Question:

Given the vectors $\vec{a} = \mathbf{i}+2\mathbf{j}-\sqrt{2}\mathbf{k}$ and $\vec{b} = \sqrt{2}\mathbf{i}-\mathbf{j}+\sqrt{2}\mathbf{k}$. We are asked to find $|\vec{c}|$ where $\vec{c} = \vec{a} \times (\vec{a} \times \vec{b})$. Step 1: Apply the vector triple product identity. The vector triple product identity states that $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$. Thus, $\vec{c} = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$. Step 2: Calculate the necessary dot products. First, calculate $\vec{a} \cdot \vec{a}$: $$ \vec{a} \cdot \vec{a} = |\vec{a}|^2 = (1)^2 + (2)^2 + (-\sqrt{2})^2 = 1 + 4 + 2 = 7 $$ Next, calculate $\vec{a} \cdot \vec{b}$: $$ \vec{a} \cdot \vec{b} = (1)(\sqrt{2}) + (2)(-1) + (-\sqrt{2})(\sqrt{2}) = \sqrt{2} - 2 - 2 = \sqrt{2} - 4 $$ Step 3: Substitute these values into the expression for $\vec{c}$. $$ \vec{c} = (\sqrt{2}-4)\vec{a} - 7\vec{b} $$ Step 4: Calculate the magnitude $|\vec{c}|$. To find $|\vec{c}|$, we calculate $|\vec{c}|^2$ using the formula $|\alpha\vec{a} + \beta\vec{b}|^2 = \alpha^2|\vec{a}|^2 + \beta^2|\vec{b}|^2 + 2\alpha\beta(\vec{a} \cdot \vec{b})$. Here, $\alpha = \sqrt{2}-4$ and $\beta = -7$. We already have $|\vec{a}|^2 = 7$ and $\vec{a} \cdot \vec{b} = \sqrt{2}-4$. We need to calculate $|\vec{b}|^2$: $$ |\vec{b}|^2 = (\sqrt{2})^2 + (-1)^2 + (\sqrt{2})^2 = 2 + 1 + 2 = 5 $$ Now substitute all values into the magnitude formula: $$ |\vec{c}|^2 = (\sqrt{2}-4)^2 |\vec{a}|^2 + (-7)^2 |\vec{b}|^2 + 2(\sqrt{2}-4)(-7)(\vec{a} \cdot \vec{b}) $$ $$ |\vec{c}|^2 = (\sqrt{2}-4)^2 (7) + 49 (5) - 14(\sqrt{2}-4)(\sqrt{2}-4) $$ $$ |\vec{c}|^2 = 7(\sqrt{2}-4)^2 + 245 - 14(\sqrt{2}-4)^2 $$ $$ |\vec{c}|^2 = -7(\sqrt{2}-4)^2 + 245 $$ Calculate $(\sqrt{2}-4)^2$: $$ (\sqrt{2}-4)^2 = (\sqrt{2})^2 - 2(4)(\sqrt{2}) + 4^2 = 2 - 8\sqrt{2} + 16 = 18 - 8\sqrt{2} $$ Substitute this back into the expression for $|\vec{c}|^2$: $$ |\vec{c}|^2 = -7(18 - 8\sqrt{2}) + 245 $$ $$ |\vec{c}|^2 = -126 + 56\sqrt{2} + 245 $$ $$ |\vec{c}|^2 = 119 + 56\sqrt{2} $$ Therefore, the magnitude of $\vec{c}$ is: $$ |\vec{c}| = \sqrt{119 + 56\sqrt{2}} $$