G12 Chapter 4: Examples

Vectors and 3D Geometry — Comprehensive Chapter Notes

📐 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: $\overrightarrow{QP} = -\overrightarrow{PQ}$.

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.

3. The Dot Product (Scalar Product)

  • Definition: The dot product of two vectors yields a scalar value and is defined as:
    \[ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\vec{a}||\vec{b}|\cos\theta \] where $\theta$ is the angle between the two vectors ($0 \le \theta \le \pi$).
  • Angle Between Vectors: The angle $\theta$ can be derived using:
    \[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \]
  • Perpendicular (Orthogonal) Vectors: Two vectors are perpendicular ($\vec{a} \perp \vec{b}$) if and only if their dot product is zero ($\vec{a} \cdot \vec{b} = 0$).
  • Algebraic Properties:
    • $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ (Commutative)
    • $\vec{a} \cdot \vec{a} = |\vec{a}|^2$

4. The Cross Product (Vector Product)

  • Definition: Unlike the dot product, the cross product results in a vector that is perpendicular to both originating vectors. It is evaluated using a $3 \times 3$ matrix determinant:
    \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
  • Magnitude and Geometric Interpretation:
    \[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \]
    • The magnitude $|\vec{a} \times \vec{b}|$ equals the area of the parallelogram formed by vectors $\vec{a}$ and $\vec{b}$.
    • The area of a triangle with adjacent sides $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is given by $\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$.
  • Algebraic Properties:
    • $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$ (Anti-commutative)
    • $\vec{a} \times \vec{a} = \vec{0}$
  • Perpendicular Unit Vector: A unit vector $\hat{n}$ orthogonal to both $\vec{a}$ and $\vec{b}$ is computed via:
    \[ \hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \]

5. Equations of Lines and Planes in 3D Space

A. Straight Lines

  • Vector Equation: A line passing through a point with position vector $\vec{a}$ and moving parallel to a direction vector $\vec{d}$ is represented as:
    \[ \vec{r} = \vec{a} + t\vec{d} \]
  • Cartesian (Symmetric) Equation: Eliminating the parameter $t$ from the parametric equations yields:
    \[ \frac{x - x_0}{d_1} = \frac{y - y_0}{d_2} = \frac{z - z_0}{d_3} \]
  • Point Verification: To determine if an arbitrary point lies on a given line, substitute its coordinates into the line's Cartesian equation and verify if the resulting ratios are equal.

B. Planes

  • Vector Equation (using direction vectors): A plane passing through a point $\vec{a}$ and containing two non-parallel direction vectors $\vec{u}$ and $\vec{v}$ is:
    \[ \vec{r} = \vec{a} + \lambda\vec{u} + \mu\vec{v} \]
  • Vector Equation (using a normal vector): A plane containing a point $\vec{a}$ with a perpendicular normal vector $\vec{n}$ is:
    \[ \vec{r} \cdot \vec{n} = d \quad \text{where } d = \vec{a} \cdot \vec{n} \]
  • Cartesian Equation: Written in the linear form:
    \[ ax + by + cz = d \] where $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ represents the components of the normal vector $\vec{n}$.
  • Coplanarity of Points: A set of four points $A, B, C,$ and $D$ are coplanar (lie in the same plane) if the point $D$ satisfies the unique Cartesian equation created by the other three points ($A, B,$ and $C$).

📌 Summary of Key Formulas

Dot Product

$\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$

$\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$

$\vec{a}\perp\vec{b} \iff \vec{a}\cdot\vec{b}=0$

Cross Product

$|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$

Area (parallelogram) $= |\vec{a}\times\vec{b}|$

Area (triangle) $= \frac{1}{2}|\vec{a}\times\vec{b}|$

Line (Vector)

$\vec{r} = \vec{a} + t\vec{d}$

Cartesian: $\dfrac{x-x_0}{d_1} = \dfrac{y-y_0}{d_2} = \dfrac{z-z_0}{d_3}$

Plane

$\vec{r} = \vec{a} + \lambda\vec{u} + \mu\vec{v}$

$\vec{r}\cdot\vec{n} = d$   (normal form)

