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Chapter 6: Problem 26

Using coordinates, show that if \(s\) and \(t\) are scalars and \(\mathbf{u}\) is a vector, then $$ (s t) \mathbf{u}=s(t \mathbf{u}) $$.

Short Answer

Expert verified
To prove that \((s t) \mathbf{u}=s(t \mathbf{u})\), we first define the vector \(\mathbf{u}\) with components \((u_x, u_y, u_z)\) and the scalars \(s\) and \(t\). Next, calculate the LHS and RHS of the equation, finding that \((s t) \mathbf{u} = (st u_x, st u_y, st u_z)\) and \(s(t \mathbf{u}) = (st u_x, st u_y, st u_z)\). As the components are equal for both sides, it follows that \((s t) \mathbf{u}=s(t \mathbf{u})\), proving the exercise.

Step by step solution

01

Define the vector and scalars

Let the vector \(\mathbf{u}\) have components \((u_x, u_y, u_z)\) and the scalars be \(s\) and \(t\).
02

Calculate the Left-Hand Side (LHS) of the equation

We need to find \((s t) \mathbf{u}\), which means scalar multiplication of \(\mathbf{u}\) with the product of the scalars \(s\) and \(t\). Using the definition of scalar multiplication, we can find that the components of the new vector \((s t) \mathbf{u}\) are: \[ (s t) \mathbf{u} = (st u_x, st u_y, st u_z) \]
03

Calculate the Right-Hand Side (RHS) of the equation

We need to find \(s(t \mathbf{u})\), which means scalar multiplication of \(\mathbf{u}\) with the scalar \(t\), then with the scalar \(s\). Using the definition of scalar multiplication, we can find: \[ t \mathbf{u} = (t u_x, t u_y, t u_z) \] Now, we can multiply the new vector \(t \mathbf{u}\) by the scalar \(s\), giving us: \[ s(t \mathbf{u}) = s(t u_x, t u_y, t u_z) = (st u_x, st u_y, st u_z) \]
04

Compare the LHS and RHS

Now that we have both the LHS and RHS of the equation, we can easily compare their components. From what we have found: \[ (s t) \mathbf{u} = (st u_x, st u_y, st u_z) \] and \[ s(t \mathbf{u}) = (st u_x, st u_y, st u_z) \] Since \((st u_x, st u_y, st u_z)\) is equal for both the LHS and RHS, we can conclude that: \[ (s t) \mathbf{u} = s(t \mathbf{u}) \] Hence, the exercise is proved.

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