Dot product vs cross product
Both products combine two vectors, but they produce very different outputs.
Dot product a ⃗ ⋅ b ⃗ \vec{a}\cdot\vec{b} a ⋅ b Cross product a ⃗ × b ⃗ \vec{a}\times\vec{b} a × b Output scalar vector Dimension any R n \mathbb{R}^n R n only R 3 \mathbb{R}^3 R 3 Key trig ∥ a ⃗ ∥ ∥ b ⃗ ∥ cos θ \|\vec{a}\|\|\vec{b}\|\cos\theta ∥ a ∥∥ b ∥ cos θ ∥ a ⃗ ∥ ∥ b ⃗ ∥ sin θ \|\vec{a}\|\|\vec{b}\|\sin\theta ∥ a ∥∥ b ∥ sin θ Zero when perpendicular (θ = 90 ° \theta=90° θ = 90° parallel (θ = 0 ° \theta=0° θ = 0° 180 ° 180° 180°
The dot product measures how much two vectors align ; the cross product measures how much they span — the result is a new vector perpendicular to both.
A geometric question
Given two non-parallel vectors a ⃗ \vec{a} a b ⃗ \vec{b} b R 3 \mathbb{R}^3 R 3 n ⃗ \vec{n} n
That is, we want n ⃗ ≠ 0 ⃗ \vec{n} \neq \vec{0} n = 0
n ⃗ ⋅ a ⃗ = 0 and n ⃗ ⋅ b ⃗ = 0. \vec{n}\cdot\vec{a}=0 \quad\text{and}\quad \vec{n}\cdot\vec{b}=0. n ⋅ a = 0 and n ⋅ b = 0. Writing n ⃗ = ( n 1 , n 2 , n 3 ) \vec{n}=(n_1,n_2,n_3) n = ( n 1 , n 2 , n 3 ) 2 × 3 2\times 3 2 × 3 cross product provides exactly such a formula.
The graph below shows two 3D vectors and their cross product. Notice how a ⃗ × b ⃗ \vec{a}\times\vec{b} a × b
a × b is perpendicular to both a and b a⃗ × b⃗ = (1, -2, 4)|a⃗ × b⃗| = 4.58 θ = 66.42° Area of ▱ = 4.58
Drag to rotate · Scroll to zoom · Edit the inputs above to change the vectors.
Definition of cross product
For a ⃗ = ( a 1 , a 2 , a 3 ) \vec{a}=(a_1,a_2,a_3) a = ( a 1 , a 2 , a 3 ) b ⃗ = ( b 1 , b 2 , b 3 ) \vec{b}=(b_1,b_2,b_3) b = ( b 1 , b 2 , b 3 ) cross product is defined by the symbolic 3 × 3 3\times 3 3 × 3
a ⃗ × b ⃗ = ∣ i ⃗ j ⃗ k ⃗ a 1 a 2 a 3 b 1 b 2 b 3 ∣ , \vec{a}\times\vec{b}
=
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{vmatrix}, a × b = i a 1 b 1 j a 2 b 2 k a 3 b 3 , where i ⃗ = ( 1 , 0 , 0 ) \vec{i}=(1,0,0) i = ( 1 , 0 , 0 ) j ⃗ = ( 0 , 1 , 0 ) \vec{j}=(0,1,0) j = ( 0 , 1 , 0 ) k ⃗ = ( 0 , 0 , 1 ) \vec{k}=(0,0,1) k = ( 0 , 0 , 1 )
Expanding the determinant
We expand along the top row (cofactor expansion):
a ⃗ × b ⃗ = i ⃗ ∣ a 2 a 3 b 2 b 3 ∣ − j ⃗ ∣ a 1 a 3 b 1 b 3 ∣ + k ⃗ ∣ a 1 a 2 b 1 b 2 ∣ . \vec{a}\times\vec{b}
= \vec{i}\begin{vmatrix}a_2&a_3\\b_2&b_3\end{vmatrix}
- \vec{j}\begin{vmatrix}a_1&a_3\\b_1&b_3\end{vmatrix}
+ \vec{k}\begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix}. a × b = i a 2 b 2 a 3 b 3 − j a 1 b 1 a 3 b 3 + k a 1 b 1 a 2 b 2 . Computing each 2 × 2 2\times 2 2 × 2
a ⃗ × b ⃗ = ( a 2 b 3 − a 3 b 2 ) i ⃗ − ( a 1 b 3 − a 3 b 1 ) j ⃗ + ( a 1 b 2 − a 2 b 1 ) k ⃗ . \vec{a}\times\vec{b}
=
(a_2b_3-a_3b_2)\,\vec{i}
-(a_1b_3-a_3b_1)\,\vec{j}
+(a_1b_2-a_2b_1)\,\vec{k}. a × b = ( a 2 b 3 − a 3 b 2 ) i − ( a 1 b 3 − a 3 b 1 ) j + ( a 1 b 2 − a 2 b 1 ) k . In component form:
a ⃗ × b ⃗ = ( a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 ) . \boxed{
\vec{a}\times\vec{b}
=
\bigl(a_2b_3-a_3b_2,\;
a_3b_1-a_1b_3,\;
a_1b_2-a_2b_1\bigr).
