Determinant form

✍️ Worked examples

• Example 3 Find $\vec{A}\times \vec{B}$ for $\vec{A}=(-2,3,1)$ and $\vec{B}=(1,2,4)$, and a unit vector perpendicular to both.
• Solution:
• i^ coefficient 3×4−1×2=10.
• j^ coefficient −[(−2)(4)−(1)(1)]=−(−9)=9.
• k^ coefficient (−2)(2)−(3)(1)=−7.
• $\vec{A}\times \vec{B}=(10,9,-7)$, with norm $\sqrt{100+81+49}=\sqrt{230}$.
• Unit vector perpendicular to the plane 1230(10,9,−7) (the opposite vector also works).
• Example 4 Find a unit vector perpendicular to the plane of $\vec{A}=(2,-3,1)$ and $\vec{B}=(-1,2,-3)$.
• Solution:
• i^ (−3)(−3)−(1)(2)=7.
• j^ −[(2)(−3)−(1)(−1)]=−(−5)=5.
• k^ (2)(2)−(−3)(−1)=1.
• $\vec{A}\times \vec{B}=(7,5,1)$ and its norm is $\sqrt{75}=5\sqrt{3}$.
• Unit vector 153(7,5,1).
• Check (7,5,1)⋅A⃗=14−15+1=0 and (7,5,1)⋅B⃗=−7+10−3=0.
• Example 5 For $\vec{A}=(3,2)$ and $\vec{B}=(1,4)$ in the plane, find $\vec{A}\times \vec{B}$.
• Solution:
• (3)(4)−(2)(1)=10, so A⃗×B⃗=10 k^ the product points along the positive z-axis, since B⃗ is anticlockwise from A⃗.
• Example 6:
• $ABCD$ is a square of side 12 cm, labelled anticlockwise, and $\widehat{e}$ is the unit vector perpendicular to its plane (out of the page).
• Find $\overrightarrow{AB}\times \overrightarrow{AC}$ and $\overrightarrow{AB}\times \overrightarrow{BC}$.
• Solution:
• The angle between $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is $45^{\circ }$ and $AC=12\sqrt{2}$.
• $\overrightarrow{AB}\times \overrightarrow{AC}=12\times 12\sqrt{2}\times \sin 45^{\circ }\widehat{e}=144\widehat{e}$.
• AB→⊥BC→ AB→×BC→=12×12×1 e^=144 e^.

Definition

• For three vectors $\vec{A}$, $\vec{B}$, $\vec{C}$ the scalar triple product is
  $$\vec{A}\cdot (\vec{B}\times \vec{C})$$
• The brackets are essential in the sense that $\vec{B}\times \vec{C}$ is calculated first (the other order, $(\vec{A}\cdot \vec{B})\times \vec{C}$, is meaningless since $\vec{A}\cdot \vec{B}$ is a number).
• The result is a number.

• If $\vec{A}=(A_x,A_y,A_z)$, $\vec{B}=(B_x,B_y,B_z)$, $\vec{C}=(C_x,C_y,C_z)$, then
  $$\vec{A}\cdot (\vec{B}\times \vec{C})=\mid \begin{array}{ccc}{A}_x& A_y& A_z\\ B_x& B_y& B_z\\ C_x& C_y& C_z\end{array}\mid$$
  $$=A_x(B_y{C}_z-B_z{C}_y)-A_y(B_x{C}_z-B_z{C}_x)+A_z(B_x{C}_y-B_y{C}_x)$$

• Cyclic permutation keeps the value A⃗⋅(B⃗×C⃗)=B⃗⋅(C⃗×A⃗)=C⃗⋅(A⃗×B⃗).
• Interchanging two vectors changes the sign (two rows of the determinant swap).
• $\vec{A}\cdot (\vec{B}\times \vec{C})=(\vec{A}\times \vec{B})\cdot \vec{C}$, so the dot and the cross may be exchanged.
• If two of the vectors are equal or parallel, the product is zero.
• If the three vectors lie in one plane (coplanar), the product is zero, and conversely.

✍️ Worked examples

• Example 13 Calculate $\vec{A}\cdot (\vec{B}\times \vec{C})$ for $\vec{A}=(2,1,3)$, $\vec{B}=(-1,3,2)$, $\vec{C}=(1,1,-2)$.
• Solution:
• Determinant 2[3(−2)−2(1)]−1[(−1)(−2)−2(1)]+3[(−1)(1)−3(1)].
• $=2(-8)-1(0)+3(-4)=-16-0-12=-28$.
• Example 14 Show that $\vec{A}=(1,2,3)$, $\vec{B}=(0,1,4)$, $\vec{C}=(1,3,7)$ are coplanar.
• Solution:
• Note that $\vec{C}=\vec{A}+\vec{B}$.
• Determinant 1[1×7−4×3]−2[0×7−4×1]+3[0×3−1×1]=−5+8−3=0.
• The triple product is zero, so the vectors are coplanar.