Dot Product, Cross Product, Scalar Triple Product

TAKEAWAYS:

• The dot product of two vectors gives a scalar (real number) and measures how much one vector points in the direction of another.
• The cross product of two vectors gives a vector perpendicular to both and is used to find areas and torques.
• The scalar triple product combines three vectors to give a scalar and is useful in calculating volumes of parallelepipeds.
• Dot and cross products have unique algebraic formulas involving components and trigonometric functions.
• These products are essential in physics (work, torque, magnetic fields) and engineering (forces, rotations).

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What is the Product of Vectors?

Vectors are quantities with both magnitude and direction. The product of vectors refers to operations that combine two or more vectors to produce a scalar (dot product) or another vector (cross product). These products help us solve problems in physics, engineering, and computer graphics.

Types of Vector Products

There are three main types of vector products:

1. Dot Product (Scalar Product)
2. Cross Product (Vector Product)
3. Scalar Triple Product

We will study the first two in detail and briefly discuss the third.

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1. Dot Product (Scalar Product)

The dot product of two vectors A and B is a scalar (a real number). It is calculated using the formula:

$$\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}||\mathbf{B}|\cos \theta$$

where:

• $|\mathbf{A}|$ and $|\mathbf{B}|$ are the magnitudes of vectors A and B.
• $\theta$ is the angle between the two vectors.

Geometric Interpretation

The dot product measures how much one vector points in the direction of another. If the angle between them is:

• 0° (parallel): $\cos 0°=1$, so $\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}||\mathbf{B}|$.
• 90° (perpendicular): $\cos 90°=0$, so $\mathbf{A}\cdot \mathbf{B}=0$.
• 180° (anti-parallel): $\cos 180°=-1$, so $\mathbf{A}\cdot \mathbf{B}=-|\mathbf{A}||\mathbf{B}|$.

Algebraic Formula (Component Form)

If two vectors are given in component form: $$\mathbf{A}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$$ $$\mathbf{B}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$$ then the dot product is: $$\mathbf{A}\cdot \mathbf{B}=a_1{b}_1+a_2{b}_2+a_3{b}_3$$

Properties of Dot Product

1. Commutative: $\mathbf{A}\cdot \mathbf{B}=\mathbf{B}\cdot \mathbf{A}$
2. Distributive: $\mathbf{A}\cdot (\mathbf{B}+\mathbf{C})=\mathbf{A}\cdot \mathbf{B}+\mathbf{A}\cdot \mathbf{C}$
3. Dot product with itself: $\mathbf{A}\cdot \mathbf{A}=|\mathbf{A}{|}^2$

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Solved Example 1: Dot Product Using Magnitude and Angle

Problem: Find the dot product of vectors A and B, where $|\mathbf{A}|=5$, $|\mathbf{B}|=3$, and the angle between them is $60°$.

Solution: $$\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}||\mathbf{B}|\cos \theta =5\times 3\times \cos 60°=15\times 0.5=7.5$$

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Solved Example 2: Dot Product Using Components

Problem: Given vectors $\mathbf{A}=2\mathbf{i}+3\mathbf{j}-\mathbf{k}$ and $\mathbf{B}=-\mathbf{i}+4\mathbf{j}+2\mathbf{k}$, find $\mathbf{A}\cdot \mathbf{B}$.

Solution: $$\mathbf{A}\cdot \mathbf{B}=(2)(-1)+(3)(4)+(-1)(2)=-2+12-2=8$$

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2. Cross Product (Vector Product)

The cross product of two vectors A and B is a vector that is perpendicular to both A and B. Its magnitude is given by: $$|\mathbf{A}\times \mathbf{B}|=|\mathbf{A}||\mathbf{B}|\sin \theta$$ where $\theta$ is the angle between A and B.

Geometric Interpretation

The cross product gives a vector whose:

• Direction is perpendicular to the plane containing A and B (found using the right-hand rule).
• Magnitude is equal to the area of the parallelogram formed by A and B.

Right-Hand Rule

To find the direction of $\mathbf{A}\times \mathbf{B}$:

1. Point your index finger in the direction of A.
2. Point your middle finger in the direction of B.
3. Your thumb points in the direction of $\mathbf{A}\times \mathbf{B}$.

