Key Concepts and Formulas

• Lagrange's Identity: For any two vectors $\vec{u}$ and $\vec{v}$, $|\vec{u} \times \vec{v}|^2 = |\vec{u}|^2 |\vec{v}|^2 - (\vec{u} \cdot \vec{v})^2$.
• Scalar Triple Product (STP) and its Determinant Form: The scalar triple product of three vectors $\vec{u}, \vec{v}, \vec{w}$ is given by $[\vec{u}, \vec{v}, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$. Its square can be expressed as a determinant: $[\vec{u}, \vec{v}, \vec{w}]^2 = \left| \begin{matrix} \vec{u} \cdot \vec{u} & \vec{u} \cdot \vec{v} & \vec{u} \cdot \vec{w} \\ \vec{v} \cdot \vec{u} & \vec{v} \cdot \vec{v} & \vec{v} \cdot \vec{w} \\ \vec{w} \cdot \vec{u} & \vec{w} \cdot \vec{v} & \vec{w} \cdot \vec{w} \end{matrix} \right|$
• Vector Magnitude and Dot Product: For a vector $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$, its magnitude squared is $|\vec{v}|^2 = v_1^2 + v_2^2 + v_3^2$. The dot product of two vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ is $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$.

Step-by-Step Solution

Step 1: Calculate initial vector properties and simplify given conditions. We are given $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$. We need to find $|\vec{a} \times \vec{c}|^2$. First, let's compute the magnitudes squared of $\vec{a}$ and $\vec{b}$, and their dot product.

• $|\vec{a}|^2 = (1)^2 + (2)^2 + (3)^2 = 1 + 4 + 9 = 14$.
• $|\vec{b}|^2 = (1)^2 + (1)^2 + (-1)^2 = 1 + 1 + 1 = 3$. Thus, $|\vec{b}| = \sqrt{3}$.
• $\vec{a} \cdot \vec{b} = (1)(1) + (2)(1) + (3)(-1) = 1 + 2 - 3 = 0$. This implies $\vec{a}$ and $\vec{b}$ are orthogonal.

Now, let's simplify the given conditions:

• $\vec{a} \cdot \vec{c} = 11$ (given).
• $\vec{b} \cdot (\vec{a} \times \vec{c}) = 27$ (given). This is the scalar triple product $[\vec{b}, \vec{a}, \vec{c}]$.
• $\vec{b} \cdot \vec{c} = -\sqrt{3}|\vec{b}|$. Substituting $|\vec{b}| = \sqrt{3}$, we get $\vec{b} \cdot \vec{c} = -\sqrt{3}(\sqrt{3}) = -3$.

Step 2: Use the Scalar Triple Product determinant form to find $|\vec{c}|^2$. The square of the scalar triple product $[\vec{b}, \vec{a}, \vec{c}]$ is given by the determinant: $[\vec{b}, \vec{a}, \vec{c}]^2 = \left| \begin{matrix} \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{c} \\ \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{c} \\ \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{c} \end{matrix} \right|$ We know $[\vec{b}, \vec{a}, \vec{c}] = 27$, so $[\vec{b}, \vec{a}, \vec{c}]^2 = 27^2 = 729$. We have the following dot products:

• $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 3$
• $\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b} = 0$
• $\vec{b} \cdot \vec{c} = -3$
• $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 14$
• $\vec{a} \cdot \vec{c} = 11$
• $\vec{c} \cdot \vec{b} = \vec{b} \cdot \vec{c} = -3$
• $\vec{c} \cdot \vec{a} = \vec{a} \cdot \vec{c} = 11$
• $\vec{c} \cdot \vec{c} = |\vec{c}|^2$ (this is what we need to find)

Substituting these values into the determinant equation: $729 = \left| \begin{matrix} 3 & 0 & -3 \\ 0 & 14 & 11 \\ -3 & 11 & |\vec{c}|^2 \end{matrix} \right|$ Expand the determinant: $729 = 3(14|\vec{c}|^2 - 11 \cdot 11) - 0(\dots) + (-3)(0 \cdot 11 - 14 \cdot (-3))$ $729 = 3(14|\vec{c}|^2 - 121) - 3(42)$ $729 = 42|\vec{c}|^2 - 363 - 126$ $729 = 42|\vec{c}|^2 - 489$ Now, solve for $42|\vec{c}|^2$: $42|\vec{c}|^2 = 729 + 489$ $42|\vec{c}|^2 = 1218$ $|\vec{c}|^2 = \frac{1218}{42} = \frac{203}{7} = 29$

Step 3: Use Lagrange's Identity to find $|\vec{a} \times \vec{c}|^2$. Lagrange's Identity states: $|\vec{a} \times \vec{c}|^2 = |\vec{a}|^2 |\vec{c}|^2 - (\vec{a} \cdot \vec{c})^2$. We have:

• $|\vec{a}|^2 = 14$
• $|\vec{c}|^2 = 29$
• $\vec{a} \cdot \vec{c} = 11$

Substitute these values into Lagrange's Identity: $|\vec{a} \times \vec{c}|^2 = (14)(29) - (11)^2$ $|\vec{a} \times \vec{c}|^2 = 406 - 121$ $|\vec{a} \times \vec{c}|^2 = 285$

Common Mistakes & Tips

• Arithmetic Errors: Be extremely careful with calculations involving squares, dot products, and determinant expansion, as small errors can lead to incorrect final answers.
• Misapplication of Identities: Ensure you are using the correct vector identities (Lagrange's, STP determinant form) and substituting the correct vector components.
• Orthogonality Simplification: Recognize when vectors are orthogonal ($\vec{a} \cdot \vec{b} = 0$) as this significantly simplifies many calculations, especially in determinant expansions.

Summary The problem was solved by first calculating the necessary magnitudes and dot products of the given vectors $\vec{a}$ and $\vec{b}$, and simplifying the given conditions. The key to finding $|\vec{c}|^2$ was utilizing the determinant form of the scalar triple product $[\vec{b}, \vec{a}, \vec{c}]$, which allowed us to form an equation involving $|\vec{c}|^2$. After solving for $|\vec{c}|^2$, Lagrange's Identity was applied to directly compute $|\vec{a} \times \vec{c}|^2$ using the previously found values of $|\vec{a}|^2$, $|\vec{c}|^2$, and $\vec{a} \cdot \vec{c}$.

The final answer is $\boxed{285}$.