Physics Lab

Vector (Cross) Product

Rotational quantities — torque, angular momentum, angular velocity — are all defined using the cross product of vectors. Unlike the dot product (which gives a scalar), the cross product gives another vector, perpendicular to both inputs. It encodes both magnitude and sense of rotation.

Concept

The vector (cross) product of two vectors a\vec{a} and b\vec{b} is

a×b=absinθn^\vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}

where θ\theta is the angle between them (0θπ0 \le \theta \le \pi) and n^\hat{n} is a unit vector perpendicular to both, in the direction given by the right-hand rule: curl the fingers of the right hand from a\vec{a} to b\vec{b}; the thumb points along n^\hat{n}.

Key properties:

  • Anticommutative: a×b=b×a\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}.
  • Distributive: a×(b+c)=a×b+a×c\vec{a}\times(\vec{b}+\vec{c}) = \vec{a}\times\vec{b} + \vec{a}\times\vec{c}.
  • a×a=0\vec{a}\times\vec{a} = \vec{0}.
  • a×b|\vec{a}\times\vec{b}| equals the area of the parallelogram with sides a\vec{a}, b\vec{b}.

In Cartesian components with unit vectors i^\hat{i}, j^\hat{j}, k^\hat{k}:

i^×j^=k^,j^×k^=i^,k^×i^=j^\hat{i}\times\hat{j} = \hat{k}, \quad \hat{j}\times\hat{k} = \hat{i}, \quad \hat{k}\times\hat{i} = \hat{j}

For a=axi^+ayj^+azk^\vec{a} = a_x\hat{i}+a_y\hat{j}+a_z\hat{k} and similarly b\vec{b}:

a×b=i^j^k^axayazbxbybz\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_x & a_y & a_z\\ b_x & b_y & b_z\end{vmatrix}

=(aybzazby)i^(axbzazbx)j^+(axbyaybx)k^= (a_y b_z - a_z b_y)\hat{i} - (a_x b_z - a_z b_x)\hat{j} + (a_x b_y - a_y b_x)\hat{k}

Use in rotation:

  • Torque: τ=r×F\vec{\tau} = \vec{r}\times\vec{F}.
  • Angular momentum: L=r×p\vec{L} = \vec{r}\times\vec{p}.
  • Velocity of rotating particle: v=ω×r\vec{v} = \vec{\omega}\times\vec{r}.

Derivation

Area of parallelogram interpretation. Two vectors a\vec{a} and b\vec{b} at angle θ\theta form a parallelogram. Its base is a|\vec{a}| and height bsinθ|\vec{b}|\sin\theta, so area =absinθ=a×b= |\vec{a}||\vec{b}|\sin\theta = |\vec{a}\times\vec{b}|.

v=ω×r\vec{v} = \vec{\omega}\times\vec{r}. A particle at position r\vec{r} from the axis, with angular velocity ω\vec{\omega} along the axis, moves with linear velocity perpendicular to both: magnitude ωrsinθ\omega r \sin\theta (where θ\theta is angle between ω\vec{\omega} and r\vec{r}, and rsinθr\sin\theta is the perpendicular distance from the axis). Direction is tangential, given by the right-hand rule.

Worked Example

Find a×b\vec{a}\times\vec{b} for a=2i^+3j^k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}, b=i^j^+2k^\vec{b} = \hat{i} - \hat{j} + 2\hat{k}.

Solution:

a×b=i^j^k^231112\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 2&3&-1\\ 1&-1&2\end{vmatrix}

=i^(32(1)(1))j^(22(1)(1))+k^(2(1)3(1))= \hat{i}(3 \cdot 2 - (-1)(-1)) - \hat{j}(2 \cdot 2 - (-1)(1)) + \hat{k}(2(-1) - 3(1))

=i^(61)j^(4+1)+k^(23)=5i^5j^5k^= \hat{i}(6 - 1) - \hat{j}(4 + 1) + \hat{k}(-2 - 3) = 5\hat{i} - 5\hat{j} - 5\hat{k}

Check perpendicularity: (5)(2)+(5)(3)+(5)(1)=1015+5=0(5)(2) + (-5)(3) + (-5)(-1) = 10 - 15 + 5 = 0. ✓ (a×b\vec{a}\times\vec{b} is perpendicular to a\vec{a}.)

Common Confusions

  • Cross product is not commutative. a×b=b×a\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}. Order matters.
  • Sin, not cos. Dot product uses cos\cos; cross uses sin\sin. Parallel vectors have zero cross product; perpendicular vectors have maximum cross product magnitude.
  • Right-hand rule is the convention for direction; left-handed would flip all signs.
  • Magnitude not enough. A cross product needs both magnitude and direction; always determine the direction.

Key Takeaways

  • a×b=absinθn^\vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}; direction from right-hand rule.
  • Anticommutative: a×b=b×a\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}.
  • a×b|\vec{a}\times\vec{b}| = area of parallelogram with sides a\vec{a}, b\vec{b}.
  • Components: determinant of the 3×33\times 3 matrix.
  • Underpins torque, angular momentum, and rotational velocity in physics.

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