Key Concepts and Formulas

• Distributive Property of Dot Product: For any vectors $\vec{x}, \vec{y}, \vec{z}$ and scalar $k$, $\vec{x} \cdot (\vec{y} + \vec{z}) = \vec{x} \cdot \vec{y} + \vec{x} \cdot \vec{z}$ and $\vec{x} \cdot (k\vec{y}) = k(\vec{x} \cdot \vec{y})$.
• Dot Product of a Vector with Itself: $\vec{x} \cdot \vec{x} = |\vec{x}|^2$.
• Scalar Triple Product Property: The scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w})$ is zero if any two of the vectors are identical or parallel. Specifically, $\vec{u} \cdot (\vec{v} \times \vec{u}) = 0$.

Step-by-Step Solution

We are given $|\vec{a}|=1$, $|\vec{b}|=4$, and $\vec{a} \cdot \vec{b}=2$. We need to find $\vec{b} \cdot \vec{c}$, where $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$.

Step 1: Substitute the expression for $\vec{c}$ into the dot product. Our goal is to compute $\vec{b} \cdot \vec{c}$. We substitute the given expression for $\vec{c}$: $\vec{b} \cdot \vec{c} = \vec{b} \cdot ((2 \vec{a} \times \vec{b}) - 3 \vec{b})$

• Why: This step directly sets up the calculation we need to perform.

Step 2: Apply the distributive property of the dot product. We distribute $\vec{b}$ over the terms within the parentheses: $\vec{b} \cdot \vec{c} = \vec{b} \cdot (2 \vec{a} \times \vec{b}) - \vec{b} \cdot (3 \vec{b})$

• Why: This breaks down the problem into two simpler dot product calculations.

Step 3: Evaluate the first term: $\vec{b} \cdot (2 \vec{a} \times \vec{b})$. We can factor out the scalar $2$ from the dot product: $\vec{b} \cdot (2 \vec{a} \times \vec{b}) = 2 (\vec{b} \cdot (\vec{a} \times \vec{b}))$ Now, we examine the scalar triple product $\vec{b} \cdot (\vec{a} \times \vec{b})$. In this scalar triple product, the vector $\vec{b}$ appears twice. According to the property of scalar triple products, if two of the vectors are identical, the value is zero. $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ Therefore, the first term is: $2 \times 0 = 0$

• Why: This step utilizes a key property of scalar triple products to simplify the expression significantly.

Step 4: Evaluate the second term: $- \vec{b} \cdot (3 \vec{b})$. We can factor out the scalar $3$: $- \vec{b} \cdot (3 \vec{b}) = -3 (\vec{b} \cdot \vec{b})$ Now, we use the property that the dot product of a vector with itself is the square of its magnitude: $\vec{b} \cdot \vec{b} = |\vec{b}|^2$. $-3 (\vec{b} \cdot \vec{b}) = -3 |\vec{b}|^2$ We are given that $|\vec{b}|=4$. Substituting this value: $-3 (4)^2 = -3 \times 16 = -48$

• Why: This step uses the definition of the dot product with itself and the given magnitude of $\vec{b}$ to find the value of the second term.

Step 5: Combine the results of the two terms. Now we add the results from Step 3 and Step 4: $\vec{b} \cdot \vec{c} = 0 + (-48)$ $\vec{b} \cdot \vec{c} = -48$

• Why: This is the final step where we sum the simplified values of the two parts of the original expression.

Common Mistakes & Tips

• Scalar Triple Product Property: Always remember that if a scalar triple product has repeated vectors (e.g., $\vec{u} \cdot (\vec{v} \times \vec{u})$), its value is zero. This is a powerful shortcut.
• Distributive Property of Dot Product: Ensure you correctly apply the distributive property of the dot product, not the cross product, when dealing with expressions like $\vec{x} \cdot (\vec{y} - \vec{z})$.
• Unnecessary Calculations: Notice that $|\vec{a}|$ and $\vec{a} \cdot \vec{b}$ were not needed for this specific calculation. Avoid calculating vector components if vector properties can directly solve the problem.

Summary

To find $\vec{b} \cdot \vec{c}$, we substituted the expression for $\vec{c}$ and applied the distributive property of the dot product. The first term, involving a scalar triple product with repeated vectors, simplified to zero. The second term was evaluated using the dot product of a vector with itself and the given magnitude of $\vec{b}$. Combining these results yielded the final answer.

The final answer is $\boxed{\text{-48}}$.