Vectors & Vector Calculus: Dot/Cross Products, Planes, Derivatives, Tangents

1. Introduction to Vectors in 3D Space

1.1 Definition and Representation

A vector in 3D space is a quantity with both magnitude and direction, represented as: $$\vec{v}=a\vec{i}+b\vec{j}+c\vec{k}\text{or}\vec{v}=\left(\begin{array}{c}a\\ b\\ c\end{array}\right)$$ where $a,b,c$ are components along the $x,y,z$-axes, respectively.

Key Properties

• Magnitude (Length): $$\Vert \vec{v}\Vert =\sqrt{a^2+b^2+c^2}$$
• Unit Vector: $$\hat{v}=\frac{\vec{v}}{\Vert \vec{v}\Vert }$$
• Vector Addition/Subtraction: Component-wise: $\vec{u}\pm \vec{v}=(u_1\pm v_1)\vec{i}+(u_2\pm v_2)\vec{j}+(u_3\pm v_3)\vec{k}$.

Worked Example 1

Given $\vec{a}=2\vec{i}-\vec{j}+3\vec{k}$ and $\vec{b}=-\vec{i}+4\vec{j}-2\vec{k}$:

1. Find $\vec{a}+\vec{b}$: $$\vec{a}+\vec{b}=(2-1)\vec{i}+(-1+4)\vec{j}+(3-2)\vec{k}=\vec{i}+3\vec{j}+\vec{k}$$
2. Find $\Vert \vec{a}\Vert$: $$\Vert \vec{a}\Vert =\sqrt{2^2+(-1{)}^2+3^2}=\sqrt{14}$$

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2. Dot Product and Cross Product

2.1 Dot Product (Scalar Product)

Definition: $$\vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3=\Vert \vec{a}\Vert \Vert \vec{b}\Vert \cos \theta$$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Properties

Worked Example 2

Find the angle between $\vec{a}=(2,2,-1)$ and $\vec{b}=(1,3,2)$: $$\vec{a}\cdot \vec{b}=2(1)+2(3)+(-1)(2)=2+6-2=6$$ $$\Vert \vec{a}\Vert =\sqrt{4+4+1}=3,\Vert \vec{b}\Vert =\sqrt{1+9+4}=\sqrt{14}$$ $$\cos \theta =\frac{6}{3\sqrt{14}}=\frac{2}{\sqrt{14}}\Rightarrow \theta ={\cos }^{-1}\left(\frac{2}{\sqrt{14}}\right)$$

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2.2 Cross Product (Vector Product)

Definition: For $\vec{a}=a_1\vec{i}+a_2\vec{j}+a_3\vec{k}$ and $\vec{b}=b_1\vec{i}+b_2\vec{j}+b_3\vec{k}$, $$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\vec{i}& \vec{j}& \vec{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|=(a_2{b}_3-a_3{b}_2)\vec{i}-(a_1{b}_3-a_3{b}_1)\vec{j}+(a_1{b}_2-a_2{b}_1)\vec{k}$$

Properties

Worked Example 3

Given $\vec{a}=\vec{i}+3\vec{j}+4\vec{k}$ and $\vec{b}=2\vec{i}+7\vec{j}-5\vec{k}$, find $\vec{b}\times \vec{a}$: $$\vec{b}\times \vec{a}=\left|\begin{array}{ccc}\vec{i}& \vec{j}& \vec{k}\\ 2& 7& -5\\ 1& 3& 4\end{array}\right|=(7\cdot 4-(-5)\cdot 3)\vec{i}-(2\cdot 4-(-5)\cdot 1)\vec{j}+(2\cdot 3-7\cdot 1)\vec{k}$$ $$=(28+15)\vec{i}-(8+5)\vec{j}+(6-7)\vec{k}=43\vec{i}-13\vec{j}-\vec{k}$$ Note: $\vec{a}\times \vec{b}=-(\vec{b}\times \vec{a})=-43\vec{i}+13\vec{j}+\vec{k}$.