$ax+by+cz = d$

💡 Tip: The cross product $\vec{a}\times\vec{b}$ gives a vector perpendicular to both $\vec{a}$ and $\vec{b}$. It is essential for finding normal vectors to planes and computing areas.
— Comprehensive reference for vectors, lines, and planes in 3D space —
=========================== သင်္ချာပုစ္ဆာများ၏ အသေးစိတ်တွက်ချက်ပုံအဖြေလွှာ
Example 1

Example 1

Illustrate the points (a) $A(0,3,0)$   (b) $B(4,0,2)$   (c) $C(-1,2,2)$
  • (a) $A(0,3,0)$: $y$-ဝင်ရိုး (y-axis) ပေါ်တွင် တည်ရှိပါသည်။
  • (b) $B(4,0,2)$: $xz$-ပြင်ညီ (xz-plane) ပေါ်တွင် တည်ရှိပါသည်။
  • (c) $C(-1,2,2)$: $x=-1, y=2, z=2$ နေရာတွင် ရှိပါသည်။
Example 2

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) $|\overrightarrow{QP}|$

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

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:
(a) $\vec{p} + \vec{q}$   (b) $\vec{p} - \frac{1}{2}\vec{q}$   (c) $\frac{3}{2}\vec{q} - \vec{p}$

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

Example 4

Example 4

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

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 $v = 2$, $u = 1$.

$u = 1,\; v = 2$

Example 5

Example 5

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

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

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

$C(6,0,1)$

$\overrightarrow{AB}$ $\overrightarrow{DC}$ A B D C
Example 6

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 $\vec{a}$.

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

Example 7

Example 7

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

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

Thus $A,B,C$ are collinear.

A B C $\overrightarrow{AB}$ $\overrightarrow{BC}=2\overrightarrow{AB}$
Example 8

Example 8

Angle between $\begin{pmatrix}3\\4\end{pmatrix}$ and $\begin{pmatrix}5\\-12\end{pmatrix}$.

\[ \cos\theta = \frac{-33}{5\cdot 13} = -\frac{33}{65} \implies \theta \approx 120.5^\circ \]

Example 9

Example 9

Given $P(1,0,-1), Q(2,4,1), R(3,5,6)$, find $\angle QPR$.

\[ \overrightarrow{PQ} = \begin{pmatrix}1\\4\\2\end{pmatrix}, \quad \overrightarrow{PR} = \begin{pmatrix}2\\5\\7\end{pmatrix} \]

\[ \cos\theta = \frac{36}{\sqrt{21}\sqrt{78}} \implies \theta \approx 27.2^\circ \]

$\overrightarrow{PQ}$ $\overrightarrow{PR}$ P Q R $\theta$
Example 10

Example 10

$\vec{a}\perp\vec{b}$, $|\vec{a}|=3$, $|\vec{b}|=1$. Evaluate $(\vec{a}-\vec{b})\cdot(\vec{a}+5\vec{b})$.

\[ (\vec{a}-\vec{b})\cdot(\vec{a}+5\vec{b}) = |\vec{a}|^2 - 5|\vec{b}|^2 = 9 - 5 = 4 \]

$4$

Example 11

Example 11

$\vec{a} = k\begin{pmatrix}2\\-1\\1\end{pmatrix}, \vec{b}=\begin{pmatrix}3\\2\\-2\end{pmatrix}, \vec{c}=\begin{pmatrix}1\\1\\4\end{pmatrix}$.
(a) $\overrightarrow{BC}$   (b) $\overrightarrow{AB}$ in terms of $k$   (c) $k$ such that $\overrightarrow{AB}\perp\overrightarrow{BC}$

(a) $\overrightarrow{BC} = \begin{pmatrix}-2\\-1\\6\end{pmatrix}$

(b) $\overrightarrow{AB} = \begin{pmatrix}3-2k\\2+k\\-2-k\end{pmatrix}$

(c) $\overrightarrow{AB}\cdot\overrightarrow{BC}=0 \implies k=-\frac{20}{3}$

Example 12

Example 12

Area of parallelogram determined by $\vec{a}=\begin{pmatrix}1\\2\\3\end{pmatrix}$, $\vec{b}=\begin{pmatrix}1\\4\\-1\end{pmatrix}$.

