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Chapter 6: Problem 19

Show that the scalar product obeys the distributive law: \(\vec{A} \cdot(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}\)

Short Answer

Expert verified
The distributive law of scalar product has been proven: \(\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}\). This was done by performing vector addition, performing the resulting scalar product and finally distributing \(\vec{A}\) over the addition operation.

Step by step solution

01

Vector Addition

First, add the vectors \(\vec{B}\) and \(\vec{C}\). This operation can be performed component-wise, i.e., \(B_x + C_x, B_y + C_y, B_z + C_z\). Let's denote the result as \(\vec{D} = \vec{B} + \vec{C}\).
02

Perform Scalar Product with \(\vec{A}\) and \(\vec{D}\)

Next, perform the scalar product between vectors \(\vec{A}\) and \(\vec{D}\). By the definition of scalar product, this equals to \(A_x \cdot D_x + A_y \cdot D_y + A_z \cdot D_z\). Substituting \(\vec{D}\) from Step 1 gives \(\vec{A} \cdot (\vec{B} + \vec{C}) = A_x \cdot (B_x + C_x) + A_y \cdot (B_y + C_y) + A_z \cdot (B_z + C_z)\).
03

Distribute \(\vec{A}\)

Now distribute \(\vec{A}\) over the addition operation. This gives, \(A_x \cdot B_x + A_x \cdot C_x + A_y \cdot B_y + A_y \cdot C_y + A_z \cdot B_z + A_z \cdot C_z\). If we group these terms back together, we get \((A_x \cdot B_x + A_y \cdot B_y + A_z \cdot B_z) + (A_x \cdot C_x + A_y \cdot C_y + A_z \cdot C_z)\) - which are just the scalar products of \(\vec{A}\) with \(\vec{B}\) and \(\vec{C}\) respectively.

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