} a × b = ( a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 ) . A worked example
Compute ( 1 , 2 , − 1 ) × ( 3 , 0 , 2 ) (1,2,-1)\times(3,0,2) ( 1 , 2 , − 1 ) × ( 3 , 0 , 2 )
Step 1. Write the determinant:
∣ i ⃗ j ⃗ k ⃗ 1 2 − 1 3 0 2 ∣ \begin{vmatrix}
\vec{i}&\vec{j}&\vec{k}\\
1&2&-1\\
3&0&2
\end{vmatrix} i 1 3 j 2 0 k − 1 2 Step 2. Expand:
i ⃗ ∣ 2 − 1 0 2 ∣ − j ⃗ ∣ 1 − 1 3 2 ∣ + k ⃗ ∣ 1 2 3 0 ∣ \vec{i}\begin{vmatrix}2&-1\\0&2\end{vmatrix}
-\vec{j}\begin{vmatrix}1&-1\\3&2\end{vmatrix}
+\vec{k}\begin{vmatrix}1&2\\3&0\end{vmatrix} i 2 0 − 1 2 − j 1 3 − 1 2 + k 1 3 2 0 = i ⃗ ( 4 − 0 ) − j ⃗ ( 2 − ( − 3 ) ) + k ⃗ ( 0 − 6 ) = 4 i ⃗ − 5 j ⃗ − 6 k ⃗ . =\vec{i}(4-0)-\vec{j}(2-(-3))+\vec{k}(0-6)
=4\vec{i}-5\vec{j}-6\vec{k}. = i ( 4 − 0 ) − j ( 2 − ( − 3 )) + k ( 0 − 6 ) = 4 i − 5 j − 6 k . So ( 1 , 2 , − 1 ) × ( 3 , 0 , 2 ) = ( 4 , − 5 , − 6 ) (1,2,-1)\times(3,0,2)=(4,-5,-6) ( 1 , 2 , − 1 ) × ( 3 , 0 , 2 ) = ( 4 , − 5 , − 6 )
▼ Exercise 1: Compute a basic cross product Compute ( 1 , 0 , 0 ) × ( 0 , 1 , 0 ) (1,0,0)\times(0,1,0) ( 1 , 0 , 0 ) × ( 0 , 1 , 0 )
Answer:
∣ i ⃗ j ⃗ k ⃗ 1 0 0 0 1 0 ∣ = i ⃗ ( 0 ⋅ 0 − 0 ⋅ 1 ) − j ⃗ ( 1 ⋅ 0 − 0 ⋅ 0 ) + k ⃗ ( 1 ⋅ 1 − 0 ⋅ 0 ) = ( 0 , 0 , 1 ) . \begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&0&0\\0&1&0\end{vmatrix}
=\vec{i}(0\cdot0-0\cdot1)-\vec{j}(1\cdot0-0\cdot0)+\vec{k}(1\cdot1-0\cdot0)
=(0,0,1). i 1 0 j 0 1 k 0 0 = i ( 0 ⋅ 0 − 0 ⋅ 1 ) − j ( 1 ⋅ 0 − 0 ⋅ 0 ) + k ( 1 ⋅ 1 − 0 ⋅ 0 ) = ( 0 , 0 , 1 ) . This is expected: i ⃗ × j ⃗ = k ⃗ \vec{i}\times\vec{j}=\vec{k} i × j = k
▼ Exercise 2: Compute (2, −1, 3) × (1, 2, 0) ∣ i ⃗ j ⃗ k ⃗ 2 − 1 3 1 2 0 ∣ \begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\2&-1&3\\1&2&0\end{vmatrix} i 2 1 j − 1 2 k 3 0 = i ⃗ ( ( − 1 ) ( 0 ) − ( 3 ) ( 2 ) ) − j ⃗ ( ( 2 ) ( 0 ) − ( 3 ) ( 1 ) ) + k ⃗ ( ( 2 ) ( 2 ) − ( − 1 ) ( 1 ) ) =\vec{i}((-1)(0)-(3)(2))-\vec{j}((2)(0)-(3)(1))+\vec{k}((2)(2)-(-1)(1)) = i (( − 1 ) ( 0 ) − ( 3 ) ( 2 )) − j (( 2 ) ( 0 ) − ( 3 ) ( 1 )) + k (( 2 ) ( 2 ) − ( − 1 ) ( 1 )) = i ⃗ ( − 6 ) − j ⃗ ( − 3 ) + k ⃗ ( 5 ) = ( − 6 , 3 , 5 ) . =\vec{i}(-6)-\vec{j}(-3)+\vec{k}(5)
=(-6,3,5). = i ( − 6 ) − j ( − 3 ) + k ( 5 ) = ( − 6 , 3 , 5 ) . ▼ Exercise 3: Find k so the cross product has a given component Find k k k ( k , 1 , − 1 ) × ( 2 , 0 , 1 ) (k,1,-1)\times(2,0,1) ( k , 1 , − 1 ) × ( 2 , 0 , 1 ) z z z − 2 -2 − 2
Answer:
The z z z
k ⋅ 0 − 1 ⋅ 2 = − 2. k\cdot0-1\cdot2 = -2. k ⋅ 0 − 1 ⋅ 2 = − 2. This gives − 2 = − 2 -2=-2 − 2 = − 2 k k k k k k z z z − 2 -2 − 2 k k k
(The z z z k ⃗ \vec{k} k k z = a 1 b 2 − a 2 b 1 = k ⋅ 0 − 1 ⋅ 2 = − 2 k_z=a_1b_2-a_2b_1=k\cdot0-1\cdot2=-2 k z = a 1 b 2 − a 2 b 1 = k ⋅ 0 − 1 ⋅ 2 = − 2
Key algebraic properties
a ⃗ × a ⃗ = 0 ⃗ \vec{a}\times\vec{a}=\vec{0} a × a = 0 A determinant with two identical rows equals zero:
a ⃗ × a ⃗ = ∣ i ⃗ j ⃗ k ⃗ a 1 a 2 a 3 a 1 a 2 a 3 ∣ = 0 ⃗ . \vec{a}\times\vec{a}
