Notes on Chapter 12 - The Dot Product and Related Concepts

12.3 The Dot Product

• Definition of the dot product in 3D

  • For vectors \mathbf{u} = (u1, u2, u3) and \mathbf{v} = (v1, v2, v3), the dot product is the scalar
    $\mathbf{u} \cdot \mathbf{v} = u<em>1 v</em>1 + u<em>2 v</em>2 + u<em>3 v</em>3.$

  • In two dimensions, for \mathbf{u} = (u1, u2) and \mathbf{v} = (v1, v2), the dot product is
    $\mathbf{u} \cdot \mathbf{v} = u<em>1 v</em>1 + u<em>2 v</em>2.$

  • The dot product is a scalar (not a vector).

• Geometric interpretation: angle between two nonzero vectors

  • If the angle between nonzero \mathbf{u} and \mathbf{v} is \theta (0 ≤ \theta ≤ π), then the cosine of the angle is given by
    $\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| \, |\mathbf{v}|}.$

  • The magnitude of the projection of a force (or vector) F in the direction of another vector v is |F| cos \theta, which is the scalar component in the direction of v.

  • The dot product provides a direct way to compute this angle and the projection without geometric construction.

• Magnitudes used with the dot product

  • The magnitude (length) of a vector \mathbf{u} is
    $|\mathbf{u}| = \sqrt{u<em>1^2 + u</em>2^2 + u_3^2}.$

  • A 2D vectors, |\mathbf{u}| = \sqrt{u1^2 + u2^2}.

• Theorem 1 (Angle Between Two Vectors)

  • The angle between two nonzero vectors \mathbf{u} = (u1, u2, u3) and \mathbf{v} = (v1, v2, v3) is given by
    $\theta = \cos^{-1} \left( \, \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| \cdot |\mathbf{v}|} \, \right).$

  • Sketch/idea of proof via the Law of Cosines: if \mathbf{w} = \mathbf{u} - \mathbf{v}, then
    $|\mathbf{w}|^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2 - 2 |\mathbf{u}| |\mathbf{v}| \cos \theta,$
    and since \mathbf{w} = \mathbf{u} - \mathbf{v}, \mathbf{w}\cdot \mathbf{w} = (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) = |\mathbf{u}|^2 + |\mathbf{v}|^2 - 2 (\mathbf{u} \cdot \mathbf{v}).$</p></li><li><p>Equating the two expressions and simplifying yields<br>$ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}. $</p></li></ul></li><li><p>Key example computations (illustrative; values may reflect the transcript or standard practice):</p><ul><li><p>Example 1 (3D dot product):</p></li><li><p>Compute$ (1, \, 2, \, -1) \cdot (-6, \, 2, \, -3) = (1)(-6) + (2)(2) + (-1)(-3) = -6 + 4 + 3 = 1. $</p></li><li><p>Example 2 (angle between u = (1, -2, -2) and v = (6, 3, 2)):</p></li><li><p>$ u \cdot v = 1\cdot 6 + (-2)\cdot 3 + (-2)\cdot 2 = 6 - 6 - 4 = -4. $</p></li><li><p>$ |u| = \sqrt{1^2 + (-2)^2 + (-2)^2} = 3, \quad |v| = \sqrt{6^2 + 3^2 + 2^2} = 7. $</p></li><li><p>$ \cos \theta = \frac{-4}{3 \cdot 7} = -\frac{4}{21} \approx -0.1905. $</p></li><li><p>$ \theta \approx \cos^{-1}(-\frac{4}{21}) \approx 1.76 \text{ radians} \approx 100.98^ ext{\circ}. $</p></li><li><p>Example 3 (angle in triangle with A=(0,0), B=(3,5), C=(5,2)):</p></li><li><p>Vectors: \overrightarrow{CA} = (-5, -2), \overrightarrow{CB} = (-2, 3).</p></li><li><p>Dot:$ (-5)(-2) + (-2)(3) = 10 - 6 = 4. $</p></li><li><p>Magnitudes:$ |\overrightarrow{CA}| = \sqrt{29}, \quad |\overrightarrow{CB}| = \sqrt{13}. $</p></li><li><p>$ \cos \theta = \frac{4}{\sqrt{29}\sqrt{13}} = \frac{4}{\sqrt{377}} \approx 0.206. $</p></li><li><p>$ \theta \approx \cos^{-1}(0.206) \approx 1.36 \text{ radians} \approx 78.1^ ext{\circ}. $</p></li></ul></li><li><p>Orthogonality (perpendicular vectors)</p><ul><li><p>Definition: Two nonzero vectors are orthogonal if<br>$ \mathbf{u} \cdot \mathbf{v} = 0. $</p></li><li><p>Examples:</p></li><li><p>(a) \mathbf{u} = (3, -2), \mathbf{v} = (4, 6) \Rightarrow \mathbf{u} \cdot \mathbf{v} = 3\cdot 4 + (-2)\cdot 6 = 12 - 12 = 0.$

