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8.5 Vectors in 2D and 3D, Dot Product, Cross Product & Scalar Triple Product

Key Takeaways

  • A vector in 3D Euclidean space is represented as v=xi+yj+zk\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} with magnitude ∣v∣=x2+y2+z2|\mathbf{v}| = \sqrt{x^2+y^2+z^2}; its direction cosines satisfy l2+m2+n2=1l^2+m^2+n^2=1 and sin⁡2α+sin⁡2β+sin⁡2γ=2\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2.

  • The dot product u⋅v=∣u∣∣v∣cos⁡θ=u1v1+u2v2+u3v3\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta = u_1 v_1 + u_2 v_2 + u_3 v_3 evaluates orthogonality (u⋅v=0\mathbf{u} \cdot \mathbf{v} = 0), projection u⋅v∣v∣\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|}, and work done W=F⋅dW = \mathbf{F} \cdot \mathbf{d}.

  • The cross product u×v=(∣u∣∣v∣sin⁡θ)n^\mathbf{u} \times \mathbf{v} = (|\mathbf{u}||\mathbf{v}|\sin\theta)\hat{\mathbf{n}} is evaluated via 3×33\times 3 determinant, evaluates parallelism (u×v=0\mathbf{u} \times \mathbf{v} = \mathbf{0}), and determines parallelogram area ∣u×v∣|\mathbf{u} \times \mathbf{v}| and triangle area 12∣u×v∣\frac{1}{2}|\mathbf{u} \times \mathbf{v}|.

  • The scalar triple product [u  v  w]=u⋅(v×w)[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) computes the volume of a parallelepiped (volume of tetrahedron is 16∣[u  v  w]∣\frac{1}{6}|[\mathbf{u}\;\mathbf{v}\;\mathbf{w}]|), with [u  v  w]=0[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = 0 establishing coplanarity.

  • The vector triple product satisfies the non-associative BAC-CAB identity u×(v×w)=(u⋅w)v−(u⋅v)w\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u}\cdot\mathbf{w})\mathbf{v} - (\mathbf{u}\cdot\mathbf{v})\mathbf{w}, and connects with dot and cross products via Lagrange's identity ∣u×v∣2=∣u∣2∣v∣2−(u⋅v)2|\mathbf{u}\times\mathbf{v}|^2 = |\mathbf{u}|^2|\mathbf{v}|^2 - (\mathbf{u}\cdot\mathbf{v})^2.

Last updated: October 2026

Vectors in 2D and 3D, Dot Product, Cross Product & Scalar Triple Product

Vector algebra forms the mathematical infrastructure connecting pure analytic geometry with classical mechanics in the Pakistan ECAT syllabus (Intermediate Mathematics Part II and Physics Part I). In the ECAT examination, questions routinely bridge vector operations with physical concepts such as mechanical work, torque, projections, and 3D spatial geometry. Candidates must be fluent in component forms, determinant evaluations of cross and triple products, and algebraic shortcuts that bypass multi-line matrix expansions.


1. Vector Representation in Euclidean Space and Magnitudes

A vector v\mathbf{v} in three-dimensional space R3\mathbb{R}^3 represents a directed line segment characterized by both magnitude and direction. Given an initial point A(x1,y1,z1)A(x_1, y_1, z_1) and a terminal point B(x2,y2,z2)B(x_2, y_2, z_2), the displacement vector AB⃗\vec{AB} is:

v=AB⃗=(x2−x1)i+(y2−y1)j+(z2−z1)k=⟨x,y,z⟩\mathbf{v} = \vec{AB} = (x_2 - x_1)\mathbf{i} + (y_2 - y_1)\mathbf{j} + (z_2 - z_1)\mathbf{k} = \langle x, y, z \rangle

where i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} represent the standard right-handed orthonormal unit basis vectors directed along the positive xx-, yy-, and zz-axes, respectively.