\[ \vec{a}\times\vec{b} = \begin{pmatrix}-14\\4\\2\end{pmatrix},\quad |\vec{a}\times\vec{b}| = \sqrt{216} = 6\sqrt{6} \]

$6\sqrt{6}$ sq. units

Example 13

Example 13

Area of triangle $ABC$: $A(1,-1,3), B(0,4,1), C(2,7,2)$.

\[ \overrightarrow{AB} = \begin{pmatrix}-1\\5\\-2\end{pmatrix},\quad \overrightarrow{AC} = \begin{pmatrix}1\\8\\-1\end{pmatrix} \]

\[ \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}11\\-3\\-13\end{pmatrix},\quad \text{Area} = \frac{1}{2}\sqrt{299} \]

Example 14

Example 14

(a) $\vec{a}\times\vec{b}$ for $\vec{a}=3\hat{i}+2\hat{j}+5\hat{k}$, $\vec{b}=\hat{i}-4\hat{j}+2\hat{k}$.
(b) Unit vector $\hat{n}$ perpendicular to both.

(a) $\vec{a}\times\vec{b} = \begin{pmatrix}24\\-1\\-14\end{pmatrix}$

(b) $\hat{n} = \pm \frac{1}{\sqrt{773}}\begin{pmatrix}24\\-1\\-14\end{pmatrix}$

Example 15

Example 15

$|\vec{a}|=4, |\vec{b}|=5, \vec{a}\perp\vec{b}$. Evaluate $|(2\vec{a}-\vec{b})\times(\vec{a}+3\vec{b})|$.

\[ (2\vec{a}-\vec{b})\times(\vec{a}+3\vec{b}) = 7(\vec{a}\times\vec{b}) \]

\[ |\vec{a}\times\vec{b}| = 20 \implies 7\cdot20 = 140 \]

Example 16

Example 16

Cartesian equation of line $\vec{r} = \begin{pmatrix}1\\4\\-1\end{pmatrix} + t\begin{pmatrix}3\\2\\5\end{pmatrix}$.

\[ \frac{x-1}{3} = \frac{y-4}{2} = \frac{z+1}{5} \]

Example 17

Example 17

Does $A(3,-2,2)$ lie on $\frac{x+1}{2} = \frac{4-y}{3} = \frac{2z}{3}$?

\[ \frac{3+1}{2}=2,\quad \frac{4-(-2)}{3}=2,\quad \frac{2(2)}{3}=\frac{4}{3} \]

$2=2\neq \frac{4}{3}$ so no.

Example 18

Example 18

Vector equation of plane through $M(2,2,-2), N(1,-1,3), P(4,0,2)$.

\[ \vec{r} = \begin{pmatrix}2\\2\\-2\end{pmatrix} + t_1\begin{pmatrix}-1\\-3\\5\end{pmatrix} + t_2\begin{pmatrix}2\\-2\\4\end{pmatrix} \]

Plane M N P $\vec{d}_1$ $\vec{d}_2$
Example 19

Example 19

Cartesian equation of same plane.

Normal vector $\vec{n} = \overrightarrow{MN}\times\overrightarrow{MP} = \begin{pmatrix}-2\\14\\8\end{pmatrix} \sim \begin{pmatrix}1\\-7\\-4\end{pmatrix}$.

Using $M(2,2,-2)$: $1(2) - 7(2) - 4(-2) = -4$.

\[ x - 7y - 4z = -4 \]

Example 20

Example 20

Are $A(3,-1,4), B(2,1,1), C(4,3,1), D(-3,1,4)$ coplanar?

Plane through $A,B,C$: $\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}6\\-6\\-6\end{pmatrix} \sim \begin{pmatrix}1\\-1\\-1\end{pmatrix}$.