  • (b) \mathbf{u} = (3, -2, 1), \mathbf{v} = (0, 2, 4) \Rightarrow \mathbf{u} \cdot \mathbf{v} = 0 + (-2)(2) + (1)(4) = 0. $</p></li><li><p>(c) The zero vector is orthogonal to every vector:$ 0 \cdot \mathbf{u} = 0. $</p></li></ul></li><li><p>Dot product properties (algebraic rules)</p><ul><li><p>1) Commutativity:$ \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}. $</p></li><li><p>2) Scalar compatibility:$ (c\mathbf{u}) \cdot \mathbf{v} = c (\mathbf{u} \cdot \mathbf{v}) = \mathbf{u} \cdot (c\mathbf{v}). $</p></li><li><p>3) Distributivity:$ (\mathbf{u} + \mathbf{w}) \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{v} + \mathbf{w} \cdot \mathbf{v}. $</p></li><li><p>4) Self-dot equals squared magnitude:$ \mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2. $</p></li><li><p>5) Zero vector dot product:$ 0 \cdot \mathbf{u} = 0. $</p></li><li><p>Sketch of proofs (sketches only):</p></li><li><p>1) By expanding with components, take \mathbf{u} \cdot \mathbf{v} = u<em>1 v</em>1 + u<em>2 v</em>2 + u<em>3 v</em>3 = v<em>1 u</em>1 + v<em>2 u</em>2 + v<em>3 u</em>3 = \mathbf{v} \cdot \mathbf{u}.</p></li><li><p>3) Use linearity in the first argument: \mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = u<em>1(v</em>1+w<em>1) + u</em>2(v<em>2+w</em>2) + u<em>3(v</em>3+w_3) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}.</p></li></ul></li><li><p>Vector projection and the scalar component along a direction</p><ul><li><p>Vector projection of \mathbf{u} onto a nonzero vector \mathbf{v} (the component of \mathbf{u} in the direction of \mathbf{v}):<br>$ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \right) \mathbf{v} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2} \right) \mathbf{v}. $</p></li><li><p>The unit direction along \mathbf{v}: the unit vector is<br>$ \hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}. $</p></li><li><p>Scalar component of \mathbf{u} in the direction of \mathbf{v}<br>$ u_{\parallel} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|} = \, \, \text{(signed length of the projection of } \mathbf{u} \text{ onto } \mathbf{v}). $</p></li><li><p>The relationship between the two forms<br>$ \text{proj}{\mathbf{v}} \mathbf{u} = (u{\parallel}) \, \hat{\mathbf{v}} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2} \right) \, \mathbf{v}. $</p></li><li><p>Geometric interpretation: The length of the projection is |\mathbf{u}| cos \theta, and its sign is given by the sign of cos \theta via the direction of \hat{\mathbf{v}}.</p></li></ul></li><li><p>Worked examples of projection and decomposition (illustrative)</p><ul><li><p>Example A (3D): Let \mathbf{u} = (6, 3, 2) and \mathbf{v} = (1, -2, -2). Then<br>$ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \right) \mathbf{v} = \left( \frac{-4}{9} \right) (1, -2, -2) = \left( -\tfrac{4}{9}, \tfrac{8}{9}, \tfrac{8}{9} \right). $</p></li><li><p>Scalar component along \mathbf{v}:<br>$ u_{\parallel} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|} = \frac{-4}{\sqrt{1+4+4}} = -\frac{4}{3}. $</p></li><li><p>In 2D, a similar approach applies (Example 6 in the transcript): for F = (5, 2) and \mathbf{v} = (1, -3),<br>$ \text{proj}_{\mathbf{v}} \mathbf{F} = \left( \frac{F \cdot v}{v \cdot v} \right) v = \left( \frac{5(1) + 2(-3)}{1^2 + (-3)^2}
    ight) (1, -3) = \left( \frac{-1}{10} \right) (1, -3) = \left( -\tfrac{1}{10}, \tfrac{3}{10} \right). $</p></li><li><p>Decomposition identity: for any \mathbf{u} and nonzero \mathbf{v}<br>$ \mathbf{u} = \text{proj}{\mathbf{v}} \mathbf{u} + (\mathbf{u} - \text{proj}{\mathbf{v}} \mathbf{u}), $<br>and the two summands are orthogonal:<br>$ (\mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u}) \cdot \mathbf{v} = 0. $</p></li></ul></li><li><p>Work done by a constant force along a displacement</p><ul><li><p>If a constant force \mathbf{F} acts through a displacement \mathbf{D}, the work is<br>$ W = \mathbf{F} \cdot \mathbf{D}. $</p></li><li><p>If the force makes an angle \theta with the displacement, then<br>$ W = |\mathbf{F}| \, |\mathbf{D}| \, \cos \theta = \mathbf{F} \cdot \mathbf{D}. $</p></li><li><p>Example: If F = 40 N, |D| = 3 m, and \theta = 60^ ext{o},<br>$ W = (40)(3) \cos 60^ ext{o} = 60 \, \text{J}. $</p></li><li><p>Note: Chapter 16 extends these ideas to variable forces along general paths.