Magnitude (Norm) of a Vector

The Euclidean length or magnitude of v=xi+yj+zk\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} is given by:

∣v∣=∥v∥=x2+y2+z2|\mathbf{v}| = \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2}

The Unit Vector

A unit vector v^\hat{\mathbf{v}} parallel to v\mathbf{v} with unit magnitude (∣v^∣=1|\hat{\mathbf{v}}| = 1) is obtained by dividing v\mathbf{v} by its magnitude:

v^=v∣v∣=xx2+y2+z2i+yx2+y2+z2j+zx2+y2+z2k\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{x}{\sqrt{x^2+y^2+z^2}}\mathbf{i} + \frac{y}{\sqrt{x^2+y^2+z^2}}\mathbf{j} + \frac{z}{\sqrt{x^2+y^2+z^2}}\mathbf{k}


2. Direction Angles, Direction Cosines, and Direction Ratios

Let α,β,γ∈[0,π]\alpha, \beta, \gamma \in [0, \pi] denote the angles that non-zero vector v=xi+yj+zk\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} forms with the positive xx-, yy-, and zz-axes, respectively.

Direction Cosines (l,m,nl, m, n)

The cosines of these directional angles are the direction cosines of v\mathbf{v}:

l=cos⁡α=x∣v∣,m=cos⁡β=y∣v∣,n=cos⁡γ=z∣v∣l = \cos\alpha = \frac{x}{|\mathbf{v}|}, \quad m = \cos\beta = \frac{y}{|\mathbf{v}|}, \quad n = \cos\gamma = \frac{z}{|\mathbf{v}|}

Notice that the unit vector is composed directly of these direction cosines: v^=li+mj+nk\hat{\mathbf{v}} = l\mathbf{i} + m\mathbf{j} + n\mathbf{k}.

Core Direction Cosine Identities

  1. Sum of Squares of Cosines:

l2+m2+n2=cos⁡2α+cos⁡2β+cos⁡2γ=x2+y2+z2∣v∣2=1l^2 + m^2 + n^2 = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = \frac{x^2 + y^2 + z^2}{|\mathbf{v}|^2} = 1

  1. Sum of Squares of Sines: Using sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta:

sin⁡2α+sin⁡2β+sin⁡2γ=(1−cos⁡2α)+(1−cos⁡2β)+(1−cos⁡2γ)=3−(cos⁡2α+cos⁡2β+cos⁡2γ)=3−1=2\sin^2\alpha + \sin^2\beta + \sin^2\gamma = (1 - \cos^2\alpha) + (1 - \cos^2\beta) + (1 - \cos^2\gamma) = 3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma) = 3 - 1 = 2

ECAT Trap Alert: Questions frequently ask for sin⁡2α+sin⁡2β+sin⁡2γ\sin^2\alpha + \sin^2\beta + \sin^2\gamma or cos⁡(2α)+cos⁡(2β)+cos⁡(2γ)\cos(2\alpha) + \cos(2\beta) + \cos(2\gamma). Recall cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta) = 2\cos^2\theta - 1, so ∑cos⁡(2α)=2(l2+m2+n2)−3=2(1)−3=−1\sum \cos(2\alpha) = 2(l^2 + m^2 + n^2) - 3 = 2(1) - 3 = -1.

Direction Ratios

Any triad of real numbers a,b,ca, b, c proportional to the direction cosines l,m,nl, m, n are termed direction ratios (a:b:c=l:m:na : b : c = l : m : n). Direction cosines are recovered via:

l=a±a2+b2+c2,m=b±a2+b2+c2,n=c±a2+b2+c2l = \frac{a}{\pm\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\pm\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\pm\sqrt{a^2 + b^2 + c^2}}


