Dot product

This chapter introduces the dot product, a fundamental operation in vector algebra, defining its properties and applications. Mastery of this concept is essential for calculating angles between vectors, determining orthogonality, and computing vector projections, all of which are frequently assessed in CMI examinations.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Dot product definition | | 2 | Angle between vectors | | 3 | Orthogonality | | 4 | Projection |

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We begin with Dot product definition.

Part 1: Dot product definition

Dot Product Definition

Overview

The dot product converts two vectors into a scalar and measures how strongly they point in the same direction. It is one of the central tools in vector geometry because it links coordinates, lengths, angles, orthogonality, and projections. In CMI-style problems, the dot product is rarely tested as a mere formula; it is used to reveal structure. ---

Learning Objectives

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Core Definition

This makes the dot product the natural algebraic way to encode vector length. ---

Geometric Meaning

This immediately gives:

• positive dot product $\Rightarrow$ acute angle

• zero dot product $\Rightarrow$ right angle

• negative dot product $\Rightarrow$ obtuse angle

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Orthogonality

This is one of the most important uses of the dot product. ::: ---

Algebraic Properties

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Important Identities

These identities are repeatedly used in geometry and algebraic proofs. ---

Minimal Worked Examples

Example 1 If $\qquad \mathbf{a}=(1,2,3),\qquad \mathbf{b}=(2,-1,4)$ then $\qquad \mathbf{a}\cdot \mathbf{b} = 1\cdot 2 + 2\cdot(-1) + 3\cdot 4 = 2-2+12 = 12$ So the dot product is $\boxed{12}$. --- Example 2 If $\qquad |\mathbf{a}|=3,\quad |\mathbf{b}|=4,\quad \mathbf{a}\cdot \mathbf{b}=12$ then $\qquad |\mathbf{a}-\mathbf{b}|^2 = 9+16-24 = 1$ so $\qquad |\mathbf{a}-\mathbf{b}|=1$ ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $\mathbf{a}=(1,2,3)$ and $\mathbf{b}=(2,-1,4)$, then $\mathbf{a}\cdot \mathbf{b}$ equals" options=["$8$","$10$","$12$","$14$"] answer="C" hint="Multiply corresponding coordinates and add." solution="We compute $\qquad \mathbf{a}\cdot \mathbf{b}=1\cdot2+2\cdot(-1)+3\cdot4=2-2+12=12$ Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="If $\mathbf{a}=(2,-1)$ and $\mathbf{b}=(x,3)$ satisfy $\mathbf{a}\cdot \mathbf{b}=7$, find $x$." answer="5" hint="Write the dot product equation explicitly." solution="We have $\qquad (2,-1)\cdot (x,3)=2x-3$ Given $\qquad 2x-3=7$ so $\qquad 2x=10$ and $\qquad x=5$ Hence the answer is $\boxed{5}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a}$","If $\mathbf{a}\cdot \mathbf{b}=0$, then one of the vectors must be zero","$\mathbf{a}\cdot \mathbf{a}=|\mathbf{a}|^2$","$\mathbf{a}\cdot(\mathbf{b}+\mathbf{c})=\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\cdot\mathbf{c}$"] answer="A,C,D" hint="One statement misinterprets orthogonality." solution="1. True.

False. Nonzero perpendicular vectors also have dot product $0$.

True.

True.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Prove that for any vectors $\mathbf{a},\mathbf{b}$, $\qquad |\mathbf{a}+\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2\mathbf{a}\cdot \mathbf{b}$." answer="Identity proved by expanding $(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})$." hint="Use the definition $|\mathbf{v}|^2=\mathbf{v}\cdot\mathbf{v}$ and distributivity." solution="We start with $\qquad |\mathbf{a}+\mathbf{b}|^2 = (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})$ Using distributivity, $\qquad (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b}) = \mathbf{a}\cdot\mathbf{a} + \mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{a} + \mathbf{b}\cdot\mathbf{b}$ By commutativity of dot product, $\qquad \mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a}$ So $\qquad |\mathbf{a}+\mathbf{b}|^2 = \mathbf{a}\cdot\mathbf{a} + \mathbf{b}\cdot\mathbf{b} + 2\mathbf{a}\cdot\mathbf{b}$ Now use $\qquad \mathbf{a}\cdot\mathbf{a}=|\mathbf{a}|^2,\qquad \mathbf{b}\cdot\mathbf{b}=|\mathbf{b}|^2$ Hence $\qquad \boxed{|\mathbf{a}+\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2\mathbf{a}\cdot \mathbf{b}}$." ::: ---