</p></li></ul></li><li><p>Additional context and connections</p><ul><li><p>The dot product is a foundational tool for geometric and physical calculations in both space and the plane.</p></li><li><p>It connects algebraic operations with geometric quantities: angle, projection, work, and decompositions into parallel and perpendicular components.</p></li></ul></li><li><p>Selected problems from Chapter 12 (12.2/12.3 level, as listed in the transcript)</p><ul><li><p>Find the vector from the origin to the point of intersection of the medians of triangle with vertices A(1, -1, 2), B(2, 1, 3), C(-1, 2, -1).</p></li><li><p>The medians intersect at the centroid, which is at the average of the vertex coordinates:<br>$ ext{Centroid} = \left( \frac{1+2-1}{3}, \frac{-1+1+2}{3}, \frac{2+3-1}{3} \right) = \left( \frac{2}{3}, \frac{2}{3}, \frac{4}{3} \right). $</p></li><li><p>Vector from the origin to the centroid is the same as the centroid coordinates:$ \left( \tfrac{2}{3}, \tfrac{2}{3}, \tfrac{4}{3} \right). $</p></li><li><p>Let ABCD be a general, not necessarily planar, quadrilateral in space. Show that the two segments joining the midpoints of opposite sides of ABCD bisect each other.</p></li><li><p>Let A, B, C, D be position vectors. The midpoints are:</p><ul><li><p>M<em>AB = (A + B)/2, M</em>CD = (C + D)/2 for the segment joining AB and CD midpoints.</p></li><li><p>M<em>BC = (B + C)/2, M</em>AD = (A + D)/2 for the segment joining BC and AD midpoints.</p></li></ul></li><li><p>The midpoint of the segment joining M<em>AB and M</em>CD is<br>$ rac{M{AB} + M{CD}}{2} = \frac{A + B + C + D}{4}. $</p></li><li><p>The midpoint of the segment joining M<em>BC and M</em>AD is<br>$ \frac{M{BC} + M{AD}}{2} = \frac{A + B + C + D}{4}. $</p></li><li><p>Therefore the two segments share the same midpoint and bisect each other.</p></li><li><p>Vectors drawn from the center of a regular n-sided polygon in the plane to the vertices: show the sum of the vectors is zero.</p></li><li><p>Reason: rotational symmetry. Rotating the polygon by 360°/n permutes the vertex vectors; the sum must be invariant under this rotation, which forces the sum to be the zero vector.</p></li><li><p>If A, B, C are vertices of a triangle and a, b, c are respectively the midpoints of the opposite sides, show that \mathbf{A}\mathbf{a} + \mathbf{B}\mathbf{b} + \mathbf{C}\mathbf{c} = 0.</p></li><li><p>Let a be the midpoint of BC, b of CA, c of AB. Then a = (B + C)/2, etc. Compute:<br>$ \mathbf{A} \mathbf{a} + \mathbf{B} \mathbf{b} + \mathbf{C} \mathbf{c} = (A - A) + (B - B) + (C - C) = 0, $</p></li><li><p>or more explicitly by writing a - A = (B + C)/2 - A, etc., and summing.</p></li><li><p>Unit vectors in the plane: show that a unit vector can be expressed as$ \mathbf{u} = (\cos \theta) \mathbf{i} + (\sin \theta) \mathbf{j}, $obtained by rotating the standard basis vector \mathbf{i} through an angle \theta in the counterclockwise direction.</p></li><li><p>This parameterization covers every direction in the plane as \theta ranges over [0, 2π). The length is 1 because \cos^2 \theta + \sin^2 \theta = 1.</p></li></ul></li><li><p>Practical takeaway</p><ul><li><p>The dot product is the primary tool for converting angle and projection questions into algebraic computations via component sums.</p></li><li><p>It provides a direct route to work calculations via F \cdot D instead of resolving F into components along a path.</p></li></ul></li><li><p>Summary of key formulas to memorize</p><ul><li><p>Dot product (3D):$ \mathbf{u} \cdot \mathbf{v} = u1 v1 + u2 v2 + u3 v3. $</p></li><li><p>Magnitude:$ |\mathbf{u}| = \sqrt{u1^2 + u2^2 + u_3^2}. $</p></li><li><p>Angle between vectors:$ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| \, |\mathbf{v}|}. $</p></li><li><p>Vector projection onto v:$ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \right) \mathbf{v}. $</p></li><li><p>Scalar component along v:$ u_{\parallel} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|} = \mathbf{u} \\cdot \hat{\mathbf{v}}. $</p></li><li><p>Decomposition:$ \mathbf{u} = \text{proj}{\mathbf{v}} \mathbf{u} + (\mathbf{u} - \text{proj}{\mathbf{v}} \mathbf{u}), \quad (\mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u}) \cdot \mathbf{v} = 0. $</p></li><li><p>Work done by constant force:$ W = \mathbf{F} \cdot \mathbf{D} = |\mathbf{F}| \, |\mathbf{D}| \, \cos \theta. $$