3. The Dot Product (Scalar Product)

Geometric and Component Definitions

The scalar (dot) product of two vectors u\mathbf{u} and v\mathbf{v} separated by angle θ∈[0,π]\theta \in [0, \pi] is defined as:

u⋅v=∣u∣∣v∣cos⁡θ\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos\theta

In Cartesian components where u=u1i+u2j+u3k\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k} and v=v1i+v2j+v3k\mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k}:

u⋅v=u1v1+u2v2+u3v3\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3

Fundamental Properties of the Dot Product

  • Commutative Law: u⋅v=v⋅u\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}.
  • Distributive Law: u⋅(v+w)=u⋅v+u⋅w\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}.
  • Self Dot Product: u⋅u=∣u∣2  ⟹  ∣u∣=u⋅u\mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2 \implies |\mathbf{u}| = \sqrt{\mathbf{u} \cdot \mathbf{u}}.
  • Orthonormal Basis Relations:
    • i⋅i=j⋅j=k⋅k=1\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1
    • i⋅j=j⋅k=k⋅i=0\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0

Condition for Orthogonality

Two non-zero vectors u\mathbf{u} and v\mathbf{v} are mutually perpendicular (orthogonal, θ=90∘\theta = 90^\circ) if and only if their scalar product vanishes:

u⊥v  ⟺  u⋅v=0  ⟺  u1v1+u2v2+u3v3=0\mathbf{u} \perp \mathbf{v} \iff \mathbf{u} \cdot \mathbf{v} = 0 \iff u_1 v_1 + u_2 v_2 + u_3 v_3 = 0

Angle Between Two Vectors

cos⁡θ=u⋅v∣u∣∣v∣=u1v1+u2v2+u3v3u12+u22+u32v12+v22+v32\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|} = \frac{u_1 v_1 + u_2 v_2 + u_3 v_3}{\sqrt{u_1^2 + u_2^2 + u_3^2} \sqrt{v_1^2 + v_2^2 + v_3^2}}

Scalar and Vector Projections

  • Scalar Projection (Component) of u\mathbf{u} onto v\mathbf{v}: The directed length of the shadow of u\mathbf{u} along v\mathbf{v} is:

projvu=∣u∣cos⁡θ=u⋅v∣v∣=u⋅v^\text{proj}_{\mathbf{v}}\mathbf{u} = |\mathbf{u}|\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|} = \mathbf{u} \cdot \hat{\mathbf{v}}

  • Vector Projection of u\mathbf{u} onto v\mathbf{v}: The vector component parallel to v\mathbf{v} is:

u∥v=(u⋅v∣v∣2)v\mathbf{u}_{\parallel \mathbf{v}} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2} \right) \mathbf{v}

Physical Application: Work Done

The mechanical work WW performed by a constant force vector F\mathbf{F} moving a body through displacement d\mathbf{d} is:

W=F⋅d=∣F∣∣d∣cos⁡θW = \mathbf{F} \cdot \mathbf{d} = |\mathbf{F}||\mathbf{d}|\cos\theta


4. The Cross Product (Vector Product)

Geometric and Component Definitions

The vector (cross) product of two vectors u\mathbf{u} and v\mathbf{v} is a vector perpendicular to the plane containing u\mathbf{u} and v\mathbf{v}:

u×v=(∣u∣∣v∣sin⁡θ)n^\mathbf{u} \times \mathbf{v} = (|\mathbf{u}| |\mathbf{v}| \sin\theta) \hat{\mathbf{n}}

where θ∈[0,π]\theta \in [0, \pi] is the interior angle and n^\hat{\mathbf{n}} is a unit normal vector whose direction is determined by the right-hand rule. In component form, the cross product is evaluated via the 3×33 \times 3 determinant:

u×v=∣ijku1u2u3v1v2v3∣=(u2v3−u3v2)i−(u1v3−u3v1)j+(u1v2−u2v1)k\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = (u_2 v_3 - u_3 v_2)\mathbf{i} - (u_1 v_3 - u_3 v_1)\mathbf{j} + (u_1 v_2 - u_2 v_1)\mathbf{k}

