Understanding the Vector Dot Product Problem

The question asks us to find the magnitude of vector \( \mathbf{b} \), denoted by \( | \mathbf{b} | \), given two conditions involving vectors \( \mathbf{a} \) and \( \mathbf{b} \). The conditions are: \( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = 27 \) and \( | \mathbf{a} | = 2 | \mathbf{b} | \).

Expanding the Dot Product Expression

Let's start by expanding the dot product expression \( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) \). We can use the distributive property of the dot product, similar to how we multiply binomials in algebra:

\( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = \mathbf{a} \cdot (\mathbf{a} + \mathbf{b}) - \mathbf{b} \cdot (\mathbf{a} + \mathbf{b}) \)

Further expanding:

\( = \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{b} - \mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} \)

Using Properties of the Dot Product

We know the following properties of the dot product:

• The dot product of a vector with itself is the square of its magnitude: \( \mathbf{v} \cdot \mathbf{v} = | \mathbf{v} |^2 \). So, \( \mathbf{a} \cdot \mathbf{a} = | \mathbf{a} |^2 \) and \( \mathbf{b} \cdot \mathbf{b} = | \mathbf{b} |^2 \).
• The dot product is commutative: \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \).

Substituting these properties into our expanded expression:

\( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = | \mathbf{a} |^2 + \mathbf{a} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{b} - | \mathbf{b} |^2 \)

The terms \( + \mathbf{a} \cdot \mathbf{b} \) and \( - \mathbf{a} \cdot \mathbf{b} \) cancel each other out:

\( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = | \mathbf{a} |^2 - | \mathbf{b} |^2 \)

Setting up the Equation

We are given that \( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = 27 \). So, we have the equation:

\( | \mathbf{a} |^2 - | \mathbf{b} |^2 = 27 \quad \text{(Equation 1)} \)

Using the Second Condition

We are also given that \( | \mathbf{a} | = 2 | \mathbf{b} | \quad \text{(Equation 2)} \). We can use this to substitute for \( | \mathbf{a} | \) in Equation 1.

Substitute \( | \mathbf{a} | = 2 | \mathbf{b} | \) into \( | \mathbf{a} |^2 - | \mathbf{b} |^2 = 27 \):

\( (2 | \mathbf{b} |)^2 - | \mathbf{b} |^2 = 27 \)

Squaring \( 2 | \mathbf{b} | \):

\( 4 | \mathbf{b} |^2 - | \mathbf{b} |^2 = 27 \)

Solving for \( | \mathbf{b} | \)

Now, we combine the terms involving \( | \mathbf{b} |^2 \):

\( 3 | \mathbf{b} |^2 = 27 \)

Divide both sides by 3:

\( | \mathbf{b} |^2 = \frac{27}{3} \)

\( | \mathbf{b} |^2 = 9 \)

To find \( | \mathbf{b} | \), we take the square root of both sides. Since the magnitude of a vector is always non-negative, we take the positive square root:

\( | \mathbf{b} | = \sqrt{9} \)

\( | \mathbf{b} | = 3 \)

Final Answer for Vector Magnitude

The magnitude of vector \( \mathbf{b} \) is 3.

Let's verify the answer using the given conditions:

• If \( | \mathbf{b} | = 3 \), then \( | \mathbf{a} | = 2 | \mathbf{b} | = 2 \times 3 = 6 \).
• Check \( | \mathbf{a} |^2 - | \mathbf{b} |^2 = 27 \): \( 6^2 - 3^2 = 36 - 9 = 27 \). This matches the given condition.

The value \( | \mathbf{b} | = 3 \) satisfies both given equations.

Revision Table: Key Steps in Finding Vector Magnitude

Additional Information: Dot Product and Vector Magnitude

The dot product (also known as the scalar product) of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is a scalar quantity. It is defined as \( \mathbf{u} \cdot \mathbf{v} = | \mathbf{u} | | \mathbf{v} | \cos \theta \), where \( \theta \) is the angle between the vectors. If vectors are given in component form, say \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), the dot product is \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \). An important property is that \( \mathbf{v} \cdot \mathbf{v} = | \mathbf{v} |^2 \), because the angle between a vector and itself is 0, and \( \cos 0^\circ = 1 \).

The magnitude of a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) is its length, calculated as \( | \mathbf{v} | = \sqrt{v_1^2 + v_2^2 + v_3^2} \). Thus, \( | \mathbf{v} |^2 = v_1^2 + v_2^2 + v_3^2 \). This confirms why \( \mathbf{v} \cdot \mathbf{v} = | \mathbf{v} |^2 \).

The identity \( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = | \mathbf{a} |^2 - | \mathbf{b} |^2 \) is a very useful result in vector algebra, analogous to the difference of squares formula in scalar algebra.