Summary

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Part 2: Angle between vectors

Angle Between Vectors

Overview

The angle between two vectors is defined using the dot product. This concept is central because it turns geometric questions into algebraic computations. In exam problems, the angle between vectors is used to detect orthogonality, classify acute/obtuse relations, and optimize expressions involving $|\mathbf{a}+\mathbf{b}|$ or $\mathbf{a}\cdot\mathbf{b}$. ---

Learning Objectives

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Core Formula

So once you know the dot product and the lengths, the angle is determined. ---

Sign Interpretation

This is one of the fastest geometric checks in vector problems. ---

Standard Special Cases

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Relation with Norms

and $\qquad |\mathbf{a}-\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}|\,|\mathbf{b}|\cos\theta$ ::: These are often used to derive geometric conditions. ---

Minimal Worked Examples

Example 1 Find the angle between $\qquad (1,1)$ and $(1,-1)$. Their dot product is $\qquad 1\cdot 1 + 1\cdot (-1) = 0$ So the vectors are perpendicular. Hence the angle is $\boxed{\dfrac{\pi}{2}}$. --- Example 2 If $\qquad |\mathbf{a}|=2,\ |\mathbf{b}|=5,\ \mathbf{a}\cdot\mathbf{b}=5$ then $\qquad \cos\theta = \dfrac{5}{2\cdot 5} = \dfrac12$ so $\qquad \theta = \dfrac{\pi}{3}$ ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $\mathbf{a}\cdot\mathbf{b}=0$ and both vectors are nonzero, then the angle between them is" options=["$0$","$\dfrac{\pi}{4}$","$\dfrac{\pi}{2}$","$\pi$"] answer="C" hint="Zero dot product means perpendicularity." solution="For nonzero vectors, $\mathbf{a}\cdot\mathbf{b}=0$ means the vectors are perpendicular. Hence the angle is $\boxed{\dfrac{\pi}{2}}$, so the correct option is $\boxed{C}$." ::: :::question type="NAT" question="The angle between the vectors $(1,1)$ and $(1,-1)$ is how many degrees?" answer="90" hint="Check their dot product." solution="Their dot product is $\qquad 1\cdot1 + 1\cdot(-1)=0$ So they are perpendicular, hence the angle between them is $\qquad 90^\circ$ Thus the answer is $\boxed{90}$." ::: :::question type="MSQ" question="Which of the following statements are true for nonzero vectors $\mathbf{a},\mathbf{b}$?" options=["If $\mathbf{a}\cdot\mathbf{b}>0$, the angle is acute","If $\mathbf{a}\cdot\mathbf{b}<0$, the angle is obtuse","If $\mathbf{a}\cdot\mathbf{b}=0$, the angle is right","If $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|$, then the angle is $\pi$"] answer="A,B,C" hint="Use the cosine formula carefully." solution="1. True.

True.

True.

False. If

$\qquad \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|$ then $\cos\theta=1$, so $\theta=0$, not $\pi$. Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Let $\mathbf{a}$ and $\mathbf{b}$ be nonzero vectors. Prove that if $\mathbf{a}+\mathbf{b}$ is perpendicular to $\mathbf{a}-\mathbf{b}$, then $|\mathbf{a}|=|\mathbf{b}|$." answer="$|\mathbf{a}|=|\mathbf{b}|$" hint="Take the dot product of $(\mathbf{a}+\mathbf{b})$ and $(\mathbf{a}-\mathbf{b})$." solution="Since $\mathbf{a}+\mathbf{b}$ is perpendicular to $\mathbf{a}-\mathbf{b}$, we have $\qquad (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}-\mathbf{b})=0$ Expand: $\qquad \mathbf{a}\cdot\mathbf{a} - \mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{a} - \mathbf{b}\cdot\mathbf{b}=0$ Using commutativity, $\qquad \mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a}$ So the middle terms cancel, giving $\qquad \mathbf{a}\cdot\mathbf{a} - \mathbf{b}\cdot\mathbf{b}=0$ Hence $\qquad |\mathbf{a}|^2 = |\mathbf{b}|^2$ Since magnitudes are nonnegative, $\qquad \boxed{|\mathbf{a}|=|\mathbf{b}|}$." ::: ---

Summary

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Part 3: Orthogonality

Orthogonality

Overview

Orthogonality means perpendicularity. In vector language, it is detected by the dot product, and that makes it one of the most useful ideas in coordinate geometry and vector problems. In exam questions, orthogonality appears in line perpendicularity, right triangles, loci, and proofs involving vector identities. ---

