Key Concepts and Formulas

• Vector Triple Product Identity: For any three vectors $\vec{P}, \vec{Q}, \vec{R}$, the vector triple product is given by: $(\vec{P} \times \vec{Q}) \times \vec{R} = (\vec{P} \cdot \vec{R})\vec{Q} - (\vec{Q} \cdot \vec{R})\vec{P}$
• Properties of Cross Product:
  • $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$ (Anti-commutativity)
  • $\vec{A} \times \vec{A} = \vec{0}$
  • The cross product $\vec{A} \times \vec{B}$ is orthogonal to both $\vec{A}$ and $\vec{B}$. This implies $(\vec{A} \times \vec{B}) \cdot \vec{A} = 0$ and $(\vec{A} \times \vec{B}) \cdot \vec{B} = 0$.
• Properties of Dot Product:
  • $(\vec{A}+\vec{B}) \cdot (\vec{A}-\vec{B}) = |\vec{A}|^2 - |\vec{B}|^2$
• Magnitude of a Scalar Multiple of a Vector: $|k\vec{V}|^2 = k^2 |\vec{V}|^2$ for a scalar $k$ and vector $\vec{V}$.
• Cross Product of Sums/Differences: $(\vec{A}+\vec{B}) \times (\vec{A}-\vec{B}) = -2(\vec{A} \times \vec{B})$.

Step-by-Step Solution

Step 1: Simplify the given vector equation using vector identities. We are given the equation $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$. Let $\vec{P} = \vec{a}+\vec{b}$, $\vec{Q} = \vec{a} \times \vec{b}$, and $\vec{R} = \vec{a}-\vec{b}$. The expression becomes $(\vec{P} \times \vec{Q}) \times \vec{R}$. Using the vector triple product identity: $(\vec{P} \times \vec{Q}) \times \vec{R} = (\vec{P} \cdot \vec{R})\vec{Q} - (\vec{Q} \cdot \vec{R})\vec{P}$ Let's evaluate the dot product terms:

• $\vec{P} \cdot \vec{R} = (\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = |\vec{a}|^2 - |\vec{b}|^2$.
• $\vec{Q} \cdot \vec{R} = (\vec{a} \times \vec{b}) \cdot (\vec{a}-\vec{b}) = (\vec{a} \times \vec{b}) \cdot \vec{a} - (\vec{a} \times \vec{b}) \cdot \vec{b}$. Since $\vec{a} \times \vec{b}$ is orthogonal to both $\vec{a}$ and $\vec{b}$, we have $(\vec{a} \times \vec{b}) \cdot \vec{a} = 0$ and $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$. Therefore, $\vec{Q} \cdot \vec{R} = 0$. Substituting these back into the triple product identity: $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b}) = (|\vec{a}|^2 - |\vec{b}|^2)(\vec{a} \times \vec{b}) - (0)(\vec{a}+\vec{b})$ $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b}) = (|\vec{a}|^2 - |\vec{b}|^2)(\vec{a} \times \vec{b})$

Step 2: Calculate the magnitudes of $\vec{a}$ and $\vec{b}$ and their difference of squares. Given $\vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}$. $|\vec{a}|^2 = (\lambda)^2 + (2)^2 + (-3)^2 = \lambda^2 + 4 + 9 = \lambda^2 + 13$ $|\vec{b}|^2 = (1)^2 + (-\lambda)^2 + (2)^2 = 1 + \lambda^2 + 4 = \lambda^2 + 5$ Now, calculate the difference: $|\vec{a}|^2 - |\vec{b}|^2 = (\lambda^2 + 13) - (\lambda^2 + 5) = 13 - 5 = 8$

Step 3: Calculate the cross product $\vec{a} \times \vec{b}$. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2 \end{vmatrix}$ $= \hat{i}((2)(2) - (-3)(-\lambda)) - \hat{j}((\lambda)(2) - (-3)(1)) + \hat{k}((\lambda)(-\lambda) - (2)(1))$ $= \hat{i}(4 - 3\lambda) - \hat{j}(2\lambda + 3) + \hat{k}(-\lambda^2 - 2)$ $= (4 - 3\lambda)\hat{i} - (2\lambda + 3)\hat{j} - (\lambda^2 + 2)\hat{k}$

