### Related Formula vecx cdot (vecy times vecz) = [vecx \ vecy \ vecz] = (vecx times vecy) cdot vecz$\vec{x} \cdot (\vec{y} \times \vec{z}) = [\vec{x} \ \vec{y} \ \vec{z}] = (\vec{x} \times \vec{y}) \cdot \vec{z}$ ### Core Logic The expression to evaluate is: E = veca cdot ((vecc times vecb) - vecb - vecc)$E = \vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c})$ Distributing the dot product over the terms gives: E = veca cdot (vecc times vecb) - veca cdot vecb - veca cdot vecc$E = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$ ### Step 1: Evaluating the Scalar Triple Product The first term is a scalar triple product: veca cdot (vecc times vecb) = (veca times vecc) cdot vecb$\vec{a} \cdot (\vec{c} \times \vec{b}) = (\vec{a} \times \vec{c}) \cdot \vec{b}$ We are given that veca times vecc = vecb$\vec{a} \times \vec{c} = \vec{b}$. Substitute this in: (vecb) cdot vecb = |vecb|^2$(\vec{b}) \cdot \vec{b} = |\vec{b}|^2$ Given vecb = 3hati - 3hatj + 3hatk$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$, its magnitude squared is: |vecb|^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27$|\vec{b}|^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27$ ### Step 2: Evaluating the remaining Dot Products For the second term, calculate veca cdot vecb$\vec{a} \cdot \vec{b}$: veca = 1hati + 2hatj + 1hatk$\vec{a} = 1\hat{i} + 2\hat{j} + 1\hat{k}$ vecb = 3hati - 3hatj + 3hatk$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$ veca cdot vecb = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$ For the third term, we are explicitly given: veca cdot vecc = 3$\vec{a} \cdot \vec{c} = 3$ ### Step 3: Final Output Calculation Substitute all individual values back into the expanded expression: E = 27 - 0 - 3 = 24$E = 27 - 0 - 3 = 24$ ### Pattern Recognition When asked to evaluate complex vector expressions containing veca cdot (dots times dots)$\vec{a} \cdot (\dots \times \dots)$, immediately distribute and convert them into Scalar Triple Products [veca \ vecb \ vecc]$[\vec{a} \ \vec{b} \ \vec{c}]$. Cyclic permutations and given cross-product relationships will rapidly collapse the expression into simple magnitudes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra