Question:

Step 1: Determine the relationship between the vectors using the scalar triple product. Given the magnitudes of the vectors are $|\vec{a}| = 2$, $|\vec{b}| = 3$, and $|\vec{c}| = 5$. The product of their magnitudes is $|\vec{a}||\vec{b}||\vec{c}| = 2 \times 3 \times 5 = 30$. We are also given that the absolute value of the scalar triple product is $|[\vec{a}, \vec{b}, \vec{c}]| = |\vec{a} \cdot (\vec{b} \times \vec{c})| = 30$. Since $|[\vec{a}, \vec{b}, \vec{c}]| = |\vec{a}||\vec{b}||\vec{c}|$, the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ must be mutually perpendicular. Therefore, their dot products are: $\vec{a} \cdot \vec{b} = 0$ $\vec{b} \cdot \vec{c} = 0$ $\vec{c} \cdot \vec{a} = 0$ Step 2: Expand the vector expression $(\vec{a} \times \vec{c}) \times (\vec{a} - \vec{c})$. Using the distributive property of the cross product: $$(\vec{a} \times \vec{c}) \times (\vec{a} - \vec{c}) = (\vec{a} \times \vec{c}) \times \vec{a} - (\vec{a} \times \vec{c}) \times \vec{c}$$ We apply the vector triple product identity $(\vec{X} \times \vec{Y}) \times \vec{Z} = (\vec{X} \cdot \vec{Z})\vec{Y} - (\vec{Y} \cdot \vec{Z})\vec{X}$. For the first term, $(\vec{a} \times \vec{c}) \times \vec{a}$: Let $\vec{X} = \vec{a}$, $\vec{Y} = \vec{c}$, $\vec{Z} = \vec{a}$. $$(\vec{a} \times \vec{c}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{c} - (\vec{c} \cdot \vec{a})\vec{a}$$ Since $\vec{a} \perp \vec{c}$, we have $\vec{c} \cdot \vec{a} = 0$. $$(\vec{a} \times \vec{c}) \times \vec{a} = |\vec{a}|^2 \vec{c} - 0 = 2^2 \vec{c} = 4\vec{c}$$ For the second term, $(\vec{a} \times \vec{c}) \times \vec{c}$: Let $\vec{X} = \vec{a}$, $\vec{Y} = \vec{c}$, $\vec{Z} = \vec{c}$. $$(\vec{a} \times \vec{c}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{c} - (\vec{c} \cdot \vec{c})\vec{a}$$ Since $\vec{a} \perp \vec{c}$, we have $\vec{a} \cdot \vec{c} = 0$. $$(\vec{a} \times \vec{c}) \times \vec{c} = 0 - |\vec{c}|^2 \vec{a} = -5^2 \vec{a} = -25\vec{a}$$ Substituting these results back into the expanded expression: $$(\vec{a} \times \vec{c}) \times (\vec{a} - \vec{c}) = 4\vec{c} - (-25\vec{a}) = 25\vec{a} + 4\vec{c}$$ Step 3: Simplify the expression for the dot product. The second vector in the dot product is $(\vec{a} \times \vec{c}) \times (\vec{a} - \vec{c}) + \vec{b}$. Substituting the result from Step 2: $$(\vec{a} \times \vec{c}) \times (\vec{a} - \vec{c}) + \vec{b} = 25\vec{a} + 4\vec{c} + \vec{b} = 25\vec{a} + \vec{b} + 4\vec{c}$$ Step 4: Compute the dot product $k$. We need to calculate $k = (2\vec{a} + \vec{b} + \vec{c}) \cdot (25\vec{a} + \vec{b} + 4\vec{c})$. Since $\vec{a}$, $\vec{b}$, and $\vec{c}$ are mutually perpendicular, the dot product of any two distinct vectors among them is zero. Therefore, only terms involving the dot product of a vector with itself will be non-zero. $$k = (2\vec{a} \cdot 25\vec{a}) + (\vec{b} \cdot \vec{b}) + (\vec{c} \cdot 4\vec{c})$$ $$k = 50(\vec{a} \cdot \vec{a}) + (\vec{b} \cdot \vec{b}) + 4(\vec{c} \cdot \vec{c})$$ $$k = 50|\vec{a}|^2 + |\vec{b}|^2 + 4|\vec{c}|^2$$ Substitute the given magnitudes: $|\vec{a}| = 2$, $|\vec{b}| = 3$, $|\vec{c}| = 5$. $$k = 50(2^2) + (3^2) + 4(5^2)$$ $$k = 50(4) + 9 + 4(25)$$ $$k = 200 + 9 + 100$$ $$k = 309$$ Step 5: Calculate the final value. The value to be determined is $\left(\dfrac{k}{103}\right)$. $$\frac{k}{103} = \frac{309}{103} = 3$$