Fundamental Properties of the Cross Product

  • Anti-Commutative: u×v=−(v×u)\mathbf{u} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{u}).
  • Distributive Law: u×(v+w)=u×v+u×w\mathbf{u} \times (\mathbf{v} + \mathbf{w}) = \mathbf{u} \times \mathbf{v} + \mathbf{u} \times \mathbf{w}.
  • Self Cross Product: u×u=0\mathbf{u} \times \mathbf{u} = \mathbf{0}.
  • Cyclic Basis Identities:
    • i×j=k,j×k=i,k×i=j\mathbf{i} \times \mathbf{j} = \mathbf{k}, \quad \mathbf{j} \times \mathbf{k} = \mathbf{i}, \quad \mathbf{k} \times \mathbf{i} = \mathbf{j}
    • j×i=−k,k×j=−i,i×k=−j\mathbf{j} \times \mathbf{i} = -\mathbf{k}, \quad \mathbf{k} \times \mathbf{j} = -\mathbf{i}, \quad \mathbf{i} \times \mathbf{k} = -\mathbf{j}
    • i×i=j×j=k×k=0\mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = \mathbf{0}

Collinearity / Parallelism Condition

Two non-zero vectors u\mathbf{u} and v\mathbf{v} are collinear (parallel or anti-parallel, θ=0∘\theta = 0^\circ or 180∘180^\circ) if and only if:

u×v=0  ⟺  u1v1=u2v2=u3v3\mathbf{u} \times \mathbf{v} = \mathbf{0} \iff \frac{u_1}{v_1} = \frac{u_2}{v_2} = \frac{u_3}{v_3}

Lagrange's Identity

The magnitude of the cross product is linked to the dot product by Lagrange's Identity:

∣u×v∣2=∣u∣2∣v∣2−(u⋅v)2|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 - (\mathbf{u} \cdot \mathbf{v})^2

ECAT Speed Trick: If asked for ∣u×v∣|\mathbf{u} \times \mathbf{v}| when ∣u∣,∣v∣,|\mathbf{u}|, |\mathbf{v}|, and u⋅v\mathbf{u} \cdot \mathbf{v} are known, never compute angles or trigonometric tables. Apply Lagrange's Identity directly: ∣u×v∣=∣u∣2∣v∣2−(u⋅v)2|\mathbf{u} \times \mathbf{v}| = \sqrt{|\mathbf{u}|^2 |\mathbf{v}|^2 - (\mathbf{u} \cdot \mathbf{v})^2}.

Geometric Applications of the Cross Product

  1. Area of a Parallelogram: With adjacent side vectors u\mathbf{u} and v\mathbf{v}:

Area=∣u×v∣\text{Area} = |\mathbf{u} \times \mathbf{v}|

If the diagonals of a parallelogram are vectors d1\mathbf{d}_1 and d2\mathbf{d}_2:

Area=12∣d1×d2∣\text{Area} = \frac{1}{2} |\mathbf{d}_1 \times \mathbf{d}_2|

  1. Area of a Triangle: With adjacent sides u\mathbf{u} and v\mathbf{v}:

Area=12∣u×v∣\text{Area} = \frac{1}{2} |\mathbf{u} \times \mathbf{v}|

  1. Unit Normal Vector to a Plane: A unit vector normal to the plane containing u\mathbf{u} and v\mathbf{v} is:

n^=±u×v∣u×v∣\hat{\mathbf{n}} = \pm \frac{\mathbf{u} \times \mathbf{v}}{|\mathbf{u} \times \mathbf{v}|}

  1. Physical Application: Torque (Moment of a Force): The moment of a force F\mathbf{F} acting at position r\mathbf{r} relative to pivot OO is:

τ=r×F\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}


5. The Scalar Triple Product (Box Product)

Definition and Determinant Representation

The scalar triple product of three vectors u,v,w\mathbf{u}, \mathbf{v}, \mathbf{w} is defined as the dot product of u\mathbf{u} with the cross product of v\mathbf{v} and w\mathbf{w}, denoted by [u  v  w][\mathbf{u} \; \mathbf{v} \; \mathbf{w}]:

