Let veca = 2hati - 3hatj + hatk$\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$, vecb = 3hati + 2hatj + 5hatk$\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$ and a vector vecc$\vec{c}$ be such that (veca - vecc) times vecb = -18hati - 3hatj + 12hatk$(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$ and veca cdot vecc = 3$\vec{a} \cdot \vec{c} = 3$. If vecb times vecc = vecd$\vec{b} \times \vec{c} = \vec{d}$, then |veca cdot vecd|$|\vec{a} \cdot \vec{d}|$ is equal to:

### Related Formula textVector Cross product distributes over subtraction: (veca - vecc) times vecb = veca times vecb - vecc times vecb$$\text{Vector Cross product distributes over subtraction: } (\vec{a} - \vec{c}) \times \vec{b} = \vec{a} \times \vec{b} - \vec{c} \times \vec{b}$$ textScalar Triple Product cyclic identity: veca cdot (vecb times vecc) = (veca times vecb) cdot vecc$$\text{Scalar Triple Product cyclic identity: } \vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$$ textAntisymmetry: vecc times vecb = - vecb times vecc$$\text{Antisymmetry: } \vec{c} \times \vec{b} = - \vec{b} \times \vec{c}$$ ### Core Logic Instead of solving for the individual coordinates of vector vecc$\vec{c}$, we apply vector algebraic identities to compute the target scalar triple product directly. ### Step 1: Expand and rewrite the cross product Given (veca - vecc) times vecb = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk$(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}$: veca times vecb - vecc times vecb = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk$$\vec{a} \times \vec{b} - \vec{c} \times \vec{b} = -18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}$$ veca times vecb + vecb times vecc = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk$$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = -18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}$$ vecb times vecc = (-18hatmathrmi - 3hatmathrmj + 12hatmathrmk) - (veca times vecb) quad text--- (1)$$\vec{b} \times \vec{c} = (-18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}) - (\vec{a} \times \vec{b}) \quad \text{--- (1)}$$ ### Step 2: Calculate a x b Evaluate the cross product: veca times vecb = beginvmatrix hatmathrmi & hatmathrmj & hatmathrmk \\ 2 & -3 & 1 \\ 3 & 2 & 5 endvmatrix$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -3 & 1 \\ 3 & 2 & 5 \end{vmatrix}$$ veca times vecb = hatmathrmi(-15 - 2) - hatmathrmj(10 - 3) + hatmathrmk(4 - (-9)) = -17hatmathrmi - 7hatmathrmj + 13hatmathrmk$$\vec{a} \times \vec{b} = \hat{\mathrm{i}}(-15 - 2) - \hat{\mathrm{j}}(10 - 3) + \hat{\mathrm{k}}(4 - (-9)) = -17\hat{\mathrm{i}} - 7\hat{\mathrm{j}} + 13\hat{\mathrm{k}}$$ ### Step 3: Solve for the vector d Substitute veca times vecb$\vec{a} \times \vec{b}$ back into equation (1): vecd = vecb times vecc = (-18hatmathrmi - 3hatmathrmj + 12hatmathrmk) - (-17hatmathrmi - 7hatmathrmj + 13hatmathrmk)$$\vec{d} = \vec{b} \times \vec{c} = (-18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}) - (-17\hat{\mathrm{i}} - 7\hat{\mathrm{j}} + 13\hat{\mathrm{k}})$$ vecd = -hatmathrmi + 4hatmathrmj - hatmathrmk$$\vec{d} = -\hat{\mathrm{i}} + 4\hat{\mathrm{j}} - \hat{\mathrm{k}}$$ ### Step 4: Compute the final dot product Now compute the requested dot product: veca cdot vecd = (2hatmathrmi - 3hatmathrmj + hatmathrmk) cdot (-hatmathrmi + 4hatmathrmj - hatmathrmk)$$\vec{a} \cdot \vec{d} = (2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + \hat{\mathrm{k}}) \cdot (-\hat{\mathrm{i}} + 4\hat{\mathrm{j}} - \hat{\mathrm{k}})$$ veca cdot vecd = 2(-1) + (-3)(4) + 1(-1) = -2 - 12 - 1 = -15$$\vec{a} \cdot \vec{d} = 2(-1) + (-3)(4) + 1(-1) = -2 - 12 - 1 = -15$$ left| veca cdot vecd right| = 15$$\left| \vec{a} \cdot \vec{d} \right| = 15$$ ### Pattern Recognition Scalar triple product shortcut: Recognizing that veca cdot vecd = veca cdot (vecb times vecc) = [ veca \, vecb \, vecc ]$\vec{a} \cdot \vec{d} = \vec{a} \cdot (\vec{b} \times \vec{c}) = [ \vec{a} \, \vec{b} \, \vec{c} ]$ allows you to find the scalar value through simple determinants and linear equations instead of solving for the vector components. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra