Show that: (a⃗ × b⃗)² = |a⃗ · a⃗ a⃗ · b⃗; a⃗ · b⃗ b⃗ · b⃗|

Show that: $(\vec{a} \times \vec{b})^2 = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} \end{vmatrix}$

Given: $(\vec{a} \times \vec{b})^2$

To show: $\begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} \end{vmatrix}$

Open in the 3D geometry and vectors calculator

Answer

Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$. Then $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2\sin^2\theta$. $= |\vec{a}|^2|\vec{b}|^2(1 - \cos^2\theta) = |\vec{a}|^2|\vec{b}|^2 - (|\vec{a}||\vec{b}|\cos\theta)^2$ $= (\vec{a}\cdot\vec{a})(\vec{b}\cdot\vec{b}) - (\vec{a}\cdot\vec{b})^2$ $= \begin{vmatrix} \vec{a}\cdot\vec{a} & \vec{a}\cdot\vec{b} \\ \vec{a}\cdot\vec{b} & \vec{b}\cdot\vec{b} \end{vmatrix}$. Hence proved.

$(\vec{a}\times\vec{b})^2 = |\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2\sin^2\theta$, by the magnitude of a cross product.
$= |\vec{a}|^2|\vec{b}|^2(1-\cos^2\theta)$, using $\sin^2\theta = 1-\cos^2\theta$.
$= |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2$, since $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$.
$= (\vec{a}\cdot\vec{a})(\vec{b}\cdot\vec{b}) - (\vec{a}\cdot\vec{b})(\vec{a}\cdot\vec{b})$, since $\vec{a}\cdot\vec{a} = |\vec{a}|^2$.
$= \begin{vmatrix} \vec{a}\cdot\vec{a} & \vec{a}\cdot\vec{b} \\ \vec{a}\cdot\vec{b} & \vec{b}\cdot\vec{b} \end{vmatrix}$, by the definition of a $2\times2$ determinant.