[u  v  w]=u⋅(v×w)[\mathbf{u} \; \mathbf{v} \; \mathbf{w}] = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})

In terms of Cartesian components:

[u  v  w]=∣u1u2u3v1v2v3w1w2w3∣[\mathbf{u} \; \mathbf{v} \; \mathbf{w}] = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}

Algebraic Properties of the Box Product

  • Cyclic Invariance: The value is unchanged under cyclic permutation of the vectors:

[u  v  w]=[v  w  u]=[w  u  v][\mathbf{u} \; \mathbf{v} \; \mathbf{w}] = [\mathbf{v} \; \mathbf{w} \; \mathbf{u}] = [\mathbf{w} \; \mathbf{u} \; \mathbf{v}]

  • Anti-Cyclic Sign Reversal: Swapping any two adjacent vectors negates the sign:

[u  w  v]=−[u  v  w][\mathbf{u} \; \mathbf{w} \; \mathbf{v}] = -[\mathbf{u} \; \mathbf{v} \; \mathbf{w}]

  • Dot and Cross Interchangeability:

u⋅(v×w)=(u×v)⋅w\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}

  • Vanishing Condition: If any two vectors are identical or linearly dependent (parallel), the box product is zero: [u  u  v]=0[\mathbf{u} \; \mathbf{u} \; \mathbf{v}] = 0.

Geometric Applications: Volumes and Coplanarity

  1. Volume of a Parallelepiped: The volume VV of a parallelepiped having coterminous edges defined by u,v,w\mathbf{u}, \mathbf{v}, \mathbf{w} is:

V=∣[u  v  w]∣=∣u⋅(v×w)∣V = \bigl| [\mathbf{u} \; \mathbf{v} \; \mathbf{w}] \bigr| = \bigl| \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \bigr|

  1. Volume of a Tetrahedron: The volume of a tetrahedron formed by coterminous edges u,v,w\mathbf{u}, \mathbf{v}, \mathbf{w} is:

Vtetrahedron=16∣[u  v  w]∣V_{\text{tetrahedron}} = \frac{1}{6} \bigl| [\mathbf{u} \; \mathbf{v} \; \mathbf{w}] \bigr|

  1. Coplanarity Condition: Three vectors u,v,w\mathbf{u}, \mathbf{v}, \mathbf{w} are coplanar (lie in the same geometric plane) if and only if their scalar triple product vanishes:

[u  v  w]=0  ⟺  ∣u1u2u3v1v2v3w1w2w3∣=0[\mathbf{u} \; \mathbf{v} \; \mathbf{w}] = 0 \iff \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} = 0


6. The Vector Triple Product

The vector triple product of three vectors u,v,w\mathbf{u}, \mathbf{v}, \mathbf{w} is defined as:

u×(v×w)\mathbf{u} \times (\mathbf{v} \times \mathbf{w})

Because (v×w)(\mathbf{v} \times \mathbf{w}) is normal to the plane of v\mathbf{v} and w\mathbf{w}, taking the cross product with u\mathbf{u} produces a vector that lies entirely within the plane of v\mathbf{v} and w\mathbf{w}.

The BAC-CAB Expansion Rule

The vector triple product expands into a linear combination of v\mathbf{v} and w\mathbf{w} via the celebrated BAC-CAB identity:

u×(v×w)=(u⋅w)v−(u⋅v)w\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{u} \cdot \mathbf{v})\mathbf{w}

ECAT Trap Alert: The vector cross product is not associative: u×(v×w)≠(u×v)×w\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) \neq (\mathbf{u} \times \mathbf{v}) \times \mathbf{w} Expanding the right-hand expression using BAC-CAB gives: (u×v)×w=−w×(u×v)=−[(w⋅v)u−(w⋅u)v]=(u⋅w)v−(v⋅w)u(\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = -\mathbf{w} \times (\mathbf{u} \times \mathbf{v}) = -[(\mathbf{w} \cdot \mathbf{v})\mathbf{u} - (\mathbf{w} \cdot \mathbf{u})\mathbf{v}] = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{v} \cdot \mathbf{w})\mathbf{u}