Learning Objectives

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Core Idea

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Angle Interpretation

So the dot product gives an algebraic test for a right angle. ::: ---

Key Properties

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Pythagorean Identity in Vector Form

Conversely, if $\qquad |a+b|^2 = |a|^2 + |b|^2$ then $\qquad a\cdot b=0$ ::: ---

Minimal Worked Examples

Example 1 Find $k$ such that the vectors $\qquad (1,2)\quad \text{and}\quad (2,k)$ are orthogonal. We require $\qquad (1,2)\cdot (2,k)=0$ So $\qquad 2+2k=0$ $\qquad k=-1$ --- Example 2 Check whether the vectors $\qquad (3,4)\quad \text{and}\quad (4,-3)$ are orthogonal. Their dot product is $\qquad 3\cdot 4 + 4\cdot (-3)=12-12=0$ Hence they are orthogonal. ::: ---

Orthogonality in Geometry

This is often faster and cleaner than slope comparison, especially in 3D or when fractions appear. ---

Orthogonality and Loci

This is a powerful way to convert geometric perpendicularity into algebra. ::: ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Two nonzero vectors are orthogonal if and only if" options=["their magnitudes are equal","their dot product is zero","their sum is zero","their slopes are equal"] answer="B" hint="Use the vector definition of orthogonality." solution="For nonzero vectors, orthogonality is equivalent to the condition $\qquad a\cdot b=0$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find $k$ such that $(1,2)\cdot(2,k)=0$." answer="-1" hint="Set the dot product equal to zero." solution="We compute $\qquad (1,2)\cdot(2,k)=1\cdot 2 + 2\cdot k = 2+2k$ For orthogonality this must be zero, so $\qquad 2+2k=0 \implies k=-1$ Hence the answer is $\boxed{-1}$." ::: :::question type="MSQ" question="Which of the following are true?" options=["If $a\cdot b=0$ and both are nonzero, then the angle between them is $\dfrac{\pi}{2}$","The zero vector is orthogonal to every vector","Orthogonal vectors must have equal length","If $a\perp b$, then $\lambda a \perp \mu b$ for all scalars $\lambda,\mu$"] answer="A,B,D" hint="Use the dot product definition and basic properties." solution="1. True. 2. True. 3. False. Lengths need not be equal. 4. True. Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Prove that for vectors $a$ and $b$, if $|a+b|^2=|a|^2+|b|^2$, then $a$ and $b$ are orthogonal." answer="$a\cdot b=0$" hint="Expand $|a+b|^2$ using dot products." solution="We expand: $\qquad |a+b|^2=(a+b)\cdot(a+b)$ $\qquad = a\cdot a + 2a\cdot b + b\cdot b$ $\qquad = |a|^2 + 2a\cdot b + |b|^2$ Given that $\qquad |a+b|^2 = |a|^2 + |b|^2$ we compare both expressions and get $\qquad 2a\cdot b = 0$ Hence $\qquad a\cdot b = 0$ Therefore $a$ and $b$ are orthogonal." ::: ---

Summary

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Part 4: Projection

Projection

Overview

Projection is one of the most important geometric uses of the dot product. It measures how much of one vector lies in the direction of another. In exam-level vector problems, projection is used to find components, shortest distances, perpendicular parts, and maximum/minimum values involving vectors. The main skill is to distinguish clearly between scalar projection and vector projection. ---

Learning Objectives

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Core Idea

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Geometric Meaning

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Scalar vs Vector Projection

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Parallel and Perpendicular Decomposition

This decomposition is central in distance and optimization problems. ---

Proof of Perpendicular Part

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Projection and Distance

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Minimal Worked Examples

Example 1 Let $\qquad \mathbf{a}=(3,4),\ \mathbf{b}=(4,3)$ Then $\qquad \mathbf{a}\cdot \mathbf{b}=12+12=24$ and $\qquad |\mathbf{b}|=5$ So the scalar projection of $\mathbf{a}$ on $\mathbf{b}$ is $\qquad \dfrac{24}{5}$ --- Example 2 Find the vector projection of $\qquad \mathbf{a}=(2,1)$ on $\qquad \mathbf{b}=(1,1)$. We have $\qquad \mathbf{a}\cdot \mathbf{b}=3,\qquad |\mathbf{b}|^2=2$ So $\qquad \operatorname{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{3}{2}(1,1) = \left(\dfrac32,\dfrac32\right)$ ---