Equation: $x - y - z = 0$. For $D$: $-3 -1 -4 = -8 \neq 0$.

Not coplanar

Example 21

Example 21

Vector equation of plane containing line $\vec{r} = \begin{pmatrix}-2\\1\\2\end{pmatrix} + t\begin{pmatrix}-1\\1\\1\end{pmatrix}$ and point $A(3,-1,2)$.

\[ \vec{r} = \begin{pmatrix}3\\-1\\2\end{pmatrix} + t_1\begin{pmatrix}-1\\1\\1\end{pmatrix} + t_2\begin{pmatrix}-5\\2\\0\end{pmatrix} \]

Plane A B $\vec{d}_2$ $\vec{d}_1$
Example 22

Example 22

$\vec{n} = \begin{pmatrix}2\\4\\-2\end{pmatrix}$ perpendicular to plane through $A(1,-5,2)$.
(a) $\vec{r}\cdot\vec{n}=d$   (b) Cartesian equation.

(a) $d = \vec{a}\cdot\vec{n} = 2 -20 -4 = -22$

\[ \vec{r}\cdot\begin{pmatrix}2\\4\\-2\end{pmatrix} = -22 \]

(b) $2x + 4y - 2z = -22 \implies x + 2y - z = -11$

Plane $\vec{n}$ A
============================= သင်္ချာပုစ္ဆာများ၏ အသေးစိတ်တွက်ချက်ပုံအဖြေလွှာ -----------------
Example 1

Example 1

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

အသေးစိတ်ရှင်းလင်းချက်နှင့် အဖြေ (Detailed Solution)

သုံးဖက်မြင်သြဒီနိတ်စနစ် (3D Coordinate System) တွင် အမှတ်တစ်ခုအား $(x, y, z)$ ပုံစံဖြင့် ပြပါသည်။

  • (a) $A(0,3,0)$: ဤအမှတ်သည် $x=0$, $y=3$, $z=0$ ဖြစ်သောကြောင့် $y$-ဝင်ရိုး (y-axis) ပေါ်တွင် တည်ရှိပါသည်။
  • (b) $B(4,0,2)$: ဤအမှတ်သည် $x=4$, $y=0$, $z=2$ ဖြစ်သောကြောင့် $xz$-ပြင်ညီ (xz-plane) ပေါ်တွင် တည်ရှိပါသည်။
  • (c) $C(-1,2,2)$: ဤအမှတ်သည် $x=-1$, $y=2$, $z=2$ နေရာတွင် ရှိပါသည်။
x y z O A(0,3,0) B(4,0,2) C(-1,2,2)
Example 2

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) $|\overrightarrow{QP}|$

အသေးစိတ်ရှင်းလင်းချက်နှင့် အဖြေ

အမှတ် $P(-3, 1, 2)$ နှင့် $Q(1, -1, 3)$ တို့အတွက် Position Vectors များမှာ $\overrightarrow{OP} = \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix}$ နှင့် $\overrightarrow{OQ} = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}$ ဖြစ်သည်။

(a) $\overrightarrow{OP}$: \[ \overrightarrow{OP} = -3\vec{i} + \vec{j} + 2\vec{k} = \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix} \]

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

(c) $|\overrightarrow{PQ}|$: \[ |\overrightarrow{PQ}| = \sqrt{4^2 + (-2)^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \]

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

(e) $|\overrightarrow{QP}|$: \[ |\overrightarrow{QP}| = |\overrightarrow{PQ}| = \sqrt{21} \]

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:
(a) $\vec{p} + \vec{q}$   (b) $\vec{p} - \frac{1}{2}\vec{q}$   (c) $\frac{3}{2}\vec{q} - \vec{p}$.

