Key Concepts and Formulas

1. Vector Cross Product Properties:

   • Anti-commutative Property: $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$.
   • Distributive Property: $(\vec{A} + \vec{B}) \times \vec{C} = \vec{A} \times \vec{C} + \vec{B} \times \vec{C}$.
   • Parallel Vectors: If $\vec{X} \times \vec{Y} = \vec{0}$ and $\vec{X} \neq \vec{0}$, then $\vec{Y} = k\vec{X}$ for some scalar $k$.

2. Vector Dot Product Properties:

   • Distributive Property: $\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}$.
   • Scalar Multiplication: $(k\vec{A}) \cdot \vec{B} = k(\vec{A} \cdot \vec{B})$.
   • Magnitude Squared: $|\vec{A}|^2 = \vec{A} \cdot \vec{A}$.

Step-by-Step Solution

Step 1: Simplify the Cross Product Equation to find the relationship of $\vec{c}$ with $\vec{a}$ and $\vec{b}$

We are given the equation: $(\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b})$

Using the anti-commutative property of the cross product ($\vec{c} \times \vec{Y} = -\vec{Y} \times \vec{c}$), we can rewrite the right-hand side: $(\vec{a}+\vec{b}) \times \vec{c} = -(-2 \vec{a}+3 \vec{b}) \times \vec{c}$

Rearranging the terms to one side: $(\vec{a}+\vec{b}) \times \vec{c} + (-2 \vec{a}+3 \vec{b}) \times \vec{c} = \vec{0}$

Using the distributive property of the cross product: $[(\vec{a}+\vec{b}) + (-2 \vec{a}+3 \vec{b})] \times \vec{c} = \vec{0}$

Combine the vectors inside the bracket: $[\vec{a}+\vec{b}-2 \vec{a}+3 \vec{b}] \times \vec{c} = \vec{0}$ $[-\vec{a}+4 \vec{b}] \times \vec{c} = \vec{0}$

This equation implies that the vector $(-\vec{a}+4 \vec{b})$ is parallel to $\vec{c}$. Therefore, $\vec{c}$ can be expressed as a scalar multiple of $(-\vec{a}+4 \vec{b})$: $\vec{c} = k(-\vec{a}+4 \vec{b})$ where $k$ is a scalar constant.

Step 2: Calculate necessary dot products of $\vec{a}$ and $\vec{b}$

We are given $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=11 \hat{i}-\hat{j}+\hat{k}$. We need to compute $|\vec{a}|^2$, $|\vec{b}|^2$, and $\vec{a} \cdot \vec{b}$.

• $|\vec{a}|^2$: $|\vec{a}|^2 = (4)^2 + (-1)^2 + (1)^2 = 16 + 1 + 1 = 18$

• $|\vec{b}|^2$: $|\vec{b}|^2 = (11)^2 + (-1)^2 + (1)^2 = 121 + 1 + 1 = 123$

• $\vec{a} \cdot \vec{b}$: $\vec{a} \cdot \vec{b} = (4)(11) + (-1)(-1) + (1)(1) = 44 + 1 + 1 = 46$

Step 3: Use the Dot Product Equation to find the scalar $k$

We are given the dot product equation: $(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$

Substitute $\vec{c} = k(-\vec{a}+4 \vec{b})$ into this equation: $(2 \vec{a}+3 \vec{b}) \cdot [k(-\vec{a}+4 \vec{b})] = 1670$

Move the scalar $k$ outside the dot product: $k[(2 \vec{a}+3 \vec{b}) \cdot (-\vec{a}+4 \vec{b})] = 1670$

Now, expand the dot product term $(2 \vec{a}+3 \vec{b}) \cdot (-\vec{a}+4 \vec{b})$: $(2 \vec{a}+3 \vec{b}) \cdot (-\vec{a}+4 \vec{b}) = (2\vec{a}) \cdot (-\vec{a}) + (2\vec{a}) \cdot (4\vec{b}) + (3\vec{b}) \cdot (-\vec{a}) + (3\vec{b}) \cdot (4\vec{b})$ $= -2(\vec{a} \cdot \vec{a}) + 8(\vec{a} \cdot \vec{b}) - 3(\vec{b} \cdot \vec{a}) + 12(\vec{b} \cdot \vec{b})$ Using $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, $\vec{b} \cdot \vec{b} = |\vec{b}|^2$, and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$: $= -2|\vec{a}|^2 + (8-3)(\vec{a} \cdot \vec{b}) + 12|\vec{b}|^2$ $= -2|\vec{a}|^2 + 5(\vec{a} \cdot \vec{b}) + 12|\vec{b}|^2$

Substitute the values calculated in Step 2: $= -2(18) + 5(46) + 12(123)$ $= -36 + 230 + 1476$ $= 194 + 1476$ $= 1670$

Now, substitute this result back into the equation for $k$: $k[1670] = 1670$ $k = 1$

Step 4: Calculate $|\vec{c}|^2$

We have $\vec{c} = k(-\vec{a}+4 \vec{b})$ and we found $k=1$. So, $\vec{c} = -\vec{a}+4 \vec{b}$.

To find $|\vec{c}|^2$, we use the property $|\vec{c}|^2 = \vec{c} \cdot \vec{c}$: $|\vec{c}|^2 = (-\vec{a}+4 \vec{b}) \cdot (-\vec{a}+4 \vec{b})$

Expand the dot product: $|\vec{c}|^2 = (-\vec{a}) \cdot (-\vec{a}) + (-\vec{a}) \cdot (4\vec{b}) + (4\vec{b}) \cdot (-\vec{a}) + (4\vec{b}) \cdot (4\vec{b})$ $|\vec{c}|^2 = |\vec{a}|^2 - 4(\vec{a} \cdot \vec{b}) - 4(\vec{b} \cdot \vec{a}) + 16|\vec{b}|^2$ $|\vec{c}|^2 = |\vec{a}|^2 - 8(\vec{a} \cdot \vec{b}) + 16|\vec{b}|^2$

Substitute the numerical values calculated in Step 2: $|\vec{c}|^2 = 18 - 8(46) + 16(123)$ $|\vec{c}|^2 = 18 - 368 + 1968$ $|\vec{c}|^2 = 1986 - 368$ $|\vec{c}|^2 = 1600$

This value corresponds to option (A).

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when applying the anti-commutative property of the cross product and when expanding dot products involving negative terms.
• Scalar Multiplication: Ensure that scalars are correctly factored out of dot and cross products.
• Systematic Calculation: Calculate the required $|\vec{a}|^2$, $|\vec{b}|^2$, and $\vec{a} \cdot \vec{b}$ values upfront to avoid repetitive calculations and reduce the chance of errors.

Summary

The problem was solved by first simplifying the given cross product equation to establish that the unknown vector $\vec{c}$ is parallel to a linear combination of $\vec{a}$ and $\vec{b}$. This allowed us to express $\vec{c}$ as a scalar multiple of $(-\vec{a}+4\vec{b})$. We then calculated the necessary scalar products of $\vec{a}$ and $\vec{b}$. Substituting the expression for $\vec{c}$ into the given dot product equation enabled us to solve for the scalar constant $k$. Finally, we used the determined value of $k$ and the expression for $\vec{c}$ to calculate $|\vec{c}|^2$.

The final answer is $\boxed{\text{1600}}$.