Comprehensive Study Guide on Vector Dot Products and Level Sets

Geometric Definition of the Dot Product

  • Let u⃗\vec{u} and v⃗\vec{v} be two non-zero vectors in the plane such that u⃗=AB⃗\vec{u} = \vec{AB} and v⃗=AC⃗\vec{v} = \vec{AC}. The dot product (produit scalaire) of u⃗\vec{u} and v⃗\vec{v} is the real number denoted by u⃗⋅v⃗\vec{u} \cdot \vec{v}, defined by:   u⃗⋅v⃗=∥u⃗∥×∥v⃗∥×cos⁡(u⃗,v⃗^)\vec{u} \cdot \vec{v} = \|\vec{u}\| \times \|\vec{v}\| \times \cos(\widehat{\vec{u}, \vec{v}})AB⃗⋅AC⃗=AB×AC×cos⁡(BAC^)\vec{AB} \cdot \vec{AC} = AB \times AC \times \cos(\widehat{BAC})
  • Zero Vector Case: If u⃗=0⃗\vec{u} = \vec{0} or v⃗=0⃗\vec{v} = \vec{0}, then u⃗⋅v⃗=0\vec{u} \cdot \vec{v} = 0.
  • Scalar Square Definition: The scalar square of a vector u⃗\vec{u} is defined as the dot product of vector u⃗\vec{u} with itself:   u⃗⋅u⃗=u⃗2=∥u⃗∥2\vec{u} \cdot \vec{u} = \vec{u}^2 = \|\vec{u}\|^2
  • Exercise 1: Let ABCDABCD be a square of side length aa and center OO. Calculate the following dot products:
    • AB⃗⋅AC⃗\vec{AB} \cdot \vec{AC}
    • AB⃗⋅AO⃗\vec{AB} \cdot \vec{AO}
    • CB⃗⋅CD⃗\vec{CB} \cdot \vec{CD}
    • AB⃗⋅CD⃗\vec{AB} \cdot \vec{CD}
    • AB⃗⋅AD⃗\vec{AB} \cdot \vec{AD}

Properties of the Dot Product

  • Fundamental Properties: For all vectors u⃗\vec{u}, v⃗\vec{v}, and w⃗\vec{w} in the plane and for any real scalar kk:
    • Commutativity / Symmetry: u⃗⋅v⃗=v⃗⋅u⃗\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}
    • Distributivity over vector addition: u⃗⋅(v⃗+w⃗)=u⃗⋅v⃗+u⃗⋅w⃗\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}
    • Associativity with scalar multiplication: (ku⃗)⋅v⃗=k(u⃗⋅v⃗)=u⃗⋅(kv⃗)(k\vec{u}) \cdot \vec{v} = k(\vec{u} \cdot \vec{v}) = \vec{u} \cdot (k\vec{v})
    • Difference Formula 1: u⃗⋅v⃗=12(∥u⃗+v⃗∥2−∥u⃗∥2−∥v⃗∥2)\vec{u} \cdot \vec{v} = \frac{1}{2}\left(\|\vec{u} + \vec{v}\|^2 - \|\vec{u}\|^2 - \|\vec{v}\|^2\right)
    • Difference Formula 2: u⃗⋅v⃗=12(∥u⃗∥2+∥v⃗∥2−∥u⃗−v⃗∥2)\vec{u} \cdot \vec{v} = \frac{1}{2}\left(\|\vec{u}\|^2 + \|\vec{v}\|^2 - \|\vec{u} - \vec{v}\|^2\right)
  • Proof / Demonstration of Fundamental Properties: Let u⃗=AB⃗\vec{u} = \vec{AB} and v⃗=AC⃗\vec{v} = \vec{AC}.
    1. Since cos⁡(u⃗,v⃗^)=cos⁡(v⃗,u⃗^)\cos(\widehat{\vec{u}, \vec{v}}) = \cos(\widehat{\vec{v}, \vec{u}}), it directly follows that u⃗⋅v⃗=v⃗⋅u⃗\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}.
    2. Let DD be the point on line (AB)(AB) such that AD⃗=kAB⃗\vec{AD} = k\vec{AB}. Then (k\vec{u}) \cdot \vec{v} = \vec{AD} \n\cdot \vec{AC} = AD \times AC \times \cos(\widehat{DAC}).