=
\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\a_1&a_2&a_3\\a_1&a_2&a_3\end{vmatrix}=\vec{0}. a × a = i a 1 a 1 j a 2 a 2 k a 3 a 3 = 0 . Parallel vectors: a ⃗ × b ⃗ = 0 ⃗ \vec{a}\times\vec{b}=\vec{0} a × b = 0
If b ⃗ = k a ⃗ \vec{b}=k\vec{a} b = k a k k k
a ⃗ × ( k a ⃗ ) = k ( a ⃗ × a ⃗ ) = 0 ⃗ . \vec{a}\times(k\vec{a})=k(\vec{a}\times\vec{a})=\vec{0}. a × ( k a ) = k ( a × a ) = 0 . More generally, a ⃗ × b ⃗ = 0 ⃗ \vec{a}\times\vec{b}=\vec{0} a × b = 0 a ⃗ \vec{a} a b ⃗ \vec{b} b 0 ⃗ \vec{0} 0
Perpendicularity: ( a ⃗ × b ⃗ ) ⋅ a ⃗ = 0 (\vec{a}\times\vec{b})\cdot\vec{a}=0 ( a × b ) ⋅ a = 0 ( a ⃗ × b ⃗ ) ⋅ b ⃗ = 0 (\vec{a}\times\vec{b})\cdot\vec{b}=0 ( a × b ) ⋅ b = 0
▼ Proof (using the determinant interpretation) Let n ⃗ = a ⃗ × b ⃗ \vec{n}=\vec{a}\times\vec{b} n = a × b
n ⃗ ⋅ a ⃗ = ∣ a 1 a 2 a 3 a 1 a 2 a 3 b 1 b 2 b 3 ∣ = 0 , \vec{n}\cdot\vec{a}
=
\begin{vmatrix}a_1&a_2&a_3\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}=0, n ⋅ a = a 1 a 1 b 1 a 2 a 2 b 2 a 3 a 3 b 3 = 0 , because the first two rows are identical. Similarly,
n ⃗ ⋅ b ⃗ = ∣ b 1 b 2 b 3 a 1 a 2 a 3 b 1 b 2 b 3 ∣ = 0. \vec{n}\cdot\vec{b}
=
\begin{vmatrix}b_1&b_2&b_3\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}=0. n ⋅ b = b 1 a 1 b 1 b 2 a 2 b 2 b 3 a 3 b 3 = 0. Answer to the geometric question: n ⃗ = a ⃗ × b ⃗ \vec{n}=\vec{a}\times\vec{b} n = a × b a ⃗ \vec{a} a b ⃗ \vec{b} b
▼ Exercise 1: Verify (2,3,1) × (1,−1,4) is perpendicular to both Compute the cross product and check both dot products.
Step 1. a ⃗ × b ⃗ \vec{a}\times\vec{b} a × b
( 3 ⋅ 4 − 1 ⋅ ( − 1 ) , 1 ⋅ 1 − 2 ⋅ 4 , 2 ⋅ ( − 1 ) − 3 ⋅ 1 ) = ( 13 , − 7 , − 5 ) . (3\cdot4-1\cdot(-1),\;1\cdot1-2\cdot4,\;2\cdot(-1)-3\cdot1)
=(13,-7,-5). ( 3 ⋅ 4 − 1 ⋅ ( − 1 ) , 1 ⋅ 1 − 2 ⋅ 4 , 2 ⋅ ( − 1 ) − 3 ⋅ 1 ) = ( 13 , − 7 , − 5 ) . Step 2. ( 13 , − 7 , − 5 ) ⋅ ( 2 , 3 , 1 ) = 26 − 21 − 5 = 0 (13,-7,-5)\cdot(2,3,1)=26-21-5=0 ( 13 , − 7 , − 5 ) ⋅ ( 2 , 3 , 1 ) = 26 − 21 − 5 = 0
Step 3. ( 13 , − 7 , − 5 ) ⋅ ( 1 , − 1 , 4 ) = 13 + 7 − 20 = 0 (13,-7,-5)\cdot(1,-1,4)=13+7-20=0 ( 13 , − 7 , − 5 ) ⋅ ( 1 , − 1 , 4 ) = 13 + 7 − 20 = 0
▼ Exercise 2: Cross product of parallel vectors Show ( 1 , 2 , 3 ) × ( 2 , 4 , 6 ) = 0 ⃗ (1,2,3)\times(2,4,6)=\vec{0} ( 1 , 2 , 3 ) × ( 2 , 4 , 6 ) = 0
Answer: ( 2 , 4 , 6 ) = 2 ( 1 , 2 , 3 ) (2,4,6)=2(1,2,3) ( 2 , 4 , 6 ) = 2 ( 1 , 2 , 3 ) 0 ⃗ \vec{0} 0
Direct check: ( 2 ⋅ 6 − 3 ⋅ 4 , 3 ⋅ 2 − 1 ⋅ 6 , 1 ⋅ 4 − 2 ⋅ 2 ) = ( 0 , 0 , 0 ) (2\cdot6-3\cdot4,\;3\cdot2-1\cdot6,\;1\cdot4-2\cdot2)=(0,0,0) ( 2 ⋅ 6 − 3 ⋅ 4 , 3 ⋅ 2 − 1 ⋅ 6 , 1 ⋅ 4 − 2 ⋅ 2 ) = ( 0 , 0 , 0 )
Magnitude: the sine formula
Theorem. For vectors a ⃗ , b ⃗ \vec{a},\vec{b} a , b θ \theta θ
∥ a ⃗ × b ⃗ ∥ = ∥ a ⃗ ∥ ∥ b ⃗ ∥ sin θ . \|\vec{a}\times\vec{b}\| = \|\vec{a}\|\,\|\vec{b}\|\sin\theta. ∥ a × b ∥ = ∥ a ∥ ∥ b ∥ sin θ .
