Dot product and cross product (HL) Contents Angle between vectors The angle θ \theta θ
Dot product a ⋅ b = ∣ a ∣ ∣ b ∣ cos θ \bm a \cdot \bm b = \lvert \bm a\rvert\lvert \bm b\rvert \cos\theta a ⋅ b = ∣ a ∣ ∣ b ∣ cos θ cos θ \cos\theta cos θ
( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) = x 1 x 2 + y 1 y 2 + z 1 z 2 (x_1, y_1, z_1)\cdot(x_2, y_2, z_2) = x_1x_2 + y_1y_2 + z_1z_2 ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) = x 1 x 2 + y 1 y 2 + z 1 z 2 Note, a cos θ \bm a \cos\theta a cos θ a \bm a a b \bm b b a \bm a a b \bm b b a cos θ \bm a \cos\theta a cos θ
The Cauchy-Schwartz inequality is equivalent to
∣ a ⋅ b ∣ ≤ ∣ a ∣ ∣ b ∣ \lvert\bm a \cdot \bm b\rvert \leq \lvert\bm a\rvert\lvert\bm b\rvert ∣ a ⋅ b ∣ ≤ ∣ a ∣ ∣ b ∣ Cross product The cross product a × b \bm a \times \bm b a × b a \bm a a b \bm b b
a × b = ∣ a ∣ ∣ b ∣ sin θ n ^ \bm a \times \bm b = \lvert \bm a\rvert\lvert \bm b\rvert \sin\theta \,\bm{\hat n} a × b = ∣ a ∣ ∣ b ∣ sin θ n ^ with n ^ \bm{\hat n} n ^
The standard basis consisting of unit vectors i \bm i i j \bm j j k \bm k k
i × j = k \bm i \times \bm j = \bm k i × j = k j × k = i \bm j \times \bm k = \bm i j × k = i k × i = j \bm k \times \bm i = \bm j k × i = j but also
j × i = − k \bm j \times \bm i = \bm {-k} j × i = − k k × j = − i \bm k \times \bm j = \bm {-i} k × j = − i i × k = − j \bm i \times \bm k = \bm {-j} i × k = − j By using these formulas on individual components, the cross product in components form is
a × b = ( y 1 z 2 − y 2 z 1 ) i + ( z 1 x 2 − z 2 x 1 ) j + ( x 1 y 2 − x 2 y 1 ) k \begin{align*}
&\,\bm a \times \bm b \\
=&\, (y_1z_2 - y_2z_1)\bm i \\
+&\, (z_1x_2 - z_2x_1)\bm j \\
+&\, (x_1y_2 - x_2y_1)\bm k
\end{align*} = + + a × b ( y 1 z 2 − y 2 z 1 ) i ( z 1 x 2 − z 2 x 1 ) j ( x 1 y 2 − x 2 y 1 ) k There are various ways to make sense of the formula. One way to remember is to first write each vectors twice
x 1 y 1 z 1 x 1 y 1 z 1 x 2 y 2 z 2 x 2 y 2 z 2 \begin{matrix}
\sout{\color{grey}x_1} & y_1 & z_1 & x_1 & y_1 & \sout{\color{grey}z_1} \\
\sout{\color{grey}x_2} & y_2 & z_2 & x_2 & y_2 & \sout{\color{grey}z_2}
\end{matrix} x 1 x 2 y 1 y 2 z 1 z 2 x 1 x 2 y 1 y 2 z 1 z 2 The components of the cross product, after ignoring first and last columns, are “diagonal down minus diagonal up”.
Alternatively, using the determinant of a 3 3 3 3 3 3
a × b = ∣ i j k x 1 y 1 z 1 x 2 y 2 z 2 ∣ \bm a \times \bm b = \begin{vmatrix}
\bm i & \bm j & \bm k \\
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2
\end{vmatrix} a × b = i x 1 x 2 j y 1 y 2 k z 1 z 2 You are not required to interpret or simplify a determinant.