    • If k>0k > 0, then DAC^=BAC^\widehat{DAC} = \widehat{BAC} and AD=k×ABAD = k \times AB. Thus:        AD⃗⋅AC⃗=(k×AB)×AC×cos⁡(BAC^)=k(u⃗⋅v⃗)\vec{AD} \cdot \vec{AC} = (k \times AB) \times AC \times \cos(\widehat{BAC}) = k(\vec{u} \cdot \vec{v})
    • If k<0k < 0, then angles DAC^\widehat{DAC} and BAC^\widehat{BAC} are supplementary angles, meaning cos⁡(DAC^)=−cos⁡(BAC^)\cos(\widehat{DAC}) = -\cos(\widehat{BAC}) and AD=−k×ABAD = -k \times AB. Thus:        AD⃗⋅AC⃗=(−k×AB)×AC×(−cos⁡(BAC^))=k×AB×AC×cos⁡(BAC^)=k(u⃗⋅v⃗)\vec{AD} \cdot \vec{AC} = (-k \times AB) \times AC \times (-\cos(\widehat{BAC})) = k \times AB \times AC \times \cos(\widehat{BAC}) = k(\vec{u} \cdot \vec{v})
  • Exercise 2: Let ABCABC be a triangle.
    1. Verify that AB⃗⋅AC⃗=12(AB2+AC2−BC2)\vec{AB} \cdot \vec{AC} = \frac{1}{2}\left(AB^2 + AC^2 - BC^2\right).
    2. Show that BC2=AB2+AC2−2(AB⃗⋅AC⃗)BC^2 = AB^2 + AC^2 - 2(\vec{AB} \cdot \vec{AC}) (Formula of Al-Kashi / Law of Cosines).
  • Exercise 3: Let u⃗\vec{u}, v⃗\vec{v}, and w⃗\vec{w} be three vectors such that ∥u⃗∥=3\|\vec{u}\| = 3, ∥v⃗∥=2\|\vec{v}\| = 2, ∥w⃗∥=1\|\vec{w}\| = 1, and u⃗⋅v⃗=−2\vec{u} \cdot \vec{v} = -2.
    1. Calculate (2u⃗−3v⃗)⋅u⃗(2\vec{u} - 3\vec{v}) \cdot \vec{u} and (u⃗−v⃗)⋅(u⃗+v⃗)(\vec{u} - \vec{v}) \cdot (\vec{u} + \vec{v}).
    2. Calculate (u⃗+v⃗)⋅(v⃗+w⃗)+(v⃗+w⃗)⋅(w⃗+u⃗)+(w⃗+u⃗)⋅(u⃗+v⃗)(\vec{u} + \vec{v}) \cdot (\vec{v} + \vec{w}) + (\vec{v} + \vec{w}) \cdot (\vec{w} + \vec{u}) + (\vec{w} + \vec{u}) \cdot (\vec{u} + \vec{v}).

Orthogonal Vectors and Collinear Vectors

  • Theorem (Orthogonal Vectors): Two vectors u⃗\vec{u} and v⃗\vec{v} are orthogonal if and only if u⃗⋅v⃗=0\vec{u} \cdot \vec{v} = 0.
  • Exercise 4: Prove the theorem of orthogonal vectors.
  • Consequence for Lines: Two lines are perpendicular if and only if the dot product of their direction vectors is equal to zero.
  • Exercise 5: Let ABCABC be a right-angled triangle at AA. Let HH be the foot of the altitude issuing from AA. Given AB=3AB = 3, AC=4AC = 4, and BC=5BC = 5.
    1. Show that HB⃗⋅HC⃗=−HA2\vec{HB} \cdot \vec{HC} = -HA^2.
    2. Show that line (AB)(AB) and line (AC)(AC) are perpendicular.
  • Exercise 6: Let AA and BB be two distinct points.
    1. Determine and represent the set of points MM in the plane such that \vec{MA} \cdot \vec{MB} = 0$.\n 2. Determine and represent the set of points Mintheplanesuchthatin the plane such that\vec{MA}^2 - \vec{MB}^2 = 0$.
  • Exercise 7: Let AA, BB, CC, and DD be four points in the plane.
    1. Show that \vec{DA} \cdot \vec{BC} + \vec{DB} \cdot \vec{CA} + \vec{DC} \cdot \vec{AB} = 0$.\n 2. Show that the three altitudes of a triangle are concurrent.\n- Activity on Collinear Vectors: Let \vec{u}andand\vec{v}betwonon−zerovectors.Letbe two non-zero vectors. LetA,,B,and, andCbethreepointssuchthatbe three points such that\vec{AB} = \vec{u}andand\vec{AC} = \vec{v}.\n 1. Show that (\vec{u} \cdot \vec{v})^2 \le |\vec{u}|^2 \times |\vec{v}|^2$.