Geometric meaning: area of the parallelogram
The parallelogram spanned by a ⃗ \vec{a} a b ⃗ \vec{b} b ∥ b ⃗ ∥ \|\vec{b}\| ∥ b ∥ h = ∥ a ⃗ ∥ sin θ h=\|\vec{a}\|\sin\theta h = ∥ a ∥ sin θ
Area = base × height = ∥ b ⃗ ∥ ⋅ ∥ a ⃗ ∥ sin θ = ∥ a ⃗ × b ⃗ ∥ . \text{Area} = \text{base}\times\text{height}
= \|\vec{b}\|\cdot\|\vec{a}\|\sin\theta
= \|\vec{a}\times\vec{b}\|. Area = base × height = ∥ b ∥ ⋅ ∥ a ∥ sin θ = ∥ a × b ∥. Drag the vectors below to see the parallelogram, the dashed height h = ∣ a ⃗ ∣ sin θ h=|\vec{a}|\sin\theta h = ∣ a ∣ sin θ
Area of parallelogram = |a × b| = |a||b|sin θ a⃗ = (4.00, 1.00)b⃗ = (1.00, 3.00)a⃗ · b⃗ = 7.000θ = 57.5°a⃗ × b⃗ = 11.000Area = |a⃗ × b⃗| = 11.000 h = |a⃗|sin θ = 3.479
Drag the coloured dots at the arrow tips to move each vector.
-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 a⃗+b⃗ h θ a⃗ b⃗ O ▼ Proof of the sine formula We want to show ∥ a ⃗ × b ⃗ ∥ 2 = ∥ a ⃗ ∥ 2 ∥ b ⃗ ∥ 2 sin 2 θ \|\vec{a}\times\vec{b}\|^2=\|\vec{a}\|^2\|\vec{b}\|^2\sin^2\theta ∥ a × b ∥ 2 = ∥ a ∥ 2 ∥ b ∥ 2 sin 2 θ
Using sin 2 θ = 1 − cos 2 θ \sin^2\theta=1-\cos^2\theta sin 2 θ = 1 − cos 2 θ cos θ = a ⃗ ⋅ b ⃗ ∥ a ⃗ ∥ ∥ b ⃗ ∥ \cos\theta=\frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\|\vec{b}\|} cos θ = ∥ a ∥∥ b ∥ a ⋅ b
∥ a ⃗ ∥ 2 ∥ b ⃗ ∥ 2 sin 2 θ = ∥ a ⃗ ∥ 2 ∥ b ⃗ ∥ 2 − ( a ⃗ ⋅ b ⃗ ) 2 . \|\vec{a}\|^2\|\vec{b}\|^2\sin^2\theta
=\|\vec{a}\|^2\|\vec{b}\|^2-(\vec{a}\cdot\vec{b})^2. ∥ a ∥ 2 ∥ b ∥ 2 sin 2 θ = ∥ a ∥ 2 ∥ b ∥ 2 − ( a ⋅ b ) 2 . Write ∥ a ⃗ × b ⃗ ∥ 2 = ( a 2 b 3 − a 3 b 2 ) 2 + ( a 3 b 1 − a 1 b 3 ) 2 + ( a 1 b 2 − a 2 b 1 ) 2 \|\vec{a}\times\vec{b}\|^2=(a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^2 ∥ a × b ∥ 2 = ( a 2 b 3 − a 3 b 2 ) 2 + ( a 3 b 1 − a 1 b 3 ) 2 + ( a 1 b 2 − a 2 b 1 ) 2
▼ Exercise 1: Area of a triangle Find the area of the triangle with vertices P = ( 1 , 0 , 0 ) P=(1,0,0) P = ( 1 , 0 , 0 ) Q = ( 0 , 1 , 0 ) Q=(0,1,0) Q = ( 0 , 1 , 0 ) R = ( 0 , 0 , 1 ) R=(0,0,1) R = ( 0 , 0 , 1 )
Answer:
Two edge vectors from P P P a ⃗ = Q − P = ( − 1 , 1 , 0 ) \vec{a}=Q-P=(-1,1,0) a = Q − P = ( − 1 , 1 , 0 ) b ⃗ = R − P = ( − 1 , 0 , 1 ) \vec{b}=R-P=(-1,0,1) b = R − P = ( − 1 , 0 , 1 )
a ⃗ × b ⃗ = ( 1 ⋅ 1 − 0 ⋅ 0 , 0 ⋅ ( − 1 ) − ( − 1 ) ⋅ 1 , ( − 1 ) ⋅ 0 − 1 ⋅ ( − 1 ) ) = ( 1 , 1 , 1 ) . \vec{a}\times\vec{b}=(1\cdot1-0\cdot0,\;0\cdot(-1)-(-1)\cdot1,\;(-1)\cdot0-1\cdot(-1))=(1,1,1). a × b = ( 1 ⋅ 1 − 0 ⋅ 0 , 0 ⋅ ( − 1 ) − ( − 1 ) ⋅ 1 , ( − 1 ) ⋅ 0 − 1 ⋅ ( − 1 )) = ( 1 , 1 , 1 ) . Area of triangle = 1 2 ∥ a ⃗ × b ⃗ ∥ = 1 2 3 . \text{Area of triangle}=\tfrac{1}{2}\|\vec{a}\times\vec{b}\|=\tfrac{1}{2}\sqrt{3}. Area of triangle = 2 1 ∥ a × b ∥ = 2 1 3 . ▼ Exercise 2: Use the area formula Find the area of the parallelogram spanned by a ⃗ = ( 3 , 0 , 0 ) \vec{a}=(3,0,0) a = ( 3 , 0 , 0 ) b ⃗ = ( 1 , 2 , 0 ) \vec{b}=(1,2,0) b = ( 1 , 2 , 0 )
