Notes on Cartesian Vectors: Dot Product, Cross Product, and Projections

Cartesian Vectors in Cartesian Form

A vector in three dimensions can be written in Cartesian form using the three orthogonal unit vectors \mathbf{i}, \mathbf{j}, \mathbf{k}: each vector has components along x, y, z. If
\mathbf{a} = ax \mathbf{i} + ay \mathbf{j} + az \mathbf{k} and \mathbf{b} = bx \mathbf{i} + by \mathbf{j} + bz \mathbf{k}, then:

  • Vector addition is performed componentwise:
    \n\mathbf{a} + \\mathbf{b} = (ax + bx) \mathbf{i} + (ay + by) \mathbf{j} + (az + bz) \mathbf{k}.
  • Vector subtraction is also componentwise:
    \n\mathbf{b} - \mathbf{a} = (bx - ax) \mathbf{i} + (by - ay) \mathbf{j} + (bz - az) \mathbf{k}.

Dot Product: Definition, Geometry, and Basic Properties

The dot product of two Cartesian vectors is defined as
<br/>a⋅b=a<em>xb</em>x+a<em>yb</em>y+a<em>zb</em>z.<br/><br /> \mathbf{a} \cdot \mathbf{b} = a<em>x b</em>x + a<em>y b</em>y + a<em>z b</em>z. <br />
Geometrically, it also equals the product of the magnitudes and the cosine of the angle between them:
<br/>a⋅b=∣a∣ ∣b∣ cos⁡θ.<br/><br /> \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| \, |\mathbf{b}| \, \cos \theta.<br />
This makes the dot product useful for finding angles and projections.

  • Utility: the dot product can be used to find the angle between two lines or vectors via
    cos⁡θ=fraca⋅b∣a∣ ∣b∣.\cos \theta = \\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \, |\mathbf{b}|}.
  • Parallel and perpendicular components relative to a line: the dot product helps determine components along a given direction.
  • The dot product is also called the scalar product, because the result is a scalar (not a vector).

