Key Concepts and Formulas

• Distributive Property of Dot Product: $\vec{X} \cdot (\vec{Y} - \vec{Z}) = \vec{X} \cdot \vec{Y} - \vec{X} \cdot \vec{Z}$.
• Scalar Triple Product: $[\vec{X} \ \vec{Y} \ \vec{Z}] = \vec{X} \cdot (\vec{Y} \times \vec{Z}) = (\vec{X} \times \vec{Y}) \cdot \vec{Z}$.
• Magnitude Squared: $|\vec{X}|^2 = \vec{X} \cdot \vec{X}$.
• Orthogonality from Cross Product: If $\vec{X} \times \vec{Y} = \vec{Z}$, then $\vec{Z}$ is perpendicular to both $\vec{X}$ and $\vec{Y}$, which implies $\vec{X} \cdot \vec{Z} = 0$ and $\vec{Y} \cdot \vec{Z} = 0$.
• Dot Product of Vectors: If $\vec{X} = X_x\hat{i} + X_y\hat{j} + X_z\hat{k}$ and $\vec{Y} = Y_x\hat{i} + Y_y\hat{j} + Y_z\hat{k}$, then $\vec{X} \cdot \vec{Y} = X_xY_x + X_yY_y + X_zY_z$.

Step-by-Step Solution

Step 1: Analyze the given information and the expression to be evaluated. We are given $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=3(\hat{i}-\hat{j}+\hat{k})=3\hat{i}-3\hat{j}+3\hat{k}$. We are also given two conditions involving $\vec{c}$:

1. $\vec{a} \times \vec{c}=\vec{b}$
2. $\vec{a} \cdot \vec{c}=3$ We need to find the value of $\vec{a} \cdot((\vec{c} \times \vec{b})-\vec{b}-\vec{c})$.

Step 2: Expand the expression using the distributive property of the dot product. The expression is $E = \vec{a} \cdot((\vec{c} \times \vec{b})-\vec{b}-\vec{c})$. Using the distributive property, we can break this down into three terms: $E = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$ We will evaluate each of these terms separately.

Step 3: Evaluate the first term: $\vec{a} \cdot (\vec{c} \times \vec{b})$. This is a scalar triple product. Using the property $\vec{X} \cdot (\vec{Y} \times \vec{Z}) = (\vec{X} \times \vec{Y}) \cdot \vec{Z}$, we can rewrite the term as: $\vec{a} \cdot (\vec{c} \times \vec{b}) = (\vec{a} \times \vec{c}) \cdot \vec{b}$ We are given that $\vec{a} \times \vec{c} = \vec{b}$. Substituting this into the expression: $(\vec{a} \times \vec{c}) \cdot \vec{b} = \vec{b} \cdot \vec{b}$ The dot product of a vector with itself is the square of its magnitude: $\vec{b} \cdot \vec{b} = |\vec{b}|^2$ Now, we calculate $|\vec{b}|^2$ using the components of $\vec{b} = 3\hat{i}-3\hat{j}+3\hat{k}$: $|\vec{b}|^2 = (3)^2 + (-3)^2 + (3)^2 = 9 + 9 + 9 = 27$ So, the first term is 27.

Step 4: Evaluate the second term: $\vec{a} \cdot \vec{b}$. From the given condition $\vec{a} \times \vec{c} = \vec{b}$, we know that $\vec{b}$ is perpendicular to $\vec{a}$. Therefore, their dot product is zero. $\vec{a} \cdot \vec{b} = 0$ We can also verify this by direct calculation: $\vec{a} = \hat{i}+2 \hat{j}+\hat{k}$ $\vec{b} = 3\hat{i}-3\hat{j}+3\hat{k}$ $\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$ So, the second term is 0.

Step 5: Evaluate the third term: $\vec{a} \cdot \vec{c}$. This value is directly given in the problem statement: $\vec{a} \cdot \vec{c} = 3$ So, the third term is 3.

Step 6: Combine the evaluated terms to find the value of the expression. Substitute the values of the three terms back into the expanded expression for $E$: $E = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$ $E = 27 - 0 - 3$ $E = 24$

Common Mistakes & Tips

• Misapplication of Vector Identities: Ensure correct usage of scalar triple product properties and distributive laws.
• Calculation Errors: Double-check dot product and magnitude calculations, as these are common sources of error.
• Ignoring Orthogonality: The condition $\vec{a} \times \vec{c} = \vec{b}$ is a strong hint that $\vec{a}$ and $\vec{b}$ are orthogonal, which simplifies $\vec{a} \cdot \vec{b}$ to zero.

Summary

The problem requires evaluating a vector expression involving dot and cross products. By utilizing the distributive property of the dot product, we decomposed the expression into three terms. The first term, a scalar triple product, was simplified using the given cross product condition and the property $\vec{X} \cdot (\vec{Y} \times \vec{Z}) = (\vec{X} \times \vec{Y}) \cdot \vec{Z}$, leading to $|\vec{b}|^2$. The second term, $\vec{a} \cdot \vec{b}$, was found to be zero due to the orthogonality implied by the cross product condition. The third term was given directly. Combining these results yielded the final answer.

The final answer is $\boxed{24}$.