Algebraic Formula (Component Form)

If: $$\mathbf{A}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$$ $$\mathbf{B}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$$ then the cross product is: $$\mathbf{A}\times \mathbf{B}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|=(a_2{b}_3-a_3{b}_2)\mathbf{i}-(a_1{b}_3-a_3{b}_1)\mathbf{j}+(a_1{b}_2-a_2{b}_1)\mathbf{k}$$

Properties of Cross Product

1. Anti-commutative: $\mathbf{A}\times \mathbf{B}=-(\mathbf{B}\times \mathbf{A})$
2. Distributive: $\mathbf{A}\times (\mathbf{B}+\mathbf{C})=\mathbf{A}\times \mathbf{B}+\mathbf{A}\times \mathbf{C}$
3. Cross product with itself: $\mathbf{A}\times \mathbf{A}=0$ (zero vector)
4. Perpendicularity: If $\mathbf{A}\times \mathbf{B}=0$, then A and B are parallel.

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Solved Example 3: Cross Product Using Magnitude and Angle

Problem: Find the magnitude of the cross product of vectors A and B, where $|\mathbf{A}|=4$, $|\mathbf{B}|=6$, and the angle between them is $30°$.

Solution: $$|\mathbf{A}\times \mathbf{B}|=|\mathbf{A}||\mathbf{B}|\sin \theta =4\times 6\times \sin 30°=24\times 0.5=12$$

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Solved Example 4: Cross Product Using Components

Problem: Given vectors $\mathbf{A}=2\mathbf{i}+3\mathbf{j}-\mathbf{k}$ and $\mathbf{B}=-\mathbf{i}+4\mathbf{j}+2\mathbf{k}$, find $\mathbf{A}\times \mathbf{B}$.

Solution: $$\mathbf{A}\times \mathbf{B}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 3& -1\\ -1& 4& 2\end{array}\right|$$ $$=\mathbf{i}(3\times 2-(-1)\times 4)-\mathbf{j}(2\times 2-(-1)\times (-1))+\mathbf{k}(2\times 4-3\times (-1))$$ $$=\mathbf{i}(6+4)-\mathbf{j}(4-1)+\mathbf{k}(8+3)$$ $$=10\mathbf{i}-3\mathbf{j}+11\mathbf{k}$$

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3. Scalar Triple Product

The scalar triple product of three vectors A, B, and C is given by: $$\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})$$ It represents the volume of the parallelepiped formed by the three vectors.

Geometric Interpretation

The scalar triple product gives the signed volume of the parallelepiped. If the volume is zero, the three vectors are coplanar (lie in the same plane).

Algebraic Formula

$$\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$$

Properties

1. Cyclic Permutation: $\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=\mathbf{B}\cdot (\mathbf{C}\times \mathbf{A})=\mathbf{C}\cdot (\mathbf{A}\times \mathbf{B})$
2. Zero if Coplanar: If $\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=0$, the vectors are coplanar.

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Solved Example 5: Scalar Triple Product

Problem: Given vectors $\mathbf{A}=\mathbf{i}+2\mathbf{j}+3\mathbf{k}$, $\mathbf{B}=-\mathbf{i}+\mathbf{j}+\mathbf{k}$, and $\mathbf{C}=2\mathbf{i}-\mathbf{j}+\mathbf{k}$, find $\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})$.

Solution: First, find $\mathbf{B}\times \mathbf{C}$: $$\mathbf{B}\times \mathbf{C}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -1& 1& 1\\ 2& -1& 1\end{array}\right|=\mathbf{i}(1\times 1-1\times (-1))-\mathbf{j}(-1\times 1-1\times 2)+\mathbf{k}(-1\times (-1)-1\times 2)$$ $$=\mathbf{i}(1+1)-\mathbf{j}(-1-2)+\mathbf{k}(1-2)=2\mathbf{i}+3\mathbf{j}-\mathbf{k}$$

Now, take the dot product with A: $$\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=(1)(2)+(2)(3)+(3)(-1)=2+6-3=5$$

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Comparison Table: Dot Product vs. Cross Product

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Applications of Vector Products

1. Physics:

   • Dot Product: Calculating work done ($W=\mathbf{F}\cdot \mathbf{d}$), where $\mathbf{F}$ is force and $\mathbf{d}$ is displacement.
   • Cross Product: Calculating torque ($\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$), where $\mathbf{r}$ is the position vector and $\mathbf{F}$ is force.

2. Engineering:

   • Dot Product: Projections in computer graphics.
   • Cross Product: Finding normal vectors to surfaces in 3D modeling.

3. Computer Graphics:

   • Used in lighting calculations and 3D rotations.

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Exam Tip

1. Memorize Formulas:

   • Dot product: $|\mathbf{A}||\mathbf{B}|\cos \theta$ and $a_1{b}_1+a_2{b}_2+a_3{b}_3$.
   • Cross product: Determinant formula and right-hand rule.
   • Scalar triple product: Volume of parallelepiped and determinant formula.

2. Practice Component Calculations:

   • Always write vectors in component form before applying formulas.
   • Double-check signs in cross product calculations (especially the negative sign for the j component).

