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Q6.5:
Show that (a.(b \times c)) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, (a), (b) and (c).

Solution :

Step 1: Geometric Representation of the Scalar Triple Product

To analyze the volume of a parallelepiped formed by three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, we define the vectors in a 3D coordinate system. Let the vectors be represented as:

  • $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$
  • $\vec{b} = b_x\hat{i} + b_y\hat{j} + b_z\hat{k}$
  • $\vec{c} = c_x\hat{i} + c_y\hat{j} + c_z\hat{k}$
$\vec{b}$ $\vec{a}$ $\vec{c}$

Step 2: Defining the Vector Cross Product ($\vec{b} \times \vec{c}$)

The cross product $\vec{n} = \vec{b} \times \vec{c}$ yields a vector whose magnitude $|\vec{b} \times \vec{c}|$ is equal to the area of the parallelogram formed by $\vec{b}$ and $\vec{c}$ [Per the definition of the Vector Product].

The direction of $\vec{n}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$.

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

Step 3: Calculating the Volume via Scalar Projection

The volume $V$ of a parallelepiped is given by the product of the base area and the perpendicular height ($h$):

$V = (\text{Area of Base}) \times (\text{Height})$

$V = |\vec{b} \times \vec{c}| \times h$

The height $h$ is the component of vector $\vec{a}$ along the direction of the normal vector $\hat{n}$ (where $\hat{n}$ is the unit vector of $\vec{b} \times \vec{c}$).

$h = |\vec{a}| \cos(\phi)$, where $\phi$ is the angle between $\vec{a}$ and the normal vector $\hat{n}$.

Step 4: Algebraic Derivation

By substituting the projection into the volume formula [By the Geometric Interpretation of the Scalar Product]:

$V = |\vec{b} \times \vec{c}| \cdot |\vec{a}| \cos(\phi)$

Since the dot product of two vectors $\vec{u}$ and $\vec{v}$ is defined as $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos(\gamma)$, we identify:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = |\vec{a}| |\vec{b} \times \vec{c}| \cos(\phi)$

Therefore, the magnitude of the scalar triple product is exactly equivalent to the volume $V$.

Summary of Determinant Form

In Cartesian coordinates, this relationship is confirmed by the determinant:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix}$

Final Solution: The magnitude of the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$ represents the volume of the parallelepiped because it equates to (Area of base defined by $\vec{b}$ and $\vec{c}$) $\times$ (Vertical projection of $\vec{a}$ onto the normal to the base).


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