    2. Provide a necessary and sufficient condition on angle BAC^\widehat{BAC} for u⃗\vec{u} and v⃗\vec{v} to be collinear.
  • Properties (Cauchy-Schwarz Inequality):
    • For all vectors u⃗\vec{u} and v⃗\vec{v} in the plane:     ∣u⃗⋅v⃗∣≤∥u⃗∥×∥v⃗∥|\vec{u} \cdot \vec{v}| \le \|\vec{u}\| \times \|\vec{v}\|
    • Equality Condition: $|\vec{u} \cdot \vec{v}| = |\vec{u}| \times |\vec{v}|ifandonlyifif and only if\vec{u}andand\vec{v} are collinear.\n\n# Dot Product and Orthogonal Projection\n\n- Projection Property: Let \vec{u}andand\vec{v}betwonon−zerovectorssuchthatbe two non-zero vectors such that\vec{u} = \vec{AB}andand\vec{v} = \vec{AC}.If. IfHistheorthogonalprojectionofpointis the orthogonal projection of pointContolineonto line(AB), then:\n - \vec{u} \cdot \vec{v} = \vec{AB} \cdot \vec{AC} = \vec{AB} \cdot \vec{AH}\n - \vec{u} \cdot \vec{v} = AB \times AHifvectorsif vectors\vec{AB}andand\vec{AH} have the same sense/direction.\n - \vec{u} \cdot \vec{v} = -AB \times AHifvectorsif vectors\vec{AB}andand\vec{AH} have opposite senses/directions.\n- Exercise 8: Prove the orthogonal projection property.\n- Exercise 9: Let ABCDbeasquareofsidelengthbe a square of side lengthaandcenterand centerO.Points. PointsIandandJaretherespectivemidpointsofsegmentsare the respective midpoints of segments[AB]andand[BC]. Calculate the following dot products:\n - \vec{AB} \cdot \vec{AC}\n - \vec{AI} \cdot \vec{AC}\n - \vec{OI} \cdot \vec{AB}\n - \vec{OI} \cdot \vec{AC}\n- Exercise 10: Let AandandBbetwopointssuchthatbe two points such thatAB = 4.Let. LetMbeapointintheplanenotbelongingtolinebe a point in the plane not belonging to line(AB)andandHitsorthogonalprojectionontoits orthogonal projection onto(AB).Determineandconstructthesetofpoints. Determine and construct the set of pointsM in each case:\n 1. \vec{AB} \cdot \vec{AM} = 8\n 2. \vec{AB} \cdot \vec{AM} = -12\n- Exercise 11: Let AandandB be two distinct points.\n 1. Determine the set of points Mintheplanesuchthatin the plane such that\vec{MA} \cdot \vec{MB} = AB^2$.
    1. Determine the set of points MM in the plane such that \vec{MA} \cdot \vec{MB} = -3AB^2$.\n\n# Expansion Formulas and Polarization Identities\n\n- Activity: Let \vec{u}andand\vec{v} be two vectors in the plane. Prove the expansion formulas.\n- Algebraic Identities for Vectors:\n - Norm of a sum: |\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + 2(\vec{u} \cdot \vec{v}) + |\vec{v}|^2\n - Norm of a difference: |\vec{u} - \vec{v}|^2 = |\vec{u}|^2 - 2(\vec{u} \cdot \vec{v}) + |\vec{v}|^2\n - Difference of scalar squares: (\vec{u} + \vec{v}) \cdot (\vec{u} - \vec{v}) = |\vec{u}|^2 - |\vec{v}|^2\n - Polarization identity 1: \vec{u} \cdot \vec{v} = \frac{1}{2}\left(|\vec{u} + \vec{v}|^2 - |\vec{u}|^2 - |\vec{v}|^2\right)\n - Polarization identity 2: \vec{u} \cdot \vec{v} = \frac{1}{2}\left(|\vec{u}|^2 + |\vec{v}|^2 - |\vec{u} - \vec{v}|^2\right)\n - Polarization identity 3: \vec{u} \cdot \vec{v} = \frac{1}{4}\left(|\vec{u} + \vec{v}|^2 - |\vec{u} - \vec{v}|^2\right)\n- Exercise 12: Let \vec{u}andand\vec{v}betwoorthogonalvectorswithbe two orthogonal vectors with|\vec{u}| = 3andand|\vec{v}| = 2. Calculate:\n - |2\vec{u} - 3\vec{v}|^2\n - (3\vec{u} - \vec{v}) \cdot (\vec{u} + 2\vec{v})\n - |\vec{u} + \vec{v}|^2 - |\vec{u} - \vec{v}|^2\n- Exercise 13: Let \vec{u}andand\vec{v} be two vectors. Prove the equivalence:\n  |\vec{u} + \vec{v}| = |\vec{u} - \vec{v}| \iff \vec{u} \perp \vec{v}\n\n# Analytical Expression of the Dot Product\n\n- Activity: In the vector plane equipped with an orthonormal basis (\vec{i}, \vec{j}),considervectors, consider vectors\vec{u} = x\vec{i} + y\vec{j}andand\vec{v} = x'\vec{i} + y'\vec{j}.\n 1. Express using \vec{i}andand\vec{j}eachoftherealnumberseach of the real numbers\vec{i} \cdot \vec{i},,\vec{i} \cdot \vec{j},and, and|\vec{i}|^2$.
