Wize University Linear Algebra Textbook > Products of Vectors

Cross Product

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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

a⃗×b⃗=[a2b3 − a3b2a3b1 − a1b3a1b2 − a2b1]\boxed{\quad \vec{a}\times\vec{b} = \begin{bmatrix} a_2 b_3 \ −\ a_3 b_2\\ a_3 b_1 \ −\ a_1 b_3\\ a_1 b_2 \ −\ a_2 b_1\\ \end{bmatrix} \quad}




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.




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.
Image illustrating Cross Product

Pointing Method

(See video)

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Example: Cross Product

Find a vector that is orthogonal to both u⃗=⟨1,−2,−1⟩\vec{u}=\lang 1,-2,-1 \rang and v⃗=⟨0,3,4⟩\vec{v}=\lang 0,3,4 \rang . Check your answer.


Using the Shoelace Method:

Image illustrating Example


u⃗×v⃗=[(−2)(4)  −  (−1)(3)(−1)(0) −   (1)(4)(1)(3)   −  (−2)(0)]=[−5−43]\begin{aligned} \vec{u} × \vec{v} &= \begin{bmatrix} (−2)(4) \ \ − \ \ (−1)(3)\\[0.5em] (−1)(0)\ − \ \ \ (1)(4)\\[0.5em] (1)(3) \ \ \ −\ \ (−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:

u⃗⋅(u⃗×v⃗)=[1−2−1]⋅[−5−43]=(1)(−5)+(−2)(−4)+(−1)(3)=−5+8−3=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} v⃗⋅(u⃗×v⃗)=[034]⋅[−5−43]=(0)(−5)+(3)(−4)+(4)(3)=0−12+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 u⃗=[2−13]\vec u = \begin{bmatrix} 2\\ -1\\ 3 \end{bmatrix} and v⃗=[0−1−1]\vec{v} = \begin{bmatrix} 0\\ -1\\ -1\\ \end{bmatrix}.


For extra practice, check your answer!


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Extra Practice

Dot and Cross Products: Vector Norm

If u⃗, v⃗, w⃗  ∈ R3\vec{u},\ \vec{v},\ \vec{w}\ \ \in\ R^3, which of the following operations is/are defined?

(i.e. which of the following can be calculated?)

Select all that are correct.

Practice Question: Vector Product

Practice Question: Vector Product

Given that 𝑢⃗=(1,1,−1)𝑢⃗=(1,1,−1) and 𝑣⃗=(0,−2,2),𝑣⃗=(0,−2,2), compute (𝑢⃗⋅𝑣⃗)(𝑣⃗×𝑢⃗).(𝑢⃗\cdot𝑣⃗)(𝑣⃗\times𝑢⃗).

If u=(2,0,−3), v=(−2,2,−4)u=\left(2,0,-3\right),\ v=\left(-2,2,-4\right) and w=(1,2,1)w=\left(1,2,1\right), find a vector that is orthogonal to uv⃗\vec{uv} and uw⃗\vec{uw}.

If 𝑢⃗=(2,0,−3),𝑣⃗=(−2,2,−4),𝑢⃗ = (2, 0, −3), \vec𝑣 = (−2, 2, −4), and 𝑤⃗=(1,2,1)𝑤⃗= (1,2,1), compute u⃗ × w⃗\vec{u}\ \times\ \vec{w}.

Give that 𝑢⃗𝑢⃗ and 𝑣⃗\vec𝑣 are vectors in R3\mathbb{R}^3, which of the following statements is/are always true?

i. 2𝑢⃗+3𝑣⃗=2𝑣⃗+3𝑢⃗2𝑢⃗ + 3\vec𝑣 = 2\vec𝑣 + 3𝑢⃗

ii. 2𝑢⃗.(−3𝑣⃗)=−2(3𝑢⃗.𝑣⃗)2𝑢⃗ . (−3\vec𝑣 ) = −2(3𝑢⃗ . \vec𝑣 )

If 𝑢⃗=(2,0,−3),𝑣⃗=(−2,2,−4),𝑢⃗ = (2, 0, −3), \vec𝑣 = (−2, 2, −4), and 𝑤⃗=(1,2,1)𝑤⃗= (1,2,1), compute u⃗ × w⃗\vec{u}\ \times\ \vec{w}.

