Q.Show that $\vec{a} \cdot (\vec{b} \times \vec{c})$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors, $\vec{a}$, $\vec{b}$ and $\vec{c}$.

Concept: Scalar Triple Product as Volume

The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ gives the volume of the parallelepiped spanned by $\vec{a}, \vec{b}, \vec{c}$.

Step 1: Area of the base

The cross product $\vec{b} \times \vec{c}$ is a vector whose magnitude equals the area of the parallelogram formed by $\vec{b}$ and $\vec{c}$:

$|\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin\theta$, where $\theta$ is the angle between $\vec{b}$ and $\vec{c}$.

Step 2: Height of the parallelepiped

The height perpendicular to the base is the component of $\vec{a}$ along the direction of $\vec{b} \times \vec{c}$:

$h = |\vec{a}| \cos\phi$, where $\phi$ is the angle between $\vec{a}$ and $\vec{b} \times \vec{c}$. …

The key here is the geometric meaning of the cross product and the dot product, not just algebraic manipulation. Let’s build the intuition first.

The cross product $\vec{b} \times \vec{c}$ produces a vector that is perpendicular to both $\vec{b}$ and $\vec{c}$, and its magnitude equals the area of the parallelogram formed by $\vec{b}$ and $\vec{c}$. That parallelogram is the base of the parallelepiped.

Now, the dot product of $\vec{a}$ with that perpendicular vector gives the projection of $\vec{a}$ onto the direction normal to the base. That projection is precisely the height of the parallelepiped — but only if $\vec{a}$ is not necessarily perpendicular to the base. The dot product $\vec{a} \cdot (\vec{b} \times \vec{c})$ equals $|\vec{b} \times \vec{c}|$ times the component of $\vec{a}$ along the normal. That component is the perpendicular distance from the tip of $\vec{a}$ to the base plane.

So:

Base area $\times$ perpendicular height = volume. That’s exactly what the scalar triple product gives, up to a sign (which indicates orientation).

Let’s go step by step.

1. Define the vectors and the parallelepiped.

   Take three vectors $\vec{a}, \vec{b}, \vec{c}$ emanating from the same point. They form a parallelepiped — a 3D shape with opposite faces parallel. The three edges meeting at one vertex are $\vec{a}, \vec{b}, \vec{c}$.

2. Area of the base.

   Choose the face formed by $\vec{b}$ and $\vec{c}$ as the base. The area of this parallelogram is

This is a standard result: the magnitude of the cross product equals the area of the parallelogram spanned by the two vectors.

3. Direction of the normal.

   The vector $\vec{b} \times \vec{c}$ is perpendicular to the base plane. Its direction is given by the right-hand rule.

4. Height of the parallelepiped.

   The height is the perpendicular distance from the opposite face (the one containing the tip of $\vec{a}$) to the base plane. This is the magnitude of the projection of $\vec{a}$ onto the unit normal $\hat{n} = \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}$. …

Concept: Scalar Triple Product as the Volume of a Parallelepiped

Step 1: Identify the base

Take the parallelogram formed by $\vec b$ and $\vec c$ as the base of the parallelepiped. Its area is

Step 2: Identify the direction of the height

$\vec b\times\vec c$ is perpendicular to the base plane (right-hand rule).

Step 3: Find the perpendicular height

The height is the component of $\vec a$ along the unit normal $\hat n = \dfrac{\vec b\times\vec c}{|\vec b\times\vec c|}$:

Step 4: Compute the volume …