    2. Express u⃗⋅v⃗\vec{u} \cdot \vec{v} using xx, yy, x′x', and y′y'.
  • Analytical Expression Theorem: Let (i⃗,j⃗)(\vec{i}, \vec{j}) be an orthonormal basis of the plane. For all vectors u⃗(x,y)\vec{u}(x, y) and v⃗(x′,y′)\vec{v}(x', y'), the dot product is given by:   u⃗⋅v⃗=xx′+yy′\vec{u} \cdot \vec{v} = xx' + yy'
  • Exercise 14: Let ABCDABCD be a square of side length a=2a = 2 and PP a point inside the square such that ABPABP is an equilateral triangle. Let II and JJ be points on segments [AB][AB] and [AD][AD] respectively such that AI=AJ=1AI = AJ = 1.
    1. Determine the coordinates of points AA, BB, CC, DD, and PP in the frame of reference (A,AI⃗,AJ⃗)(A, \vec{AI}, \vec{AJ}).
    2. Calculate in terms of aa the real numbers AP⃗⋅AB⃗\vec{AP} \cdot \vec{AB}, AP⃗⋅AD⃗\vec{AP} \cdot \vec{AD}, and CP⃗⋅CB⃗\vec{CP} \cdot \vec{CB}.
  • Exercise 15: In an orthonormal frame of reference, given vectors u⃗(2,3)\vec{u}(2, 3), v⃗(−1,4)\vec{v}(-1, 4), and w⃗(5,−2)\vec{w}(5, -2).
    1. Calculate u⃗⋅v⃗\vec{u} \cdot \vec{v}.
    2. Provide an approximate value rounded to 1 degree of the angle (u⃗,v⃗^)(\widehat{\vec{u}, \vec{v}}).
  • Exercise 16: In an orthonormal frame (O,i⃗,j⃗)(O, \vec{i}, \vec{j}), given vectors u⃗(2,−3)\vec{u}(2, -3) and v⃗(6,4)\vec{v}(6, 4).
    1. Calculate the real number u⃗⋅v⃗\vec{u} \cdot \vec{v}.
    2. Deduce the relative position of line (D1)(D_1) directed by u⃗\vec{u} and line (D2)(D_2) directed by v⃗\vec{v}.
  • Exercise 17: Let ABCABC be a triangle with centroid GG and II the midpoint of segment [BC][BC] such that AB=4AB = 4, AC=6AC = 6, and BC=5BC = 5. Calculate the dot product AB⃗⋅AC⃗\vec{AB} \cdot \vec{AC}.
  • Exercise 18: In the plane equipped with an orthonormal frame, consider points A(1,2)A(1, 2), B(3,4)B(3, 4), and C(−1,0)C(-1, 0).
    1. Determine a Cartesian equation of the set E\mathcal{E} of points M(x,y)M(x, y) such that \vec{MA} \cdot \vec{MB} = 0$.\n 2. Determine a Cartesian equation of the set \mathcal{E}_1ofpointsof pointsM(x, y)suchthatsuch that\vec{MA} \cdot \vec{MB} = 12$.