If u⃗=(2,0,−3), v⃗=(−2,2,−4)\vec{u}=\left(2,0,-3\right),\ \vec{v}=\left(-2,2,-4\right) and w⃗=(1,2,1)\vec{w}=\left(1,2,1\right), find a vector that is orthogonal to uv⃗\vec{uv} and uw⃗\vec{uw}.

Practice: Cross Product

Practice: Cross Product

Consider u⃗=[−2, 3, −1]\vec{u}=\left[-2,\ 3,\ -1\right] and v⃗=[1, 0, −1]\vec{v}=\left[1,\ 0,\ -1\right].

Practice: Orthogonal Vectors

Practice: Orthogonal Vectors

Find a vector orthogonal to both 𝑢⃗=(2,−1,3)𝑢⃗=(2,−1,3) and 𝑣⃗=(0,−1,−1).\vec{𝑣} = (0, −1, −1).

Practice Question: Orthogonal Vectors

Practice Question: Orthogonal Vectors

Find a vector orthogonal to both 𝑢⃗=(2,−1,3)𝑢⃗=(2,−1,3) and 𝑣⃗=(0,−1,−1).\vec{𝑣} = (0, −1, −1).

Cross Product

Practice: Orthogonal Vectors

Find a vector orthogonal to both u⃗=[2−13]\vec u = \begin{bmatrix} 2\\ -1\\ 3 \end{bmatrix} and v⃗=[0−1−1]\vec{v} = \begin{bmatrix} 0\\ -1\\ -1\\ \end{bmatrix}.

For extra practice, check your answer!

Practice: Cross Product

Practice: Cross Product

Recall that i⃗=[1, 0, 0]\vec{i}=\left[1,\ 0,\ 0\right], j⃗=[0, 1, 0]\vec{j}=\left[0,\ 1,\ 0\right], and k⃗=[0, 0, 1]\vec{k}=\left[0,\ 0,\ 1\right].

Match the following cross products with the correct result.

Evaluate 2k^×(3i^×2j^)2\hat{k}\times(3\hat{i}\times2\hat{j})

Consider three vectors A⃗=2i^+j^,  B⃗=i^−2j^\vec{A}=2\hat{i}+\hat{j},\,\,\vec{B}=\hat{i}-2\hat{j}, and C⃗=−2i^+j^\vec{C}=-2\hat{i}+\hat{j}. The angle between C⃗\vec{C} and A⃗×B⃗\vec{A}\times\vec{B} is?

Practice Question: Orthogonal Vectors

Practice Question: Orthogonal Vectors

Find a vector orthogonal to both 𝑢⃗=(2,−1,3)𝑢⃗=(2,−1,3) and 𝑣⃗=(0,−1,−1).\vec{𝑣} = (0, −1, −1).

Which of the following vectors is orthogonal to both [1,2,3][1,2,3] and [−3,1,0][-3,1,0].

Find vec form line thru pt ortho to 2 vec form lines

Find the equation of the line that passes through the point (1,2,−3)\left(1,2,-3\right) and is orthogonal to the lines with equations:

(x,y,z)=(−5,0,1)+t(1,0,−1)(x,y,z)= (-5,0,1)+t(1,0,-1) and (x,y,z)=(3,−1,6)+t(0,2,1)(x,y,z)=(3,-1,6)+t(0,2,1)

Cross Product

Practice: Orthogonal Vectors

Find a vector orthogonal to both u⃗=[2−13]\vec u = \begin{bmatrix} 2\\ -1\\ 3 \end{bmatrix} and v⃗=[0−1−1]\vec{v} = \begin{bmatrix} 0\\ -1\\ -1\\ \end{bmatrix}.

For extra practice, check your answer!