Answer:
a ⃗ × b ⃗ = ( 0 ⋅ 0 − 0 ⋅ 2 , 0 ⋅ 1 − 3 ⋅ 0 , 3 ⋅ 2 − 0 ⋅ 1 ) = ( 0 , 0 , 6 ) . \vec{a}\times\vec{b}=(0\cdot0-0\cdot2,\;0\cdot1-3\cdot0,\;3\cdot2-0\cdot1)=(0,0,6). a × b = ( 0 ⋅ 0 − 0 ⋅ 2 , 0 ⋅ 1 − 3 ⋅ 0 , 3 ⋅ 2 − 0 ⋅ 1 ) = ( 0 , 0 , 6 ) . Area = ∥ ( 0 , 0 , 6 ) ∥ = 6 =\|(0,0,6)\|=6 = ∥ ( 0 , 0 , 6 ) ∥ = 6
(Geometrically: base = 3 =3 = 3 = 2 =2 = 2 = 6 =6 = 6
▼ Exercise 3: Classify angle using cross-product magnitude a ⃗ = ( 1 , 1 , 0 ) \vec{a}=(1,1,0) a = ( 1 , 1 , 0 ) b ⃗ = ( 0 , 1 , 0 ) \vec{b}=(0,1,0) b = ( 0 , 1 , 0 )
Compute ∥ a ⃗ × b ⃗ ∥ \|\vec{a}\times\vec{b}\| ∥ a × b ∥ ∥ a ⃗ ∥ ∥ b ⃗ ∥ \|\vec{a}\|\|\vec{b}\| ∥ a ∥∥ b ∥ θ \theta θ
Answer:
a ⃗ × b ⃗ = ( 0 , 0 , 1 ) \vec{a}\times\vec{b}=(0,0,1) a × b = ( 0 , 0 , 1 ) ∥ a ⃗ × b ⃗ ∥ = 1 \|\vec{a}\times\vec{b}\|=1 ∥ a × b ∥ = 1
∥ a ⃗ ∥ = 2 \|\vec{a}\|=\sqrt{2} ∥ a ∥ = 2 ∥ b ⃗ ∥ = 1 \|\vec{b}\|=1 ∥ b ∥ = 1 sin θ = 1 2 \sin\theta=\frac{1}{\sqrt{2}} sin θ = 2 1 θ = 45 ° \theta=45° θ = 45°
Arithmetic properties
Anti-commutativity
a ⃗ × b ⃗ = − ( b ⃗ × a ⃗ ) . \vec{a}\times\vec{b} = -(\vec{b}\times\vec{a}). a × b = − ( b × a ) . Swapping two rows of a determinant negates it, so swapping a ⃗ \vec{a} a b ⃗ \vec{b} b The cross product is NOT commutative.
▼ Proof b ⃗ × a ⃗ = ∣ i ⃗ j ⃗ k ⃗ b 1 b 2 b 3 a 1 a 2 a 3 ∣ = − ∣ i ⃗ j ⃗ k ⃗ a 1 a 2 a 3 b 1 b 2 b 3 ∣ = − ( a ⃗ × b ⃗ ) . \vec{b}\times\vec{a}
=
\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\b_1&b_2&b_3\\a_1&a_2&a_3\end{vmatrix}
=-\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}
=-(\vec{a}\times\vec{b}). b × a = i b 1 a 1 j b 2 a 2 k b 3 a 3 = − i a 1 b 1 j a 2 b 2 k a 3 b 3 = − ( a × b ) . Distributivity over addition
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 × ( b + c ) = a × b + a × c . ( a ⃗ + b ⃗ ) × c ⃗ = a ⃗ × c ⃗ + b ⃗ × c ⃗ . (\vec{a}+\vec{b})\times\vec{c} = \vec{a}\times\vec{c} + \vec{b}\times\vec{c}. ( a + b ) × c = a × c + b × c . Scalar compatibility
( k a ⃗ ) × b ⃗ = k ( a ⃗ × b ⃗ ) = a ⃗ × ( k b ⃗ ) . (k\vec{a})\times\vec{b} = k(\vec{a}\times\vec{b}) = \vec{a}\times(k\vec{b}). ( k a ) × b = k ( a × b ) = a × ( k b ) . Cross product is NOT associative
In general, ( a ⃗ × b ⃗ ) × c ⃗ ≠ a ⃗ × ( b ⃗ × c ⃗ ) (\vec{a}\times\vec{b})\times\vec{c}\neq\vec{a}\times(\vec{b}\times\vec{c}) ( a × b ) × c = a × ( b × c )
Counterexample. Let a ⃗ = b ⃗ = i ⃗ = ( 1 , 0 , 0 ) \vec{a}=\vec{b}=\vec{i}=(1,0,0) a = b = i = ( 1 , 0 , 0 ) c ⃗ = j ⃗ = ( 0 , 1 , 0 ) \vec{c}=\vec{j}=(0,1,0) c = j = ( 0 , 1 , 0 )