အသေးစိတ်ရှင်းလင်းချက်နှင့် အဖြေ

(a) $\vec{p} + \vec{q}$: \[ \vec{p} + \vec{q} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix} + \begin{pmatrix} -2 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \\ 6 \end{pmatrix} \]

(b) $\vec{p} - \frac{1}{2}\vec{q}$: \[ \vec{p} - \frac{1}{2}\vec{q} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix} - \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \]

(c) $\frac{3}{2}\vec{q} - \vec{p}$: \[ \frac{3}{2}\vec{q} - \vec{p} = \begin{pmatrix} -3 \\ 0 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \\ -1 \end{pmatrix} \]

Example 4

Example 4

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

အသေးစိတ်ရှင်းလင်းချက်နှင့် အဖြေ

ဗက်တာနှစ်ခု ပြိုင်ကြလျှင် ($\vec{a} \parallel \vec{b}$) တစ်ခုသည် အခြားတစ်ခု၏ ကိန်းသေမြှောက်လဒ် (scalar multiple) ဖြစ်ရပါမည်။

\[ \frac{v}{-1} = \frac{2}{-1} = \frac{-2}{u} \]

အလယ်အချိုး $\frac{2}{-1} = -2$ ဖြစ်သောကြောင့်:

  1. $\frac{v}{-1} = -2 \implies v = 2$
  2. $\frac{-2}{u} = -2 \implies u = 1$

အဖြေ: $u = 1$, $v = 2$

Example 5

Example 5

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

အသေးစိတ်ရှင်းလင်းချက်နှင့် အဖြေ

အပြိုင်စတုဂံ (Parallelogram) $ABCD$ တွင် $\overrightarrow{AB} = \overrightarrow{DC}$ ဖြစ်သည်။

$\overrightarrow{AB}$ $\overrightarrow{DC}$ A(-1,1,1) B(2,0,-2) D(3,1,4) C(x,y,z)

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

$\overrightarrow{AB} = \overrightarrow{DC}$ ဖြစ်သောကြောင့်:

\[ x - 3 = 3 \implies x = 6,\quad y - 1 = -1 \implies y = 0,\quad z - 4 = -3 \implies z = 1 \]

အဖြေ: $C(6, 0, 1)$

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 $\vec{a}$.

အသေးစိတ်ရှင်းလင်းချက်နှင့် အဖြေ

(a) ယူနစ်ဗက်တာ: \[ |\vec{a}| = \sqrt{4+4+1} = 3,\quad \hat{a} = \frac{1}{3}\begin{pmatrix}2\\-2\\1\end{pmatrix} = \begin{pmatrix}\frac{2}{3}\\-\frac{2}{3}\\\frac{1}{3}\end{pmatrix} \]

(b) ပမာဏ 5 ရှိပြီး $\vec{a}$ နှင့် ပြိုင်သော ဗက်တာ: \[ \vec{v} = \pm 5\hat{a} = \pm \begin{pmatrix}\frac{10}{3}\\-\frac{10}{3}\\\frac{5}{3}\end{pmatrix} \]

$\vec{a}$ (mag=3) $\hat{a}$ (mag=1) $5\hat{a}$ (mag=5)
Example 7

Example 7

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

အသေးစိတ်ရှင်းလင်းချက်နှင့် အဖြေ

\[ \overrightarrow{AB} = \begin{pmatrix}4\\1\\1\end{pmatrix},\quad \overrightarrow{AC} = \begin{pmatrix}12\\3\\3\end{pmatrix} = 3\overrightarrow{AB} \]

$\overrightarrow{AC} = 3\overrightarrow{AB}$ ဖြစ်သောကြောင့် $A, B, C$ တို့သည် မျဉ်းဖြောင့်တစ်ကြောင်းတည်းပေါ်တွင် တည်ရှိသည်။

A B C $\overrightarrow{AB}$ $\overrightarrow{AC}=3\overrightarrow{AB}$
Example 8

Example 8

Find the angle between the two vectors $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\begin{pmatrix} 5 \\ -12 \end{pmatrix}$.

\[ \cos\theta = \frac{(3)(5)+(4)(-12)}{\sqrt{9+16}\sqrt{25+144}} = \frac{-33}{65} \implies \theta \approx 120.51^\circ \]

x y $\vec{u}$ $\vec{v}$ $\theta$
Example 9

Example 9

Given points $P(1,0,-1)$, $Q(2,4,1)$ and $R(3,5,6)$, find $\angle QPR$.