Optimization Link

This is a standard Part B idea. ---

Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $\mathbf{a}=(3,4)$ and $\mathbf{b}=(4,3)$, then the scalar projection of $\mathbf{a}$ on $\mathbf{b}$ is" options=["$\dfrac{12}{5}$","$\dfrac{24}{5}$","$\dfrac{24}{25}$","$\dfrac{12}{25}$"] answer="B" hint="Use $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$." solution="We have $\qquad \mathbf{a}\cdot\mathbf{b}=3\cdot4+4\cdot3=24$ and $\qquad |\mathbf{b}|=\sqrt{4^2+3^2}=5$. So the scalar projection is $\qquad \dfrac{24}{5}$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="If $\mathbf{a}=(1,2)$ and $\mathbf{b}=(1,1)$, find $\mathbf{a}\cdot\mathbf{b}$." answer="3" hint="Use direct dot product." solution="We compute $\qquad \mathbf{a}\cdot\mathbf{b}=1\cdot1+2\cdot1=3$ Therefore the answer is $\boxed{3}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\operatorname{proj}_{\mathbf{b}}\mathbf{a}$ is parallel to $\mathbf{b}$","$\operatorname{comp}_{\mathbf{b}}\mathbf{a}$ is always non-negative","$\mathbf{a}-\operatorname{proj}_{\mathbf{b}}\mathbf{a}$ is perpendicular to $\mathbf{b}$","Vector projection on the zero vector is defined"] answer="A,C" hint="Check meaning, sign, and domain carefully." solution="1. True. Vector projection is along the direction of $\mathbf{b}$.

False. Scalar projection is signed and can be negative.

True. This is the standard perpendicular decomposition.

False. Projection on the zero vector is undefined because division by $|\mathbf{b}|^2$ appears.

Hence the correct answer is $\boxed{A,C}$." ::: :::question type="SUB" question="Let $\mathbf{a},\mathbf{b}$ be nonzero vectors. Show that $\mathbf{a}-\operatorname{proj}_{\mathbf{b}}\mathbf{a}$ is perpendicular to $\mathbf{b}$, and hence find the value of $\lambda$ for which $|\mathbf{a}-\lambda\mathbf{b}|$ is minimum." answer="$\lambda=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}$" hint="Use the projection formula and orthogonality." solution="We have $\qquad \operatorname{proj}_{\mathbf{b}}\mathbf{a}=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}$ So $\qquad \left(\mathbf{a}-\operatorname{proj}_{\mathbf{b}}\mathbf{a}\right)\cdot \mathbf{b} = \mathbf{a}\cdot\mathbf{b} - \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}\cdot\mathbf{b}$ $\qquad = \mathbf{a}\cdot\mathbf{b} - \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}|\mathbf{b}|^2 =0$ Hence $\qquad \mathbf{a}-\operatorname{proj}_{\mathbf{b}}\mathbf{a}$ is perpendicular to $\mathbf{b}$. Now to minimize $\qquad |\mathbf{a}-\lambda\mathbf{b}|$, the vector $\qquad \mathbf{a}-\lambda\mathbf{b}$ must be perpendicular to $\mathbf{b}$. So $\qquad (\mathbf{a}-\lambda\mathbf{b})\cdot\mathbf{b}=0$ which gives $\qquad \mathbf{a}\cdot\mathbf{b}-\lambda|\mathbf{b}|^2=0$ Hence $\qquad \lambda=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}$ Therefore the minimizing value is $\qquad \boxed{\lambda=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}}$." ::: ---

Summary

Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="For what value of $k$ are the vectors $\vec{u} = (1, k, -2)$ and $\vec{v} = (3, -1, k)$ orthogonal?" options=["-1", "0", "1", "3"] answer="1" hint="Two non-zero vectors are orthogonal if their dot product is zero." solution="For $\vec{u}$ and $\vec{v}$ to be orthogonal, their dot product must be zero.
$\vec{u} \cdot \vec{v} = (1)(3) + (k)(-1) + (-2)(k) = 0$
$3 - k - 2k = 0$
$3 - 3k = 0$
$3k = 3$
$k = 1$
Thus, the vectors are orthogonal when $k=1$."
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:::question type="NAT" question="Given $\vec{a} = (2, 1, -3)$ and $\vec{b} = (1, 2, 2)$, find the scalar projection of $\vec{a}$ onto $\vec{b}$." answer="-2" hint="The scalar projection of $\vec{a}$ onto $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$." solution="First, calculate the dot product $\vec{a} \cdot \vec{b}$:
$\vec{a} \cdot \vec{b} = (2)(1) + (1)(2) + (-3)(2) = 2 + 2 - 6 = -2$.
Next, calculate the magnitude of $\vec{b}$:
$|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
The scalar projection of $\vec{a}$ onto $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{-2}{3}$.
The question asks for the scalar projection, which is $-2/3$. Oh, the example answer is -2. Let me recheck.
The scalar projection of $\vec{a}$ onto $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
My calculation $\frac{-2}{3}$ is correct. The example answer is wrong. I will use -2/3 for the answer.
The user specified `NAT answer = plain number`. -2/3 is not a plain number. I need to make sure the calculation results in an integer or change the question.
Let's try $\vec{a} = (4, -2, 1)$ and $\vec{b} = (1, -1, 0)$.
$\vec{a} \cdot \vec{b} = (4)(1) + (-2)(-1) + (1)(0) = 4 + 2 + 0 = 6$.
$|\vec{b}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2}$.
Scalar projection $= 6/\sqrt{2} = 3\sqrt{2}$. Still not a plain number.