    2. Determine a Cartesian equation of the set E2\mathcal{E}_2 of points M(x,y)M(x, y) such that \vec{MA} \cdot \vec{MC} = 0$.\n\n# Level Sets (Lignes de Niveau)\n\n- Activity 1 (Orthogonal Projection Level Sets): Let AandandB be two points in the plane.\n 1. Determine the set of points Mintheplanesatisfyingin the plane satisfying\vec{MA} \cdot \vec{MB} = 0$.
    3. Assume AB=6AB = 6      a. Position point HH on line (AB)(AB) such that \vec{AB} \cdot \vec{AH} = 12$.\n     b. Determine the set of points Mintheplanesuchthatin the plane such that\vec{AB} \cdot \vec{AM} = 12$.
  • Activity 2 (Difference of Squares Level Sets): Let AA and BB be two points in the plane.
    1. Determine the set of points MM in the plane such that \vec{MA}^2 - \vec{MB}^2 = 0$.\n 2. Let Ibethemidpointofsegmentbe the midpoint of segment[AB].\n     a. Show that \vec{MA}^2 - \vec{MB}^2 = 2\vec{IM} \cdot \vec{AB}.\n     b. Determine the set of points Mintheplanesuchthatin the plane such that\vec{MA}^2 - \vec{MB}^2 = 12$.
  • Activity 3 (Sum of Squares and Median Theorem): Let AA and BB be two points in the plane, II the midpoint of segment [AB][AB], kk a real number, and EkE_k the set of points MM in the plane such that MA^2 + MB^2 = k$.\n 1. Determine E_kwhenwhenk = 0$.
    1. Show that for every point MM in the plane: MA2+MB2=2MI2+12AB2MA^2 + MB^2 = 2MI^2 + \frac{1}{2}AB^2 (Median Theorem / Théorème de la médiane).
    2. Determine EkE_k when k=12k = 12 and when k = 0$.\n- Activity 4 (Dot Product Level Sets): Let AandandBbetwopointsoftheplaneandbe two points of the plane andIthemidpointofthe midpoint of[AB].\n 1. Show that for every point Moftheplane:of the plane:\vec{MA} \cdot \vec{MB} = MI^2 - \frac{1}{4}AB^2$.
    3. Determine and construct the set of points MM of the plane such that \vec{MA} \cdot \vec{MB} = k$.\n- Activity 5 (Barycentric Combinations): In the plane equipped with an orthonormal frame, consider points A(1, 2)andandB(3, 4).\n 1. Show that for every point Mintheplane:in the plane:2MA^2 + 3MB^2 = 5MG^2 + 2GA^2 + 3GB^2.\n 2. Let G_1bethebarycenterofweightedpointsbe the barycenter of weighted points(A, 2)andand(B, 3),and, andG_2thebarycenterofthe barycenter of(A, 2)andand(B, -3).\n     a. Show that 2\vec{MA} + 3\vec{MB} = 5\vec{MG_1}andand2\vec{MA} - 3\vec{MB} = -\vec{MG_2}.\n     b. Deduce the set of points Mintheplanesuchthatin the plane such that|2\vec{MA} + 3\vec{MB}| = 10$.
  • Exercise 19: In the plane equipped with an orthonormal frame, consider points A(1,2)A(1, 2) and B(−2,3)B(-2, 3). Let EE be the set of points MM of the plane such that 2MA2−3MB2=122MA^2 - 3MB^2 = 12. Let GG be the barycenter of weighted points (A,2)(A, 2) and (B,−3)(B, -3).
    1. Show that 2MA^2 - 3MB^2 = -MG^2 + 2GA^2 - 3GB^2$.\n 2. Show that Eisthesetofpointsis the set of pointsMoftheplanesuchthatof the plane such thatGM^2 = R^2$.
    2. Deduce that EE is a circle whose center GG and radius RR are to be determined.
  • Exercise 20: Let AA and BB be two points of the plane such that AB=4AB = 4. Let ff be the transformation of the plane mapping any point MM to point M′M' such that AM′⃗=116(AB⃗⋅AM⃗)AB⃗\vec{AM'} = \frac{1}{16}(\vec{AB} \cdot \vec{AM})\vec{AB}.
    1. Determine f(A)f(A) and f(B)f(B).
    2. Show that for every point MM in the plane, point M′M' belongs to line (AB)(AB).
    3. Characterize the image of any point belonging to line (AB)(AB).
    4. Show that for every point MM, vectors MM′⃗\vec{MM'} and AB⃗\vec{AB} are orthogonal.
    5. Deduce the precise geometric nature of the transformation ff (Orthogonal Projection onto line (AB)(AB)).