\[ \overrightarrow{PQ} = \begin{pmatrix}1\\4\\2\end{pmatrix},\quad \overrightarrow{PR} = \begin{pmatrix}2\\5\\7\end{pmatrix} \]

\[ \cos(\angle QPR) = \frac{36}{\sqrt{21}\sqrt{78}} \approx 0.8895 \implies \angle QPR \approx 27.19^\circ \]

Example 10

Example 10

Given that $\vec{a}$ and $\vec{b}$ are perpendicular vectors such that $|\vec{a}| = 3$ and $|\vec{b}| = 1$, evaluate $(\vec{a} - \vec{b}) \cdot (\vec{a} + 5\vec{b})$.

\[ (\vec{a} - \vec{b}) \cdot (\vec{a} + 5\vec{b}) = |\vec{a}|^2 + 4(\vec{a}\cdot\vec{b}) - 5|\vec{b}|^2 = 9 + 0 - 5 = 4 \]

တန်ဖိုးမှာ $4$

Example 11

Example 11

Points $A, B, C$ have position vectors $\vec{a} = k\begin{pmatrix}2\\-1\\1\end{pmatrix}$, $\vec{b} = \begin{pmatrix}3\\2\\-2\end{pmatrix}$, $\vec{c} = \begin{pmatrix}1\\1\\4\end{pmatrix}$.
(a) Find $\overrightarrow{BC}$. (b) Find $\overrightarrow{AB}$ in terms of $k$. (c) Find $k$ for which $\overrightarrow{AB} \perp \overrightarrow{BC}$.

(a) $\overrightarrow{BC} = \vec{c} - \vec{b} = \begin{pmatrix}-2\\-1\\6\end{pmatrix}$

(b) $\overrightarrow{AB} = \vec{b} - \vec{a} = \begin{pmatrix}3-2k\\2+k\\-2-k\end{pmatrix}$

(c) $\overrightarrow{AB}\cdot\overrightarrow{BC}=0 \implies -2(3-2k) - (2+k) + 6(-2-k)=0 \implies -3k-20=0 \implies k = -\frac{20}{3}$

$\overrightarrow{AB}$ $\overrightarrow{BC}$
Example 12

Example 12

Find the area of the parallelogram determined by $\vec{a} = \begin{pmatrix}1\\2\\3\end{pmatrix}$ and $\vec{b} = \begin{pmatrix}1\\4\\-1\end{pmatrix}$.

\[ \vec{a}\times\vec{b} = \begin{pmatrix}-14\\4\\2\end{pmatrix},\quad \text{Area} = \sqrt{196+16+4} = \sqrt{216} = 6\sqrt{6} \text{ sq. units} \]

$\vec{a}$ $\vec{b}$ Area = $6\sqrt{6}$
Example 13

Example 13

Find the area of triangle $ABC$ with vertices $A(1,-1,3)$, $B(0,4,1)$, $C(2,7,2)$.

\[ \overrightarrow{AB} = \begin{pmatrix}-1\\5\\-2\end{pmatrix},\quad \overrightarrow{AC} = \begin{pmatrix}1\\8\\-1\end{pmatrix} \]

\[ \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}11\\-3\\-13\end{pmatrix},\quad \text{Area} = \frac{1}{2}\sqrt{121+9+169} = \frac{\sqrt{299}}{2} \]

Example 14

Example 14

(a) Calculate $\vec{a} \times \vec{b}$ when $\vec{a} = 3\hat{i} + 2\hat{j} + 5\hat{k}$ and $\vec{b} = \hat{i} - 4\hat{j} + 2\hat{k}$.
(b) Find a unit vector $\hat{n}$ perpendicular to both.