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3. Equations of Planes and Normal Vectors

3.1 Plane Equation from 3 Points

Given three points $P(x_1,y_1,z_1)$, $Q(x_2,y_2,z_2)$, and $R(x_3,y_3,z_3)$:

1. Find two vectors in the plane: $$\overrightarrow{PQ}=(x_2-x_1)\vec{i}+(y_2-y_1)\vec{j}+(z_2-z_1)\vec{k}$$ $$\overrightarrow{PR}=(x_3-x_1)\vec{i}+(y_3-y_1)\vec{j}+(z_3-z_1)\vec{k}$$
2. Compute the normal vector $\vec{n}=\overrightarrow{PQ}\times \overrightarrow{PR}$.
3. The plane equation is: $$n_1(x-x_1)+n_2(y-y_1)+n_3(z-z_1)=0$$

Worked Example 4

Find a vector perpendicular to the plane through $P(1,4,6)$, $Q(-2,5,-1)$, and $R(1,-1,1)$:

1. $\overrightarrow{PQ}=(-2-1)\vec{i}+(5-4)\vec{j}+(-1-6)\vec{k}=-3\vec{i}+\vec{j}-7\vec{k}$
2. $\overrightarrow{PR}=(1-1)\vec{i}+(-1-4)\vec{j}+(1-6)\vec{k}=-5\vec{j}-5\vec{k}$
3. $\vec{n}=\overrightarrow{PQ}\times \overrightarrow{PR}$: $$\vec{n}=\left|\begin{array}{ccc}\vec{i}& \vec{j}& \vec{k}\\ -3& 1& -7\\ 0& -5& -5\end{array}\right|=(1\cdot (-5)-(-7)\cdot (-5))\vec{i}-(-3\cdot (-5)-(-7)\cdot 0)\vec{j}+(-3\cdot (-5)-1\cdot 0)\vec{k}$$ $$=(-5-35)\vec{i}-(15)\vec{j}+(15)\vec{k}=-40\vec{i}-15\vec{j}+15\vec{k}$$ Simplified: $\vec{n}=8\vec{i}+3\vec{j}-3\vec{k}$ (divided by -5).

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4. Vector-Valued Functions and Derivatives

4.1 Vector-Valued Functions

A vector-valued function $\vec{r}(t)$ maps a scalar $t$ to a vector: $$\vec{r}(t)=x(t)\vec{i}+y(t)\vec{j}+z(t)\vec{k}$$ where $x(t),y(t),z(t)$ are component functions.

Derivative of $\vec{r}(t)$

The derivative ${\vec{r}}^{\prime }(t)$ is computed component-wise: $${\vec{r}}^{\prime }(t)=x^{\prime }(t)\vec{i}+y^{\prime }(t)\vec{j}+z^{\prime }(t)\vec{k}$$

Worked Example 5

Given $\vec{r}(t)=t^2\vec{i}-t{e}^{-t}\vec{j}+\sin (2t)\vec{k}$, find ${\vec{r}}^{\prime }(t)$: $${\vec{r}}^{\prime }(t)=2t\vec{i}-\left(e^{-t}-t{e}^{-t}\right)\vec{j}+2\cos (2t)\vec{k}=2t\vec{i}+(t-1)e^{-t}\vec{j}+2\cos (2t)\vec{k}$$

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4.2 Unit Tangent Vector

The unit tangent vector $\hat{T}(t)$ is: $$\hat{T}(t)=\frac{{\vec{r}}^{\prime }(t)}{\Vert {\vec{r}}^{\prime }(t)\Vert }$$

Worked Example 6

Find the unit tangent vector at $t=0$ for $\vec{r}(t)=(1+t^2)\vec{i}-t{e}^{-t}\vec{j}+\sin (2t)\vec{k}$:

1. Compute ${\vec{r}}^{\prime }(t)$: $${\vec{r}}^{\prime }(t)=2t\vec{i}-\left(e^{-t}-t{e}^{-t}\right)\vec{j}+2\cos (2t)\vec{k}$$
2. Evaluate at $t=0$: $${\vec{r}}^{\prime }(0)=0\vec{i}-(1-0)\vec{j}+2(1)\vec{k}=-\vec{j}+2\vec{k}$$
3. Compute $\Vert {\vec{r}}^{\prime }(0)\Vert$: $$\Vert {\vec{r}}^{\prime }(0)\Vert =\sqrt{0^2+(-1{)}^2+2^2}=\sqrt{5}$$
4. Unit tangent vector: $$\hat{T}(0)=\frac{-\vec{j}+2\vec{k}}{\sqrt{5}}=\frac{-1}{\sqrt{5}}\vec{j}+\frac{2}{\sqrt{5}}\vec{k}$$