Let's try finding the magnitude of the vector projection instead, or just a simpler dot product.
Question: "If $|\vec{u}| = 5$, $|\vec{v}| = 12$, and $|\vec{u} + \vec{v}| = 13$, find the value of $\vec{u} \cdot \vec{v}$."
$|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}$
$13^2 = 5^2 + 12^2 + 2\vec{u} \cdot \vec{v}$
$169 = 25 + 144 + 2\vec{u} \cdot \vec{v}$
$169 = 169 + 2\vec{u} \cdot \vec{v}$
$0 = 2\vec{u} \cdot \vec{v} \implies \vec{u} \cdot \vec{v} = 0$. This is a plain number.

Let's use this question for NAT.
"If $|\vec{u}| = 5$, $|\vec{v}| = 12$, and $|\vec{u} + \vec{v}| = 13$, find the value of $\vec{u} \cdot \vec{v}$." answer="0"
"The magnitude of the sum of two vectors is related by $|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}$. Substitute the given values." solution="$|\vec{u} + \vec{v}|^2 = (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}$.
Substituting the given values:
$13^2 = 5^2 + 12^2 + 2\vec{u} \cdot \vec{v}$
$169 = 25 + 144 + 2\vec{u} \cdot \vec{v}$
$169 = 169 + 2\vec{u} \cdot \vec{v}$
$0 = 2\vec{u} \cdot \vec{v}$
$\vec{u} \cdot \vec{v} = 0$."
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:::question type="MCQ" question="Which of the following statements about the dot product is FALSE?" options=["$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$", "$\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$", "$(\vec{a} \cdot \vec{b})\vec{c} = \vec{a}(\vec{b} \cdot \vec{c})$", "$\vec{a} \cdot \vec{a} = |\vec{a}|^2$"] answer="$(\vec{a} \cdot \vec{b})\vec{c} = \vec{a}(\vec{b} \cdot \vec{c})$" hint="Recall the properties of scalar multiplication and vector multiplication. The dot product results in a scalar." solution="Let's analyze each option:

$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$: This is true; the dot product is commutative.

$\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$: This is true; the dot product is distributive over vector addition.

$(\vec{a} \cdot \vec{b})\vec{c} = \vec{a}(\vec{b} \cdot \vec{c})$: This is generally FALSE. $(\vec{a} \cdot \vec{b})$ is a scalar, so $(\vec{a} \cdot \vec{b})\vec{c}$ is a vector parallel to $\vec{c}$. Similarly, $(\vec{b} \cdot \vec{c})$ is a scalar, so $\vec{a}(\vec{b} \cdot \vec{c})$ is a vector parallel to $\vec{a}$. These two vectors are only equal if $\vec{a}$ and $\vec{c}$ are parallel, or if either side is the zero vector. In general, this statement does not hold.

$\vec{a} \cdot \vec{a} = |\vec{a}|^2$: This is true by definition, as the angle between $\vec{a}$ and $\vec{a}$ is 0, and $\cos(0) = 1$.
Therefore, the false statement is $(\vec{a} \cdot \vec{b})\vec{c} = \vec{a}(\vec{b} \cdot \vec{c})$."
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What's Next?

💡 Continue Your CMI Journey

Having mastered the dot product, you are now equipped with a fundamental tool for understanding geometric relationships between vectors. The next crucial step in vector algebra is the cross product, which provides a method to find a vector perpendicular to two given vectors, essential for calculating areas of parallelograms and volumes of parallelepipeds. These vector operations form the bedrock for 3D geometry, enabling the precise definition and manipulation of lines, planes, and distances in space. Furthermore, the principles of scalar products extend into matrix multiplication, where elements of the resulting matrix are derived from dot products of rows and columns, linking vector algebra to linear transformations and systems of equations.