(a) $\vec{a}\times\vec{b} = \begin{pmatrix}24\\-1\\-14\end{pmatrix}$

(b) $\hat{n} = \pm \frac{1}{\sqrt{773}}\begin{pmatrix}24\\-1\\-14\end{pmatrix}$

Example 15

Example 15

Given $|\vec{a}|=4$, $|\vec{b}|=5$, $\vec{a}\perp\vec{b}$, evaluate $|(2\vec{a}-\vec{b})\times(\vec{a}+3\vec{b})|$.

\[ (2\vec{a}-\vec{b})\times(\vec{a}+3\vec{b}) = 7(\vec{a}\times\vec{b}) \]

\[ |\vec{a}\times\vec{b}| = 4\cdot5\cdot1 = 20 \implies |7(\vec{a}\times\vec{b})| = 140 \]

Example 16

Example 16

Find the Cartesian equation of the line $\vec{r} = \begin{pmatrix}1\\4\\-1\end{pmatrix} + t\begin{pmatrix}3\\2\\5\end{pmatrix}$.

\[ x=1+3t,\; y=4+2t,\; z=-1+5t \implies \frac{x-1}{3} = \frac{y-4}{2} = \frac{z+1}{5} \]

Example 17

Example 17

Does $A(3,-2,2)$ lie on the line $\frac{x+1}{2} = \frac{4-y}{3} = \frac{2z}{3}$?

\[ \frac{3+1}{2}=2,\quad \frac{4-(-2)}{3}=2,\quad \frac{2(2)}{3}=\frac{4}{3} \]

$2 = 2 \neq \frac{4}{3}$ ဖြစ်သဖြင့် မတည်ရှိပါ။

Example 18

Example 18

Find vector equation of plane through $M(2,2,-2)$, $N(1,-1,3)$, $P(4,0,2)$.

\[ \vec{r} = \begin{pmatrix}2\\2\\-2\end{pmatrix} + \lambda\begin{pmatrix}-1\\-3\\5\end{pmatrix} + \mu\begin{pmatrix}2\\-2\\4\end{pmatrix} \]

Example 19

Example 19

Find Cartesian equation of the same plane.

\[ \vec{n} = \overrightarrow{MN}\times\overrightarrow{MP} = \begin{pmatrix}-2\\14\\8\end{pmatrix} \sim \begin{pmatrix}1\\-7\\-4\end{pmatrix} \]

\[ 1x - 7y - 4z = d,\quad \text{using } M(2,2,-2): 2 -14 +8 = -4 \]

\[ x - 7y - 4z = -4 \]

Example 20

Example 20

Determine whether $A(3,-1,4), B(2,1,1), C(4,3,1), D(-3,1,4)$ are coplanar.

\[ \overrightarrow{AB} = \begin{pmatrix}-1\\2\\-3\end{pmatrix}, \overrightarrow{AC} = \begin{pmatrix}1\\4\\-3\end{pmatrix} \]

\[ \vec{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}6\\-6\\-6\end{pmatrix} \sim \begin{pmatrix}1\\-1\\-1\end{pmatrix} \]

Plane ABC: $x - y - z = 0$. For D: $-3 -1 -4 = -8 \neq 0$ so not coplanar.

Example 21

Example 21

Find vector equation of plane containing line $\vec{r} = \begin{pmatrix}-2\\1\\2\end{pmatrix} + t\begin{pmatrix}-1\\1\\1\end{pmatrix}$ and point $A(3,-1,2)$.

\[ \vec{r} = \begin{pmatrix}-2\\1\\2\end{pmatrix} + t\begin{pmatrix}-1\\1\\1\end{pmatrix} + s\begin{pmatrix}5\\-2\\0\end{pmatrix} \]

Example 22

Example 22

$\vec{n} = \begin{pmatrix}2\\4\\-2\end{pmatrix}$ is perpendicular to the plane through $A(1,-5,2)$.
(a) Write equation $\vec{r}\cdot\vec{n}=d$. (b) Find Cartesian equation.

(a) $d = \vec{a}\cdot\vec{n} = 1(2) + (-5)(4) + 2(-2) = -22$

\[ \vec{r}\cdot\begin{pmatrix}2\\4\\-2\end{pmatrix} = -22 \]

(b) $2x + 4y - 2z = -22 \implies x + 2y - z = -11$

Plane $\vec{n}$ A

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