Properties For the following properties, terms other than parallel and perpendicular are not used on exams, but nevertheless be able to apply these properties.
properties dot product cross product parallel vectors a ⋅ b = ∣ a ∣ ∣ b ∣ a ⋅ a = ∣ a ∣ 2 \bm a \cdot \bm b = \lvert\bm a\rvert\lvert\bm b\rvert \\ \bm a \cdot \bm a = \lvert\bm a\rvert^2 a ⋅ b = ∣ a ∣ ∣ b ∣ a ⋅ a = ∣ a ∣ 2 a × b = 0 \bm a \times \bm b = \bm 0 a × b = 0 perpendicular vectors a ⋅ b = 0 \bm a \cdot \bm b = 0 a ⋅ b = 0 a × b = ∣ a ∣ ∣ b ∣ n ^ \bm a \times \bm b = \lvert\bm a\rvert\lvert\bm b\rvert \bm{\hat n} a × b = ∣ a ∣ ∣ b ∣ n ^ commutative a ⋅ b = b ⋅ a \bm a \cdot \bm b = \bm b \cdot \bm a a ⋅ b = b ⋅ a N/A anticommutative N/A a × b = − b × a \bm a \times \bm b = -\bm b \times \bm a a × b = − b × a distributive k ( a ⋅ b ) = ( k a ) ⋅ b = a ⋅ ( k b ) a ⋅ ( b + c ) = a ⋅ b + a ⋅ c k(\bm a \cdot \bm b) = (k\bm a) \cdot \bm b = \bm a \cdot (k\bm b) \\ \bm a \cdot (\bm b + \bm c) = \bm a \cdot \bm b + \bm a \cdot \bm c k ( a ⋅ b ) = ( k a ) ⋅ b = a ⋅ ( k b ) a ⋅ ( b + c ) = a ⋅ b + a ⋅ c k ( a × b ) = ( k a ) × b = a × ( k b ) a × ( b + c ) = a × b + a × c k(\bm a \times \bm b) = (k\bm a) \times \bm b = \bm a \times (k\bm b) \\ \bm a\times(\bm b + \bm c) = \bm a \times \bm b + \bm a \times \bm c k ( a × b ) = ( k a ) × b = a × ( k b ) a × ( b + c ) = a × b + a × c
Example: Given ∣ a ∣ = 3 \lvert\bm a\rvert = 3 ∣ a ∣ = 3 ∣ b ∣ = 4 \lvert\bm b\rvert = 4 ∣ b ∣ = 4 a + 2 b \bm a + 2\bm b a + 2 b 3 a − b 3\bm a - \bm b 3 a − b a \bm a a b \bm b b
( a + 2 b ) ⋅ ( 3 a − b ) = 0 3 ∣ a ∣ 2 + 5 ( a ⋅ b ) − 2 ∣ b ∣ 2 = 0 3 ( 3 ) 2 + 5 ∣ a ∣ ∣ b ∣ cos θ − 2 ( 4 ) 2 = 0 27 + 5 ( 3 ) ( 4 ) cos θ − 32 = 0 60 cos θ = 5 cos − 1 1 12 = θ ■ \begin{align*}
(\bm a + 2\bm b) \cdot (3\bm a - \bm b) &= 0 \\
3\lvert\bm a\rvert^2 + 5(\bm a \cdot \bm b) - 2\lvert\bm b\rvert^2 &= 0 \\
3(3)^2 + 5\lvert\bm a\rvert\lvert\bm b\rvert \cos\theta - 2(4)^2 &= 0 \\
27 + 5(3)(4) \cos\theta - 32 &= 0 \\
60 \cos \theta &= 5\ \\
\cos^{-1} \frac{1}{12} &= \theta \qed
\end{align*} ( a + 2 b ) ⋅ ( 3 a − b ) 3 ∣ a ∣ 2 + 5 ( a ⋅ b ) − 2 ∣ b ∣ 2 3 ( 3 ) 2 + 5 ∣ a ∣ ∣ b ∣ cos θ − 2 ( 4 ) 2 27 + 5 ( 3 ) ( 4 ) cos θ − 32 60 cos θ cos − 1 12 1 = 0 = 0 = 0 = 0 = 5 = θ ■ π − cos − 1 1 12 \pi - \cos^{-1} \frac{1}{12} π − cos − 1 12 1 π − \pi - π − but not for segments.