( i ⃗ × i ⃗ ) × j ⃗ = 0 ⃗ × j ⃗ = 0 ⃗ , (\vec{i}\times\vec{i})\times\vec{j}=\vec{0}\times\vec{j}=\vec{0}, ( i × i ) × j = 0 × j = 0 , i ⃗ × ( i ⃗ × j ⃗ ) = i ⃗ × k ⃗ = ( 0 , 0 , 1 ) × ( 0 , 0 , 1 ) . . . \vec{i}\times(\vec{i}\times\vec{j})=\vec{i}\times\vec{k}=(0,0,1)\times(0,0,1)... i × ( i × j ) = i × k = ( 0 , 0 , 1 ) × ( 0 , 0 , 1 ) ... Actually: i ⃗ × j ⃗ = k ⃗ \vec{i}\times\vec{j}=\vec{k} i × j = k i ⃗ × k ⃗ = ( 1 , 0 , 0 ) × ( 0 , 0 , 1 ) = ( 0 ⋅ 1 − 0 ⋅ 0 , 0 ⋅ 0 − 1 ⋅ 1 , 1 ⋅ 0 − 0 ⋅ 0 ) = ( 0 , − 1 , 0 ) = − j ⃗ ≠ 0 ⃗ \vec{i}\times\vec{k}=(1,0,0)\times(0,0,1)=(0\cdot1-0\cdot0,\;0\cdot0-1\cdot1,\;1\cdot0-0\cdot0)=(0,-1,0)=-\vec{j}\neq\vec{0} i × k = ( 1 , 0 , 0 ) × ( 0 , 0 , 1 ) = ( 0 ⋅ 1 − 0 ⋅ 0 , 0 ⋅ 0 − 1 ⋅ 1 , 1 ⋅ 0 − 0 ⋅ 0 ) = ( 0 , − 1 , 0 ) = − j = 0
▼ Exercise 1: Verify anti-commutativity with a concrete example Compute ( 1 , 2 , 0 ) × ( 3 , 1 , 0 ) (1,2,0)\times(3,1,0) ( 1 , 2 , 0 ) × ( 3 , 1 , 0 ) ( 3 , 1 , 0 ) × ( 1 , 2 , 0 ) (3,1,0)\times(1,2,0) ( 3 , 1 , 0 ) × ( 1 , 2 , 0 )
Answer:
( 1 , 2 , 0 ) × ( 3 , 1 , 0 ) = ( 2 ⋅ 0 − 0 ⋅ 1 , 0 ⋅ 3 − 1 ⋅ 0 , 1 ⋅ 1 − 2 ⋅ 3 ) = ( 0 , 0 , − 5 ) (1,2,0)\times(3,1,0)=(2\cdot0-0\cdot1,\;0\cdot3-1\cdot0,\;1\cdot1-2\cdot3)=(0,0,-5) ( 1 , 2 , 0 ) × ( 3 , 1 , 0 ) = ( 2 ⋅ 0 − 0 ⋅ 1 , 0 ⋅ 3 − 1 ⋅ 0 , 1 ⋅ 1 − 2 ⋅ 3 ) = ( 0 , 0 , − 5 )
( 3 , 1 , 0 ) × ( 1 , 2 , 0 ) = ( 1 ⋅ 0 − 0 ⋅ 2 , 0 ⋅ 1 − 3 ⋅ 0 , 3 ⋅ 2 − 1 ⋅ 1 ) = ( 0 , 0 , 5 ) (3,1,0)\times(1,2,0)=(1\cdot0-0\cdot2,\;0\cdot1-3\cdot0,\;3\cdot2-1\cdot1)=(0,0,5) ( 3 , 1 , 0 ) × ( 1 , 2 , 0 ) = ( 1 ⋅ 0 − 0 ⋅ 2 , 0 ⋅ 1 − 3 ⋅ 0 , 3 ⋅ 2 − 1 ⋅ 1 ) = ( 0 , 0 , 5 )
Indeed ( 0 , 0 , 5 ) = − ( 0 , 0 , − 5 ) (0,0,5)=-(0,0,-5) ( 0 , 0 , 5 ) = − ( 0 , 0 , − 5 )
▼ Exercise 2: Expand (a + b) × (a − b) Simplify ( a ⃗ + b ⃗ ) × ( a ⃗ − b ⃗ ) (\vec{a}+\vec{b})\times(\vec{a}-\vec{b}) ( a + b ) × ( a − b )
Answer:
( a ⃗ + b ⃗ ) × ( a ⃗ − b ⃗ ) = a ⃗ × a ⃗ − a ⃗ × b ⃗ + b ⃗ × a ⃗ − b ⃗ × b ⃗ . (\vec{a}+\vec{b})\times(\vec{a}-\vec{b})
=\vec{a}\times\vec{a}-\vec{a}\times\vec{b}+\vec{b}\times\vec{a}-\vec{b}\times\vec{b}. ( a + b ) × ( a − b ) = a × a − a × b + b × a − b × b . Using a ⃗ × a ⃗ = 0 ⃗ \vec{a}\times\vec{a}=\vec{0} a × a = 0 b ⃗ × b ⃗ = 0 ⃗ \vec{b}\times\vec{b}=\vec{0} b × b = 0 b ⃗ × a ⃗ = − a ⃗ × b ⃗ \vec{b}\times\vec{a}=-\vec{a}\times\vec{b} b × a = − a × b
= 0 ⃗ − a ⃗ × b ⃗ − a ⃗ × b ⃗ − 0 ⃗ = − 2 ( a ⃗ × b ⃗ ) . =\vec{0}-\vec{a}\times\vec{b}-\vec{a}\times\vec{b}-\vec{0}=-2(\vec{a}\times\vec{b}). = 0 − a × b − a × b − 0 = − 2 ( a × b ) . ▼ Exercise 3: Use distributivity to compute Given a ⃗ = ( 1 , 0 , 0 ) \vec{a}=(1,0,0) a = ( 1 , 0 , 0 ) b ⃗ = ( 0 , 1 , 0 ) \vec{b}=(0,1,0) b = ( 0 , 1 , 0 ) c ⃗ = ( 0 , 0 , 1 ) \vec{c}=(0,0,1) c = ( 0 , 0 , 1 )
( 2 a ⃗ − b ⃗ ) × c ⃗ . (2\vec{a}-\vec{b})\times\vec{c}. ( 2 a − b ) × c . Answer:
( 2 a ⃗ − b ⃗ ) × c ⃗ = 2 ( a ⃗ × c ⃗ ) − ( b ⃗ × c ⃗ ) . (2\vec{a}-\vec{b})\times\vec{c}
