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Cross Product

We saw that the dot product between two vectors produces a scalar.
Now we introduce the cross product (vector product) of two vectors in R3\reals^3 which produces a vector in R3\bm{\reals^3}.

Definition

The cross product of vectors a⃗=[a1a2a3]\vec{a}= \begin{bmatrix} a_1\\ a_2\\ a_3 \end{bmatrix} and b⃗=[b1b2b3]\vec{b}= \begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix} in R3\reals^3 is
aβƒ—Γ—bβƒ—=[a2b3 βˆ’ a3b2a3b1 βˆ’ a1b3a1b2 βˆ’ a2b1]\boxed{\quad \vec{a}\times\vec{b} = \begin{bmatrix} a_2 b_3 \ βˆ’\ a_3 b_2\\ a_3 b_1 \ βˆ’\ a_1 b_3\\ a_1 b_2 \ βˆ’\ a_2 b_1\\ \end{bmatrix} \quad}

Watch Out!
The cross product only works for vectors in R3\reals^3 (with the exception of a few higher dimensional spaces)

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Important Note
The cross product of vectors u⃗\vec u and v⃗\vec v is orthogonal to both u⃗\vec u and v⃗\vec v.

Wize Tip
Use the right hand rule to determine the direction of the vector u⃗×v⃗\vec u \times \vec v:

  1. Point the fingers of your right hand in the direction of u⃗\vec u
  2. Curl your fingers towards v⃗\vec v
  3. u⃗×v⃗\vec u \times \vec v is in the direction your thumb is pointing

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Shoelace Method

  1. Copy the column vector a⃗\vec a twice on the left, and b⃗\vec b twice on the right.
  2. Cross out the top and bottom rows, then "lace" the remaining diagonals by multiplying diagonally.
  3. Each diagonal pair makes a single component. Write a minus sign between every pair of products.

Pointing Method

(See video)
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Example: Cross Product

Find a vector that is orthogonal to both uβƒ—=⟨1,βˆ’2,βˆ’1⟩\vec{u}=\lang 1,-2,-1 \rang and vβƒ—=⟨0,3,4⟩\vec{v}=\lang 0,3,4 \rang . Check your answer.

Using the Shoelace Method:

uβƒ—Γ—vβƒ—=[(βˆ’2)(4)  βˆ’  (βˆ’1)(3)(βˆ’1)(0) βˆ’   (1)(4)(1)(3)   βˆ’  (βˆ’2)(0)]=[βˆ’5βˆ’43]\begin{aligned} \vec{u} Γ— \vec{v} &= \begin{bmatrix} (βˆ’2)(4) \ \ βˆ’ \ \ (βˆ’1)(3)\\[0.5em] (βˆ’1)(0)\ βˆ’ \ \ \ (1)(4)\\[0.5em] (1)(3) \ \ \ βˆ’\ \ (βˆ’2)(0)\\ \end{bmatrix}\\[2.5em] &= \boxed{ \begin{bmatrix} -5\\ -4\\ 3\\ \end{bmatrix} } \end{aligned}

Check that this is orthogonal to u⃗\vec u and v⃗\vec v by making sure each dot product is 0:
uβƒ—β‹…(uβƒ—Γ—vβƒ—)=[1βˆ’2βˆ’1]β‹…[βˆ’5βˆ’43]=(1)(βˆ’5)+(βˆ’2)(βˆ’4)+(βˆ’1)(3)=βˆ’5+8βˆ’3=0\begin{aligned} \vec u \cdot (\vec u \times \vec v) &= \begin{bmatrix} 1\\ -2\\ -1\\ \end{bmatrix} \cdot \begin{bmatrix} -5\\ -4\\ 3\\ \end{bmatrix}\\[1.5em] &= (1)(-5) + (-2)(-4) + (-1)(3)\\ &= -5+8-3\\ &= 0 \end{aligned} vβƒ—β‹…(uβƒ—Γ—vβƒ—)=[034]β‹…[βˆ’5βˆ’43]=(0)(βˆ’5)+(3)(βˆ’4)+(4)(3)=0βˆ’12+12=0\begin{aligned} \vec v \cdot (\vec u \times \vec v) &= \begin{bmatrix} 0\\ 3\\ 4\\ \end{bmatrix} \cdot \begin{bmatrix} -5\\ -4\\ 3\\ \end{bmatrix}\\[1.5em] &= (0)(-5) + (3)(-4) + (4)(3)\\ &= 0-12+12\\ &= 0 \end{aligned}

Practice: Orthogonal Vectors

Find a vector orthogonal to both uβƒ—=[2βˆ’13]\vec u = \begin{bmatrix} 2\\ -1\\ 3 \end{bmatrix} and vβƒ—=[0βˆ’1βˆ’1]\vec{v} = \begin{bmatrix} 0\\ -1\\ -1\\ \end{bmatrix}.

For extra practice, check your answer!

Extra Practice