7. Master Reference Table of Vector Operations

OperationAlgebraic FormulaOutput TypePrimary Physical / Geometric Meaning
Magnitude∣v∣=x2+y2+z2\lvert\mathbf{v}\rvert = \sqrt{x^2 + y^2 + z^2}Scalar ≥0\ge 0Total Euclidean length / speed
Dot Productu⋅v=∣u∣∣v∣cos⁡θ=u1v1+u2v2+u3v3\mathbf{u} \cdot \mathbf{v} = \lvert\mathbf{u}\rvert\lvert\mathbf{v}\rvert\cos\theta = u_1 v_1 + u_2 v_2 + u_3 v_3ScalarOrthogonality test (u⋅v=0\mathbf{u} \cdot \mathbf{v} = 0), Work (W=F⋅dW = \mathbf{F} \cdot \mathbf{d})
Scalar Projectionprojvu=u⋅v∣v∣\text{proj}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\lvert\mathbf{v}\rvert}ScalarDirected magnitude of shadow on target axis
Cross Productu×v=∣ijku1u2u3v1v2v3∣\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}VectorParallelism test (u×v=0\mathbf{u} \times \mathbf{v} = \mathbf{0}), Torque (τ=r×F\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F})
Parallelogram AreaA=∣u×v∣A = \lvert\mathbf{u} \times \mathbf{v}\rvertScalar ≥0\ge 0Area enclosed by adjacent edge vectors
Triangle AreaA=12∣u×v∣A = \frac{1}{2}\lvert\mathbf{u} \times \mathbf{v}\rvertScalar ≥0\ge 0Area enclosed by two adjacent sides
Scalar Triple Product[u  v  w]=∣u1u2u3v1v2v3w1w2w3∣[\mathbf{u} \; \mathbf{v} \; \mathbf{w}] = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}ScalarVolume of parallelepiped, Coplanarity ([u  v  w]=0[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = 0)
Tetrahedron VolumeV=16∣[u  v  w]∣V = \frac{1}{6}\bigl\lvert[\mathbf{u} \; \mathbf{v} \; \mathbf{w}]\bigr\rvertScalar ≥0\ge 0Volume enclosed by 4 triangular faces
Vector Triple Productu×(v×w)=(u⋅w)v−(u⋅v)w\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{u} \cdot \mathbf{v})\mathbf{w}VectorVector in plane of v,w\mathbf{v}, \mathbf{w} orthogonal to u\mathbf{u}
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Vector Operations and Applications Flowchart
Test Your Knowledge

What is the volume of the parallelepiped whose coterminous edges are formed by the vectors u=i+2j+3k\mathbf{u} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}, v=2i+3j+k\mathbf{v} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}, and w=3i+j+2k\mathbf{w} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}?

A

1212

B

1414

C

1616

D

1818

Test Your Knowledge

If two vectors u\mathbf{u} and v\mathbf{v} satisfy ∣u∣=3|\mathbf{u}| = 3, ∣v∣=4|\mathbf{v}| = 4, and u⋅v=6\mathbf{u} \cdot \mathbf{v} = 6, what is the magnitude of their vector cross product ∣u×v∣|\mathbf{u} \times \mathbf{v}|?

A

66

B

626\sqrt{2}

C

636\sqrt{3}

D

1212

Test Your Knowledge

For what real value of α\alpha are the three vectors a=2i−j+k\mathbf{a} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}, b=i+2j−3k\mathbf{b} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}, and c=3i+αj+5k\mathbf{c} = 3\mathbf{i} + \alpha\mathbf{j} + 5\mathbf{k} coplanar?

A

−4-4

B

44

C

−2-2

D

00

Sections you finish are checked off in the contents.