Key Concepts and Formulas

• Vector Cross Product Properties:
  • Distributive Property: $\overrightarrow x \times (\overrightarrow y + \overrightarrow z) = \overrightarrow x \times \overrightarrow y + \overrightarrow x \times \overrightarrow z$.
  • Cross Product of a Vector with Itself: $\overrightarrow x \times \overrightarrow x = \overrightarrow 0$.
  • Anti-Commutative Property: $\overrightarrow y \times \overrightarrow x = - (\overrightarrow x \times \overrightarrow y)$.
• Magnitude of a Scalar Multiple: $|k\overrightarrow v|^2 = k^2|\overrightarrow v|^2$.
• Lagrange's Identity for Vectors: ${\left| {\overrightarrow a \times \overrightarrow b } \right|^2} + {\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}$.

Step-by-Step Solution

We are asked to evaluate the expression ${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$ given $|\overrightarrow a | = 4$ and $|\overrightarrow b | = 3$.

Step 1: Expand and simplify the cross product term We begin by expanding the cross product $\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)$ using the distributive property: ${\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} = \overrightarrow a \times (\overrightarrow a + \overrightarrow b) - \overrightarrow b \times (\overrightarrow a + \overrightarrow b)$ $= (\overrightarrow a \times \overrightarrow a) + (\overrightarrow a \times \overrightarrow b) - (\overrightarrow b \times \overrightarrow a) - (\overrightarrow b \times \overrightarrow b)$ Using the property that the cross product of a vector with itself is the zero vector ($\overrightarrow x \times \overrightarrow x = \overrightarrow 0$): $= \overrightarrow 0 + (\overrightarrow a \times \overrightarrow b) - (\overrightarrow b \times \overrightarrow a) - \overrightarrow 0$ $= (\overrightarrow a \times \overrightarrow b) - (\overrightarrow b \times \overrightarrow a)$ Now, applying the anti-commutative property ($\overrightarrow b \times \overrightarrow a = - (\overrightarrow a \times \overrightarrow b)$): $= (\overrightarrow a \times \overrightarrow b) - (- (\overrightarrow a \times \overrightarrow b))$ $= (\overrightarrow a \times \overrightarrow b) + (\overrightarrow a \times \overrightarrow b)$ $= 2(\overrightarrow a \times \overrightarrow b)$

Step 2: Substitute the simplified cross product back into the original expression Substitute the result from Step 1 into the given expression: ${\left| {2\left( {\overrightarrow a \times \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$

Step 3: Apply the magnitude of a scalar multiple property Using the property $|k\overrightarrow v|^2 = k^2|\overrightarrow v|^2$, where $k=2$ and $\overrightarrow v = \overrightarrow a \times \overrightarrow b$: $2^2{\left| {\overrightarrow a \times \overrightarrow b } \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$ $= 4{\left| {\overrightarrow a \times \overrightarrow b } \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$

Step 4: Factor out the common term and apply Lagrange's Identity Factor out the common factor of 4: $= 4\left( {{\left| {\overrightarrow a \times \overrightarrow b } \right|^2} + {\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}} \right)$ Now, we apply Lagrange's Identity, ${\left| {\overrightarrow a \times \overrightarrow b } \right|^2} + {\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}$: $= 4{\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}$

Step 5: Substitute the given magnitudes Substitute the given values $|\overrightarrow a | = 4$ and $|\overrightarrow b | = 3$: $= 4 \times (4)^2 \times (3)^2$ $= 4 \times 16 \times 9$ $= 64 \times 9$ $= 576$

Common Mistakes & Tips

• Careless Expansion: Ensure each term in the cross product expansion is handled correctly, paying close attention to signs.
• Forgetting the Anti-Commutative Property: The simplification from $(\overrightarrow a \times \overrightarrow b) - (\overrightarrow b \times \overrightarrow a)$ to $2(\overrightarrow a \times \overrightarrow b)$ is crucial and relies on this property.
• Ignoring Lagrange's Identity: Recognizing Lagrange's identity significantly simplifies the problem. Without it, one would need to use $\sin \theta$ and $\cos \theta$ and the identity $\sin^2 \theta + \cos^2 \theta = 1$, which is more laborious.
• Irrelevant Information: The given range of $\theta$ is not required for this calculation, as Lagrange's identity is universally true.

Summary

The problem requires simplifying a vector expression involving cross products and dot products. By carefully applying the distributive, anti-commutative, and self-cross product properties, the cross product term simplifies to $2(\overrightarrow a \times \overrightarrow b)$. Squaring the magnitude of this term and adding it to $4(\overrightarrow a \cdot \overrightarrow b)^2$ leads to an expression that can be factored. The application of Lagrange's identity then reduces the expression to $4|\overrightarrow a|^2|\overrightarrow b|^2$. Substituting the given magnitudes yields the final numerical answer.

The final answer is $\boxed{576}$.