=2(\vec{a}\times\vec{c})-(\vec{b}\times\vec{c}). ( 2 a − b ) × c = 2 ( a × c ) − ( b × c ) . a ⃗ × c ⃗ = ( 1 , 0 , 0 ) × ( 0 , 0 , 1 ) = ( 0 ⋅ 1 − 0 ⋅ 0 , 0 ⋅ 0 − 1 ⋅ 1 , 1 ⋅ 0 − 0 ⋅ 0 ) = ( 0 , − 1 , 0 ) = − j ⃗ \vec{a}\times\vec{c}=(1,0,0)\times(0,0,1)=(0\cdot1-0\cdot0,\;0\cdot0-1\cdot1,\;1\cdot0-0\cdot0)=(0,-1,0)=-\vec{j} a × c = ( 1 , 0 , 0 ) × ( 0 , 0 , 1 ) = ( 0 ⋅ 1 − 0 ⋅ 0 , 0 ⋅ 0 − 1 ⋅ 1 , 1 ⋅ 0 − 0 ⋅ 0 ) = ( 0 , − 1 , 0 ) = − j
b ⃗ × c ⃗ = ( 0 , 1 , 0 ) × ( 0 , 0 , 1 ) = ( 1 , 0 , 0 ) = i ⃗ \vec{b}\times\vec{c}=(0,1,0)\times(0,0,1)=(1,0,0)=\vec{i} b × c = ( 0 , 1 , 0 ) × ( 0 , 0 , 1 ) = ( 1 , 0 , 0 ) = i
So ( 2 a ⃗ − b ⃗ ) × c ⃗ = 2 ( 0 , − 1 , 0 ) − ( 1 , 0 , 0 ) = ( − 1 , − 2 , 0 ) (2\vec{a}-\vec{b})\times\vec{c}=2(0,-1,0)-(1,0,0)=(-1,-2,0) ( 2 a − b ) × c = 2 ( 0 , − 1 , 0 ) − ( 1 , 0 , 0 ) = ( − 1 , − 2 , 0 )
Scalar triple product
The scalar triple product of three vectors is
a ⃗ ⋅ ( b ⃗ × c ⃗ ) . \vec{a}\cdot(\vec{b}\times\vec{c}). a ⋅ ( b × c ) . Formula via a 3 × 3 3\times3 3 × 3
a ⃗ ⋅ ( b ⃗ × c ⃗ ) = ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ . \vec{a}\cdot(\vec{b}\times\vec{c})
=
\begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{vmatrix}. a ⋅ ( b × c ) = a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 . This follows by expanding b ⃗ × c ⃗ \vec{b}\times\vec{c} b × c a ⃗ \vec{a} a
a ⃗ ⋅ ( b ⃗ × c ⃗ ) = a 1 ( b 2 c 3 − b 3 c 2 ) − a 2 ( b 1 c 3 − b 3 c 1 ) + a 3 ( b 1 c 2 − b 2 c 1 ) , \vec{a}\cdot(\vec{b}\times\vec{c})
=a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1), a ⋅ ( b × c ) = a 1 ( b 2 c 3 − b 3 c 2 ) − a 2 ( b 1 c 3 − b 3 c 1 ) + a 3 ( b 1 c 2 − b 2 c 1 ) , which is exactly the cofactor expansion of the 3 × 3 3\times3 3 × 3
Geometric meaning: volume of the parallelepiped
The three vectors a ⃗ , b ⃗ , c ⃗ \vec{a},\vec{b},\vec{c} a , b , c parallelepiped (a 3D solid with six parallelogram faces). Its volume is
V = ∣ a ⃗ ⋅ ( b ⃗ × c ⃗ ) ∣ . V = \bigl|\vec{a}\cdot(\vec{b}\times\vec{c})\bigr|. V = a ⋅ ( b × c ) . The base parallelogram (spanned by b ⃗ \vec{b} b c ⃗ \vec{c} c ∥ b ⃗ × c ⃗ ∥ \|\vec{b}\times\vec{c}\| ∥ b × c ∥ a ⃗ \vec{a} a b ⃗ × c ⃗ \vec{b}\times\vec{c} b × c ∣ a ⃗ ⋅ n ^ ∣ |\vec{a}\cdot\hat{n}| ∣ a ⋅ n ^ ∣ n ^ = b ⃗ × c ⃗ ∥ b ⃗ × c ⃗ ∥ \hat{n}=\frac{\vec{b}\times\vec{c}}{\|\vec{b}\times\vec{c}\|} n ^ = ∥ b × c ∥ b × c
V = ∥ b ⃗ × c ⃗ ∥ ⋅ ∣ a ⃗ ⋅ ( b ⃗ × c ⃗ ) ∣ ∥ b ⃗ × c ⃗ ∥ = ∣ a ⃗ ⋅ ( b ⃗ × c ⃗ ) ∣ . V = \|\vec{b}\times\vec{c}\|\cdot\frac{|\vec{a}\cdot(\vec{b}\times\vec{c})|}{\|\vec{b}\times\vec{c}\|}
= |\vec{a}\cdot(\vec{b}\times\vec{c})|. V = ∥ b × c ∥ ⋅ ∥ b × c ∥ ∣ a ⋅ ( b × c ) ∣ = ∣ a ⋅ ( b × c ) ∣. Rotate the 3D graph below to see the parallelepiped formed by three vectors. Edit the inputs to change the shape and watch the volume update.
