Key Concepts and Formulas

• Magnitude of Cross Product: For any two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, the magnitude of their cross product is given by $\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta$, where $\theta$ is the angle between the vectors ($0 \le \theta \le \pi$).
• Dot Product: The dot product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\overrightarrow a .\,\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta$.
• Fundamental Vector Identity: A crucial identity relating the dot and cross products is ${\left| {\overrightarrow a \times \overrightarrow b } \right|^2} + {{\left( {\overrightarrow a .\,\overrightarrow b } \right)}^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}$. This identity is derived by squaring the definitions of the cross product and dot product and using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$.

Step-by-Step Solution

Step 1: Identify the Given Information We are given the magnitudes of two vectors and the magnitude of their cross product:

• $\left| {\overrightarrow a } \right| = 2$
• $\left| {\overrightarrow b } \right| = 5$
• $\left| {\overrightarrow a \times \overrightarrow b } \right| = 8$ We need to find the magnitude of the dot product, $\left| {\overrightarrow a .\,\overrightarrow b } \right|$.

Step 2: Choose the Appropriate Formula To relate the magnitudes of the cross product and dot product with the magnitudes of the individual vectors, the fundamental vector identity is the most direct approach. This identity is: ${\left| {\overrightarrow a \times \overrightarrow b } \right|^2} + {{\left( {\overrightarrow a .\,\overrightarrow b } \right)}^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}$

• Why this step? This identity directly connects all the given information with the quantity we need to find. It avoids the intermediate step of calculating the angle $\theta$, which can sometimes introduce sign errors or unnecessary calculations.

Step 3: Substitute the Given Values into the Identity Substitute the given values into the chosen identity: ${(8)^2} + {\left( {\overrightarrow a .\,\overrightarrow b } \right)^2} = {(2)^2}{(5)^2}$

• Why this step? We are substituting the known numerical values of the vector magnitudes and the cross product magnitude into the identity to set up an equation for the unknown dot product magnitude.

Step 4: Simplify the Equation Perform the squaring and multiplication operations: $64 + {\left( {\overrightarrow a .\,\overrightarrow b } \right)^2} = 4 \times 25$ $64 + {\left( {\overrightarrow a .\,\overrightarrow b } \right)^2} = 100$

• Why this step? Simplifying the equation makes it easier to isolate the term containing the dot product.

Step 5: Isolate the Term Containing the Dot Product Subtract 64 from both sides of the equation to isolate ${\left( {\overrightarrow a .\,\overrightarrow b } \right)^2}$: ${\left( {\overrightarrow a .\,\overrightarrow b } \right)^2} = 100 - 64$ ${\left( {\overrightarrow a .\,\overrightarrow b } \right)^2} = 36$

• Why this step? This algebraic manipulation isolates the square of the dot product, preparing it for the final step of finding its magnitude.

Step 6: Find the Magnitude of the Dot Product Take the square root of both sides to find $\left| {\overrightarrow a .\,\overrightarrow b } \right|$. Since the magnitude must be non-negative, we take the positive square root: $\sqrt{{{\left( {\overrightarrow a .\,\overrightarrow b } \right)}^2}} = \sqrt{36}$ $\left| {\overrightarrow a .\,\overrightarrow b } \right| = 6$

• Why this step? The question asks for the magnitude of the dot product, $\left| {\overrightarrow a .\,\overrightarrow b } \right|$. The square root of ${\left( {\overrightarrow a .\,\overrightarrow b } \right)^2}$ is $\left| {\overrightarrow a .\,\overrightarrow b } \right|$, which is always non-negative.

Common Mistakes & Tips

• Misinterpreting the Question: Ensure you are solving for the magnitude of the dot product ($\left| {\overrightarrow a .\,\overrightarrow b } \right|$) and not the dot product itself ($\overrightarrow a .\,\overrightarrow b$). While $\overrightarrow a .\,\overrightarrow b$ could be $+6$ or $-6$, its magnitude is always $6$.
• Forgetting the Identity: The identity ${\left| {\overrightarrow a \times \overrightarrow b } \right|^2} + {{\left( {\overrightarrow a .\,\overrightarrow b } \right)}^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}$ is a powerful tool. Memorizing it can save significant time in problems involving both dot and cross products.
• Sign Errors with Angles: If one were to calculate the angle $\theta$ first, there could be sign errors. For instance, $\sin\theta = \frac{8}{2 \times 5} = \frac{8}{10} = \frac{4}{5}$. Then $\cos\theta = \pm \sqrt{1 - \sin^2\theta} = \pm \sqrt{1 - (\frac{4}{5})^2} = \pm \sqrt{1 - \frac{16}{25}} = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}$. The dot product would be $\overrightarrow a .\,\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = 2 \times 5 \times (\pm \frac{3}{5}) = \pm 6$. The magnitude $\left| {\overrightarrow a .\,\overrightarrow b } \right|$ is then $6$. Using the identity directly bypasses these potential sign issues.

Summary

This problem is efficiently solved using the fundamental vector identity that relates the magnitudes of the cross product and dot product of two vectors to the magnitudes of the individual vectors. By substituting the given values into the identity ${\left| {\overrightarrow a \times \overrightarrow b } \right|^2} + {{\left( {\overrightarrow a .\,\overrightarrow b } \right)}^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}$, we can directly compute the square of the dot product and then its magnitude. This approach is robust and avoids potential complications associated with calculating trigonometric functions of the angle between the vectors.

The final answer is \boxed{6}.