Properties of the Dot Product

  • Commutative law:
    <br/>a⋅b=b⋅a.<br/><br /> \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}.<br />
  • Distributive law:
    <br/>a⋅(b+c)=(a⋅b)+(a⋅c).<br/><br /> \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = (\mathbf{a} \cdot \mathbf{b}) + (\mathbf{a} \cdot \mathbf{c}).<br />
  • Scalar multiplication: for any scalar \lambda,
    (lambdaa)⋅b=a⋅(lambdab)=lambda(a⋅b).<br/>(\\lambda \mathbf{a}) \cdot \mathbf{b} = \mathbf{a} \cdot (\\lambda \mathbf{b}) = \\lambda (\mathbf{a} \cdot \mathbf{b}).<br />
  • Unit vectors: \mathbf{3} = \hat{} \mathbf{i}, \mathbf{j}, \mathbf{k} play the role of the standard basis. Note that
    i⋅i=1,j⋅j=1,k⋅k=1,\mathbf{i} \cdot \mathbf{i} = 1, \mathbf{j} \cdot \mathbf{j} = 1, \mathbf{k} \cdot \mathbf{k} = 1,
    and any pair of distinct unit vectors are orthogonal:
    i⋅j=j⋅i=i⋅k=k⋅i=j⋅k=k⋅j=0.\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{i} = \mathbf{i} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{j} = 0.
    </li></ul><h5id="projectionandthegeometricinterpretationofthedotproduct">ProjectionandtheGeometricInterpretationoftheDotProduct</h5><ul><li>Thedotproductprovidestheprojectionofonevectorontoanother.Theprojectionofbontoahaslength∣b∣cosθ.Sincea⋅b=∣a∣∣b∣cosθ,itfollowsthatthedotproductequalsthemagnitudeofthefirstvectortimestheprojectionofthesecondontothefirst:</li><li>Scalarprojectionofbontoa:<br/></li> </ul> <h5 id="projectionandthegeometricinterpretationofthedotproduct">Projection and the Geometric Interpretation of the Dot Product</h5> <ul> <li>The dot product provides the projection of one vector onto another. The projection of \mathbf{b} onto \mathbf{a} has length |\mathbf{b}| cos θ. Since a \cdot b = |a| |b| cos θ, it follows that the dot product equals the magnitude of the first vector times the projection of the second onto the first:</li> <li>Scalar projection of \mathbf{b} onto \mathbf{a}: <br /> a \cdot b = |\mathbf{a}| \, (\text{projection of } \mathbf{b} \text{ onto } \mathbf{a}). </li><li>Ifweusetheunitvectoralonga,a^=a/∣a∣,thenthevectorprojectionofbontoa^is<br/></li> <li>If we use the unit vector along \mathbf{a}, \hat{a} = \mathbf{a} / |\mathbf{a}|, then the vector projection of \mathbf{b} onto \hat{a} is<br /> \text{proj}_{\hat{a}} \\mathbf{b} = (\mathbf{b} \cdot \hat{a}) \hat{a}. </li><li>Inaforceproblem,thedotproducthelpsquantifyhowmuchoneforceactsalongthedirectionofanotherviaprojection.</li></ul><h3id="dotproductwiththebasisvectorsandcomponents">DotProductwiththeBasisVectorsandComponents</h3><ul><li>Thedotproductofanyvectorwiththebasisvectorsgivesitscomponents:<br/></li> <li>In a force problem, the dot product helps quantify how much one force acts along the direction of another via projection.</li> </ul> <h3 id="dotproductwiththebasisvectorsandcomponents">Dot Product with the Basis Vectors and Components</h3> <ul> <li>The dot product of any vector with the basis vectors gives its components:<br /> \,\mathbf{a} \cdot \mathbf{i} = ax, \mathbf{a} \cdot \mathbf{j} = ay, \mathbf{a} \cdot \mathbf{k} = a_z. </li><li>Forunitvectorsinteractingwiththemselvesorwitheachother:<br/></li> <li>For unit vectors interacting with themselves or with each other:<br /> \mathbf{i} \cdot \mathbf{i} = 1,
    \mathbf{j} \cdot \mathbf{j} = 1,
    \mathbf{k} \cdot \mathbf{k} = 1,
    <br/><br /> \mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{i} = \mathbf{i} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{j} = 0. </li></ul><h3id="crossproductdefinitionmagnitudeanddirection">CrossProduct:Definition,Magnitude,andDirection</h3><p>Fortwovectorsaandb,thecrossproductisanothervector:<br/></li> </ul> <h3 id="crossproductdefinitionmagnitudeanddirection">Cross Product: Definition, Magnitude, and Direction</h3> <p>For two vectors \mathbf{a} and \mathbf{b}, the cross product is another vector:<br /> \mathbf{a} \times \mathbf{b} =