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5. Applications of Vectors

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6. Common Mistakes and Exam Tips

6.1 Dot vs. Cross Product

• Dot product yields a scalar (use for angles, projections).
• Cross product yields a vector (use for normals, torque).
• Never mix them up in problems requiring perpendicularity or area.

6.2 Plane Equation Pitfalls

• Normal vector direction: $\overrightarrow{PQ}\times \overrightarrow{PR}$ gives one of two possible normals (the other is $-\vec{n}$).
• Simplify: Always simplify the normal vector (e.g., divide by GCD of components).
• Point inclusion: Ensure the plane equation satisfies all three given points.

6.3 Derivatives of Vector Functions

• Component-wise differentiation: Differentiate each $\vec{i},\vec{j},\vec{k}$ component separately.
• Chain rule: Apply carefully to composite functions (e.g., $t{e}^{-t}$).
• Unit tangent: Forgetting to normalize ${\vec{r}}^{\prime }(t)$ is a common error.

6.4 Exam Strategies

1. Show all steps: Partial credit is often given for correct intermediate steps.
2. Label vectors clearly: Use $\vec{a},\vec{b}$ or boldface $\mathbf{a}$ to avoid confusion with scalars.
3. Practice cross products: Memorize the determinant formula or use the "right-hand rule" for direction.
4. Check units: In applied problems, ensure answers have consistent units (e.g., radians for angles).
5. Time management: Allocate ~10 minutes for vector problems (they often have multiple parts).

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6.5 Past Exam Patterns

• Definition-based questions: Always define dot/cross products before solving.
• Numerical problems: Expect computations involving magnitudes, angles, or derivatives.
• Geometric interpretation: Questions may ask for physical meanings (e.g., "Find the angle between two forces").
• Vector calculus: Derivatives and unit tangents are frequently tested in combination with limits/continuity concepts from Unit 2.

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6.6 Quick Revision Checklist

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6.7 Final Worked Example (Exam-Style)

Problem: Given $\vec{a}=3\vec{i}-2\vec{j}+\vec{k}$ and $\vec{b}=\vec{i}+\vec{j}-2\vec{k}$:

1. Find the angle between $\vec{a}$ and $\vec{b}$.
2. Find a vector perpendicular to both $\vec{a}$ and $\vec{b}$.
3. Find the equation of the plane passing through the origin with normal vector $\vec{a}\times \vec{b}$.

Solution:

1. Angle: $$\vec{a}\cdot \vec{b}=3(1)+(-2)(1)+(1)(-2)=3-2-2=-1$$ $$\Vert \vec{a}\Vert =\sqrt{9+4+1}=\sqrt{14},\Vert \vec{b}\Vert =\sqrt{1+1+4}=\sqrt{6}$$ $$\cos \theta =\frac{-1}{\sqrt{14}\cdot \sqrt{6}}\Rightarrow \theta ={\cos }^{-1}\left(\frac{-1}{\sqrt{84}}\right)$$

2. Perpendicular vector: $$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\vec{i}& \vec{j}& \vec{k}\\ 3& -2& 1\\ 1& 1& -2\end{array}\right|=((-2)(-2)-(1)(1))\vec{i}-(3(-2)-1(1))\vec{j}+(3(1)-(-2)(1))\vec{k}$$ $$=(4-1)\vec{i}-(-6-1)\vec{j}+(3+2)\vec{k}=3\vec{i}+7\vec{j}+5\vec{k}$$

3. Plane equation: Using $\vec{n}=3\vec{i}+7\vec{j}+5\vec{k}$ and point $(0,0,0)$: $$3(x-0)+7(y-0)+5(z-0)=0\Rightarrow 3x+7y+5z=0$$