Step 4: Substitute the calculated terms back into the simplified vector equation and solve for $\lambda$. From Step 1, the simplified expression is $(|\vec{a}|^2 - |\vec{b}|^2)(\vec{a} \times \vec{b})$. From Step 2, $|\vec{a}|^2 - |\vec{b}|^2 = 8$. From Step 3, $\vec{a} \times \vec{b} = (4 - 3\lambda)\hat{i} - (2\lambda + 3)\hat{j} - (\lambda^2 + 2)\hat{k}$. So, the left side of the given equation is: $8 \left( (4 - 3\lambda)\hat{i} - (2\lambda + 3)\hat{j} - (\lambda^2 + 2)\hat{k} \right)$ This is equal to $8 \hat{i}-40 \hat{j}-24 \hat{k}$. $8 \left( (4 - 3\lambda)\hat{i} - (2\lambda + 3)\hat{j} - (\lambda^2 + 2)\hat{k} \right) = 8 \hat{i}-40 \hat{j}-24 \hat{k}$ Dividing both sides by 8: $(4 - 3\lambda)\hat{i} - (2\lambda + 3)\hat{j} - (\lambda^2 + 2)\hat{k} = \hat{i}-5 \hat{j}-3 \hat{k}$ Equating the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$:

• For $\hat{i}$: $4 - 3\lambda = 1 \implies 3\lambda = 3 \implies \lambda = 1$.
• For $\hat{j}$: $-(2\lambda + 3) = -5 \implies 2\lambda + 3 = 5 \implies 2\lambda = 2 \implies \lambda = 1$.
• For $\hat{k}$: $-(\lambda^2 + 2) = -3 \implies \lambda^2 + 2 = 3 \implies \lambda^2 = 1 \implies \lambda = \pm 1$. The consistent value for $\lambda$ is $1$.

Step 5: Calculate the target expression $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$. We need to find $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$. Substitute $\lambda=1$: $|1 \cdot (\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2 = |(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ Using the identity $(\vec{A}+\vec{B}) \times (\vec{A}-\vec{B}) = -2(\vec{A} \times \vec{B})$: $(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b}) = -2(\vec{a} \times \vec{b})$ So the expression becomes: $|-2(\vec{a} \times \vec{b})|^2$ Using the property $|k\vec{V}|^2 = k^2 |\vec{V}|^2$: $(-2)^2 |(\vec{a} \times \vec{b})|^2 = 4 |(\vec{a} \times \vec{b})|^2$ From Step 4, we have equated the components of $\vec{a} \times \vec{b}$ to $\hat{i}-5 \hat{j}-3 \hat{k}$ (after dividing by 8). This means that $\vec{a} \times \vec{b}$ is proportional to $\hat{i}-5 \hat{j}-3 \hat{k}$. Let's re-examine Step 4. The equation was: $8 (\vec{a} \times \vec{b}) = 8 \hat{i}-40 \hat{j}-24 \hat{k}$ Thus, $\vec{a} \times \vec{b} = \hat{i}-5 \hat{j}-3 \hat{k}$ Now, we calculate the magnitude squared of $\vec{a} \times \vec{b}$: $|(\vec{a} \times \vec{b})|^2 = (1)^2 + (-5)^2 + (-3)^2 = 1 + 25 + 9 = 35$ Finally, substitute this value into the expression for the target quantity: $4 |(\vec{a} \times \vec{b})|^2 = 4 \times 35 = 140$

Common Mistakes & Tips

• Incorrect Vector Triple Product Formula: Ensure you use the correct form of the vector triple product identity, especially the order of dot and cross products and the resulting terms.
• Forgetting Orthogonality: Recognize that $(\vec{A} \times \vec{B}) \cdot \vec{A} = 0$ and $(\vec{A} \times \vec{B}) \cdot \vec{B} = 0$. This is a very common simplification in vector problems.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with squares and signs.
• Shortcut for $(\vec{A}+\vec{B}) \times (\vec{A}-\vec{B})$: Remembering or deriving the identity $(\vec{A}+\vec{B}) \times (\vec{A}-\vec{B}) = -2(\vec{A} \times \vec{B})$ can save significant time.

Summary

The problem involves simplifying a complex vector expression using the vector triple product identity and properties of vector operations. By strategically applying these identities, we reduced the given equation to a simpler form involving the cross product of $\vec{a}$ and $\vec{b}$. We then calculated the components of $\vec{a}$ and $\vec{b}$ in terms of $\lambda$, computed their cross product, and used the given equality to solve for $\lambda$. Finally, we calculated the magnitude squared of the target expression, leveraging the identity for the cross product of sums and differences of vectors.

The final answer is $\boxed{140}$.