    \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \cr ax & ay & az \cr bx & by & bz \end{vmatrix} = (ay bz - az by) \mathbf{i} + (az bx - ax bz) \mathbf{j} + (ax by - ay bx) \mathbf{k}. </p><ul><li>Thecrossproductisavector,notascalar,anditisperpendiculartotheplanecontainingaandb.</li><li>Magnitude:</p> <ul> <li>The cross product is a vector, not a scalar, and it is perpendicular to the plane containing \mathbf{a} and \mathbf{b}.</li> <li>Magnitude: |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| \, |\mathbf{b}| \, \sin \theta. </li><li>Directionisgivenbytheright−handrule:pointthefingersfromatob,andthethumbpointsinthedirectionofa×b.</li><li>Thecrossproductisnotcommutative:<br/></li> <li>Direction is given by the right-hand rule: point the fingers from \mathbf{a} to \mathbf{b}, and the thumb points in the direction of \mathbf{a} \times \mathbf{b}.</li> <li>The cross product is not commutative: <br /> \mathbf{a} \times \mathbf{b}
    eq \mathbf{b} \times \mathbf{a}, \text{in fact } \mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a}). </li><li>Scalarmultiplicationanddistributivity:<br/></li> <li>Scalar multiplication and distributivity:<br /> (\lambda \\mathbf{a}) \times \mathbf{b} = \lambda (\mathbf{a} \times \mathbf{b}) = \\mathbf{a} \times (\lambda \\mathbf{b}), <br/><br /> \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}. </li></ul><h4id="crossproductwithunitvectorsandexamples">CrossProductwithUnitVectorsandExamples</h4><ul><li>Fundamentalcyclicrelations(right−handrule):<br/></li> </ul> <h4 id="crossproductwithunitvectorsandexamples">Cross Product with Unit Vectors and Examples</h4> <ul> <li>Fundamental cyclic relations (right-hand rule):<br /> \mathbf{i} \times \mathbf{j} = \mathbf{k}, \mathbf{j} \times \mathbf{k} = \mathbf{i}, \mathbf{k} \times \mathbf{i} = \mathbf{j}. </li><li>Antisymmetryinreversedorder:<br/></li> <li>Antisymmetry in reversed order:<br /> \mathbf{j} \times \mathbf{i} = -\mathbf{k}, \mathbf{k} \times \mathbf{j} = -\mathbf{i}, \mathbf{i} \times \mathbf{k} = -\mathbf{j}. </li><li>Crossproductsofidenticalunitvectorsvanish:<br/></li> <li>Cross products of identical unit vectors vanish:<br /> \mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = 0. </li></ul><h5id="crossproductincomponentsanddeterminants">CrossProductinComponentsandDeterminants</h5><ul><li><p>Ahandywaytocomputeaa×bisthedeterminant:<br/></li> </ul> <h5 id="crossproductincomponentsanddeterminants">Cross Product in Components and Determinants</h5> <ul> <li><p>A handy way to compute a \mathbf{a} \times \mathbf{b} is the determinant:<br /> \mathbf{a} \times \mathbf{b} = \det \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \cr ax & ay & az \cr bx & by & bz \end{pmatrix} = (ay bz - az by) \mathbf{i} + (az bx - ax bz) \mathbf{j} + (ax by - ay bx) \mathbf{k}. </p></li><li><p>Longhand(component−wise)expansionisconsistentwiththedeterminantmethodandisalsoshowninpracticeinlectures.</p></li></ul><h3id="projectionscomponentsandthreedimensionalgeometry">Projections,Components,andThree−DimensionalGeometry</h3><ul><li>Projectionofavectorontoaline(definedbyaunitvectoru^)canbeexpressedas:<br/></p></li> <li><p>Longhand (component-wise) expansion is consistent with the determinant method and is also shown in practice in lectures.</p></li> </ul> <h3 id="projectionscomponentsandthreedimensionalgeometry">Projections, Components, and Three-Dimensional Geometry</h3> <ul> <li>Projection of a vector onto a line (defined by a unit vector \hat{u}) can be expressed as:<br /> \text{Parallel component: } \mathbf{a}_{\parallel} = (\mathbf{a} \cdot \hat{u}) \hat{u}. </li><li>Thescalarprojection(lengthalongtheline)isa⋅u^.</li><li>Theperpendicularcomponentistheremainder:a<em>⊥=a−a</em>∥.</li><li>Ifyouwanttheprojectionofaontotheaxisdefinedbyanon−unitvector,use<br/></li> <li>The scalar projection (length along the line) is \mathbf{a} \cdot \hat{u}.</li> <li>The perpendicular component is the remainder: \mathbf{a}<em>{\perp} = \mathbf{a} - \mathbf{a}</em>{\parallel}.</li> <li>If you want the projection of a onto the axis defined by a non-unit vector, use<br /> \text{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \right) \mathbf{b}. </li><li>Foraunitaxisalongthex−direction,u^=i,andtheprojectionlengthisa<em>x,whiletheprojectionvectorisa</em>xi.</li><li>In3D,findingprojectionsbypuregeometryismorechallenging;thedotproductandunitvectorsprovideastraightforwardmethod.</li><li>Thesignofthedotproductindicatesdirection:iftheresultispositive,theprojectionalignswiththeunitvector;ifnegative,itpointsoppositetothataxis.