(8,2,2)$, $C(20,5,5)$ and $B(12,3,3)$ are collinear.
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Solution
Therefore $BC$ is parallel to $AB$.
Since $B$ is common to both, it follows that $A, B$ and $C$ are collinear.
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$\overrightarrow{a}$
(b) $4\overrightarrow{b}$
(c) $\begin{array}{c}\overrightarrow{a}-\overrightarrow{b}\end{array}$
(d) $\begin{array}{c}\overrightarrow{b}+\overrightarrow{c}\end{array}$
(e) 2$\overrightarrow{b} + \overrightarrow{c}$
(f) $\overrightarrow{a} - 2\overrightarrow{b}$
(g) $\overrightarrow{a} + \overrightarrow{b} - 2\overrightarrow{c}$
(h) 3$\overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c}$
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 $\overrightarrow{a}.$
(b) Find the value of scalar $k$ such that $\overrightarrow{a}+k\overrightarrow{b}$ is parallel to vector $\begin{pmatrix}0\\10\\23\end{pmatrix}.$
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 $E.$
(b) Show that $ABED$ is a parallelogram.
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