Scalar triple product = signed volume of parallelepiped b⃗ × c⃗ = (4, 0, -2)a⃗ · (b⃗ × c⃗) = 8Volume = |a⃗ · (b⃗ × c⃗)| = 8
Drag to rotate · Scroll to zoom · Edit the inputs above to change the vectors.
Coplanarity test
The vectors a ⃗ , b ⃗ , c ⃗ \vec{a},\vec{b},\vec{c} a , b , c coplanar (all lie in the same plane) if and only if
a ⃗ ⋅ ( b ⃗ × c ⃗ ) = 0. \vec{a}\cdot(\vec{b}\times\vec{c})=0. a ⋅ ( b × c ) = 0. This is because the parallelepiped collapses to a flat shape with zero volume.
▼ Exercise 1: Compute a simple triple product Compute a ⃗ ⋅ ( b ⃗ × c ⃗ ) \vec{a}\cdot(\vec{b}\times\vec{c}) a ⋅ ( b × c ) a ⃗ = ( 1 , 0 , 0 ) \vec{a}=(1,0,0) a = ( 1 , 0 , 0 ) b ⃗ = ( 0 , 1 , 0 ) \vec{b}=(0,1,0) b = ( 0 , 1 , 0 ) c ⃗ = ( 0 , 0 , 1 ) \vec{c}=(0,0,1) c = ( 0 , 0 , 1 )
Answer:
∣ 1 0 0 0 1 0 0 0 1 ∣ = 1. \begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}=1. 1 0 0 0 1 0 0 0 1 = 1. The unit cube has volume 1. ✓
▼ Exercise 2: Volume of a parallelepiped Find the volume of the parallelepiped with edge vectors a ⃗ = ( 2 , 0 , 0 ) \vec{a}=(2,0,0) a = ( 2 , 0 , 0 ) b ⃗ = ( 0 , 3 , 0 ) \vec{b}=(0,3,0) b = ( 0 , 3 , 0 ) c ⃗ = ( 0 , 0 , 4 ) \vec{c}=(0,0,4) c = ( 0 , 0 , 4 )
Answer:
∣ 2 0 0 0 3 0 0 0 4 ∣ = 2 ⋅ 3 ⋅ 4 = 24. \begin{vmatrix}2&0&0\\0&3&0\\0&0&4\end{vmatrix}=2\cdot3\cdot4=24. 2 0 0 0 3 0 0 0 4 = 2 ⋅ 3 ⋅ 4 = 24. Volume = ∣ 24 ∣ = 24 =|24|=24 = ∣24∣ = 24 (A 2 × 3 × 4 2\times3\times4 2 × 3 × 4
▼ Exercise 3: Test for coplanarity Are a ⃗ = ( 1 , 2 , 3 ) \vec{a}=(1,2,3) a = ( 1 , 2 , 3 ) b ⃗ = ( 0 , 1 , 1 ) \vec{b}=(0,1,1) b = ( 0 , 1 , 1 ) c ⃗ = ( 2 , 3 , 5 ) \vec{c}=(2,3,5) c = ( 2 , 3 , 5 )
Answer:
∣ 1 2 3 0 1 1 2 3 5 ∣ = 1 ( 1 ⋅ 5 − 1 ⋅ 3 ) − 2 ( 0 ⋅ 5 − 1 ⋅ 2 ) + 3 ( 0 ⋅ 3 − 1 ⋅ 2 ) = 2 + 4 − 6 = 0. \begin{vmatrix}1&2&3\\0&1&1\\2&3&5\end{vmatrix}
=1(1\cdot5-1\cdot3)-2(0\cdot5-1\cdot2)+3(0\cdot3-1\cdot2)
=2+4-6=0. 1 0 2 2 1 3 3 1 5 = 1 ( 1 ⋅ 5 − 1 ⋅ 3 ) − 2 ( 0 ⋅ 5 − 1 ⋅ 2 ) + 3 ( 0 ⋅ 3 − 1 ⋅ 2 ) = 2 + 4 − 6 = 0. Yes, they are coplanar (the determinant is 0).
▼ Exercise 4: Triple product with a computed cross product Let a ⃗ = ( 1 , 1 , 0 ) \vec{a}=(1,1,0) a = ( 1 , 1 , 0 ) b ⃗ = ( 0 , 1 , 1 ) \vec{b}=(0,1,1) b = ( 0 , 1 , 1 ) c ⃗ = ( 1 , 0 , 1 ) \vec{c}=(1,0,1) c = ( 1 , 0 , 1 ) a ⃗ ⋅ ( b ⃗ × c ⃗ ) \vec{a}\cdot(\vec{b}\times\vec{c}) a ⋅ ( b × c )
Answer:
∣ 1 1 0 0 1 1 1 0 1 ∣ = 1 ( 1 ⋅ 1 − 1 ⋅ 0 ) − 1 ( 0 ⋅ 1 − 1 ⋅ 1 ) + 0 = 1 + 1 = 2. \begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}
=1(1\cdot1-1\cdot0)-1(0\cdot1-1\cdot1)+0
=1+1=2. 1 0 1 1 1 0 0 1 1 = 1 ( 1 ⋅ 1 − 1 ⋅ 0 ) − 1 ( 0 ⋅ 1 − 1 ⋅ 1 ) + 0 = 1 + 1 = 2. Volume of parallelepiped = ∣ 2 ∣ = 2 =|2|=2 = ∣2∣ = 2