</li></ul><h3id="practicalnotesvectoroperationsin3dproblems">PracticalNotes:VectorOperationsin3DProblems</h3><ul><li>Theresultantofscalarprojectionsalongaxescanbeusedtoanalyzehowoneforceprojectsontoanotherinagivendirection.</li><li>Thedotproductisakeytoolforresolvingforcesandcomponentsalongspecifieddirections,especiallywhencombiningmultipleforcesorcomponentsin3Dspace.</li><li>Remembertomaintainaconsistentright−handcoordinatesystemthroughoutproblems:inthestandardsystem,ipointsalong+x,jalong+y,andkalong+z(positiveztypicallyupwards).</li><li>Inallcross−products,watchfornon−commutativity:swappingtheorderflipsthesignoftheresult;indotproductstheorderdoesnotmatter.</li></ul><h3id="quickreferenceformulas">QuickReferenceFormulas</h3><ul><li>VectorsinCartesianform:<br/></li> <li>For a unit axis along the x-direction, \hat{u} = \mathbf{i}, and the projection length is a<em>x, while the projection vector is a</em>x \mathbf{i}.</li> <li>In 3D, finding projections by pure geometry is more challenging; the dot product and unit vectors provide a straightforward method.</li> <li>The sign of the dot product indicates direction: if the result is positive, the projection aligns with the unit vector; if negative, it points opposite to that axis.</li> </ul> <h3 id="practicalnotesvectoroperationsin3dproblems">Practical Notes: Vector Operations in 3D Problems</h3> <ul> <li>The resultant of scalar projections along axes can be used to analyze how one force projects onto another in a given direction.</li> <li>The dot product is a key tool for resolving forces and components along specified directions, especially when combining multiple forces or components in 3D space.</li> <li>Remember to maintain a consistent right-hand coordinate system throughout problems: in the standard system, \mathbf{i} points along +x, \mathbf{j} along +y, and \mathbf{k} along +z (positive z typically upwards).</li> <li>In all cross-products, watch for non-commutativity: swapping the order flips the sign of the result; in dot products the order does not matter.</li> </ul> <h3 id="quickreferenceformulas">Quick Reference Formulas</h3> <ul> <li>Vectors in Cartesian form:<br /> \mathbf{a} = ax \mathbf{i} + ay \mathbf{j} + az \mathbf{k}, \quad \ \mathbf{b} = bx \mathbf{i} + by \mathbf{j} + bz \mathbf{k}. </li><li>Sumanddifference:<br/></li> <li>Sum and difference:<br /> \mathbf{a} + \mathbf{b} = (ax + bx) \mathbf{i} + (ay + by) \mathbf{j} + (az + bz) \mathbf{k}, \\mathbf{b} - \mathbf{a} = (bx - ax) \mathbf{i} + (by - ay) \mathbf{j} + (bz - az) \mathbf{k}. </li><li>Dotproduct:<br/></li> <li>Dot product:<br /> \mathbf{a} \cdot \mathbf{b} = ax bx + ay by + az bz = |\mathbf{a}| |\mathbf{b}| \, \cos \theta. </li><li>Magnitudeofavector:<br/></li> <li>Magnitude of a vector:<br /> |\mathbf{a}| = \sqrt{ax^2 + ay^2 + a_z^2}. </li><li>Anglebetweenvectors:<br/></li> <li>Angle between vectors:<br /> \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \, |\mathbf{b}|}, \\theta = \arccos\left( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \ |\mathbf{b}|} \right). </li><li>Projection(vector)ontounitdirectionu^:<br/></li> <li>Projection (vector) onto unit direction \hat{u}:<br /> \text{proj}_{\hat{u}} \mathbf{a} = (\mathbf{a} \cdot \hat{u}) \hat{u}. </li><li>Crossproduct:<br/></li> <li>Cross product:<br /> \mathbf{a} \times \mathbf{b} = (ay bz - az by) \mathbf{i} + (az bx - ax bz) \mathbf{j} + (ax by - ay bx) \mathbf{k}, <br/><br /> |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin \theta. </li><li>Determinantform(alternative):<br/></li> <li>Determinant form (alternative):<br /> \mathbf{a} \times \mathbf{b} = \det \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \cr ax & ay & az \cr bx & by & bz \end{pmatrix}. </li><li>Cyclicbasisrelations(right−handrule):<br/></li> <li>Cyclic basis relations (right-hand rule):<br /> \mathbf{i} \times \mathbf{j} = \mathbf{k}, \mathbf{j} \times \mathbf{k} = \mathbf{i}, \mathbf{k} \times \mathbf{i} = \mathbf{j}, <br/>and<br/><br /> and<br /> \mathbf{j} \times \mathbf{i} = -\mathbf{k}, \mathbf{k} \times \mathbf{j} = -\mathbf{i}, \mathbf{i} \times \mathbf{k} = -\mathbf{j}. $$
  • Consistency reminder: always use the same right-hand coordinate system to avoid confusion across problems.