3. Understand Geometric Interpretations:

   • Dot product = projection, Cross product = area/perpendicular vector.
   • Use diagrams to visualize angles and directions.

4. Common Mistakes to Avoid:

   • Forgetting the right-hand rule for cross product direction.
   • Mixing up dot and cross product formulas.
   • Ignoring the angle between vectors in magnitude calculations.

5. NEB Exam Questions:

   • Expect problems involving:
     • Finding dot/cross products given magnitudes and angles.
     • Calculating work, torque, or projections.
     • Proving vectors are parallel/perpendicular using dot/cross products.
     • Finding volumes using scalar triple products.

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NEB Board-Style Questions

Short Answer Questions

1. Define the dot product of two vectors. Give one property of the dot product.
2. What is the geometric interpretation of the cross product? How is its direction determined?
3. If $\mathbf{A}\cdot \mathbf{B}=0$, what does this tell you about vectors A and B?
4. Write the algebraic formula for the cross product of two vectors in 3D.
5. What is the scalar triple product? How is it related to the volume of a parallelepiped?

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Long Answer Questions

6. Given vectors $\mathbf{A}=3\mathbf{i}-2\mathbf{j}+\mathbf{k}$ and $\mathbf{B}=-\mathbf{i}+4\mathbf{j}-2\mathbf{k}$: a) Find $\mathbf{A}\cdot \mathbf{B}$. b) Find $\mathbf{A}\times \mathbf{B}$. c) Are the vectors parallel? Justify your answer.

7. The vectors $\mathbf{P}=2\mathbf{i}+\mathbf{j}-\mathbf{k}$, $\mathbf{Q}=-\mathbf{i}+3\mathbf{j}+\mathbf{k}$, and $\mathbf{R}=\mathbf{i}-\mathbf{j}+2\mathbf{k}$ are given. a) Find $\mathbf{P}\cdot (\mathbf{Q}\times \mathbf{R})$. b) Interpret the result geometrically.

8. Prove that the vectors $\mathbf{A}=\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ and $\mathbf{B}=2\mathbf{i}+4\mathbf{j}+6\mathbf{k}$ are parallel using the dot product.

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Problem-Solving Questions

9. A force $\mathbf{F}=4\mathbf{i}+3\mathbf{j}$ N acts on a particle displaced by $\mathbf{d}=2\mathbf{i}-\mathbf{j}$ m. Calculate the work done by the force.

10. Find the angle between vectors $\mathbf{A}=6\mathbf{i}+2\mathbf{j}-3\mathbf{k}$ and $\mathbf{B}=2\mathbf{i}-2\mathbf{j}+4\mathbf{k}$.

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Answers to NEB Board-Style Questions

6. a) $\mathbf{A}\cdot \mathbf{B}=(3)(-1)+(-2)(4)+(1)(-2)=-3-8-2=-13$ b) $\mathbf{A}\times \mathbf{B}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 3& -2& 1\\ -1& 4& -2\end{array}\right|=\mathbf{i}((-2)(-2)-(1)(4))-\mathbf{j}((3)(-2)-(1)(-1))+\mathbf{k}((3)(4)-(-2)(-1))=0\mathbf{i}-5\mathbf{j}+10\mathbf{k}$ c) The vectors are not parallel because $\mathbf{A}\times \mathbf{B}\ne 0$.

7. a) $\mathbf{Q}\times \mathbf{R}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -1& 3& 1\\ 1& -1& 2\end{array}\right|=\mathbf{i}(6-(-1))-\mathbf{j}(-2-1)+\mathbf{k}(1-(-3))=7\mathbf{i}+3\mathbf{j}+4\mathbf{k}$ $\mathbf{P}\cdot (\mathbf{Q}\times \mathbf{R})=(2)(7)+(1)(3)+(-1)(4)=14+3-4=13$ b) The scalar triple product is 13, which represents the signed volume of the parallelepiped formed by the three vectors.

8. Work done $W=\mathbf{F}\cdot \mathbf{d}=(4)(2)+(3)(-1)=8-3=5$ J.

9. $\cos \theta =\frac{\mathbf{A}\cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}=\frac{(6)(2)+(2)(-2)+(-3)(4)}{\sqrt{6^2+2^2+(-3{)}^2}\sqrt{2^2+(-2{)}^2+4^2}}=\frac{12-4-12}{\sqrt{49}\sqrt{24}}=\frac{-4}{7\times 2\sqrt{6}}=\frac{-2}{7\sqrt{6}}$ $\theta ={\cos }^{-1}\left(\frac{-2}{7\sqrt{6}}\right)\approx 98.21°$.