Let veca = hati + hatj + hatk$\vec{a} = \hat{i} + \hat{j} + \hat{k}$, vecb = 3hati + 2hatj - hatk$\vec{b} = 3\hat{i} + 2\hat{j} - \hat{k}$, vecc = lambda hatj + mu hatk$\vec{c} = \lambda \hat{j} + \mu \hat{k}$ and hatd$\hat{d}$ be a unit vector such that veca times hatd = vecb times hatd$\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$ and vecc cdot hatd = 1$\vec{c} \cdot \hat{d} = 1$[cite: 694]. If vecc$\vec{c}$ is perpendicular to veca$\vec{a}$[cite: 694], then |3lambda hatd + mu vecc|^2$|3lambda \hat{d} + \mu \vec{c}|^2$ is equal to:

### Related Formula Vector cross distribution property: (veca - vecb) times hatd = 0 implies hatd parallel (veca - vecb)$$(\vec{a} - \vec{b}) \times \hat{d} = 0 \implies \hat{d} \parallel (\vec{a} - \vec{b})$$ ### Core Logic Gather cross products to solve for collineation lines [cite: 1472, 1473]: (veca - vecb) times hatd = 0 implies hatd = t(veca - vecb)$$(\vec{a} - \vec{b}) \times \hat{d} = 0 \implies \hat{d} = t(\vec{a} - \vec{b})$$ [cite: 1473, 1474] Compute the baseline difference vector [cite: 1475]: veca - vecb = (1-3)hati + (1-2)hatj + (1 - (-1))hatk = -2hati - hatj + 2hatk$$\vec{a} - \vec{b} = (1-3)\hat{i} + (1-2)\hat{j} + (1 - (-1))\hat{k} = -2\hat{i} - \hat{j} + 2\hat{k}$$ [cite: 1475] hatd = t(-2hati - hatj + 2hatk)$$\hat{d} = t(-2\hat{i} - \hat{j} + 2\hat{k})$$ [cite: 1475] Since hatd$\hat{d}$ is a unit vector[cite: 1476]: |t| cdot sqrt(-2)^2 + (-1)^2 + 2^2 = 1 implies 3|t| = 1 implies |t| = frac13$$|t| \cdot \sqrt{(-2)^2 + (-1)^2 + 2^2} = 1 \implies 3|t| = 1 \implies |t| = \frac{1}{3}$$ [cite: 1476, 1479] ### Step 1: Applying orthogonality constraints Using orthogonal information given for vectors vecc$\vec{c}$ and veca$\vec{a}$ [cite: 1482]: vecc cdot veca = 0 implies (0)(1) + lambda(1) + mu(1) = 0 implies mu = -lambda$$\vec{c} \cdot \vec{a} = 0 \implies (0)(1) + \lambda(1) + \mu(1) = 0 \implies \mu = -\lambda$$ [cite: 1483, 1484] vecc = lambda(hatj - hatk) implies |vecc|^2 = 2lambda^2$$\vec{c} = \lambda(\hat{j} - \hat{k}) \implies |\vec{c}|^2 = 2\lambda^2$$ [cite: 1485] Use product condition vecc cdot hatd = 1$\vec{c} \cdot \hat{d} = 1$ to isolate scalar values [cite: 1486]: t(-2hati - hatj + 2hatk) cdot lambda(hatj - hatk) = 1$$t(-2\hat{i} - \hat{j} + 2\hat{k}) \cdot \lambda(\hat{j} - \hat{k}) = 1$$ [cite: 1487] tlambda(-1 - 2) = 1 implies -3tlambda = 1 implies tlambda = -frac13$$t\lambda(-1 - 2) = 1 \implies -3t\lambda = 1 \implies t\lambda = -\frac{1}{3}$$ [cite: 1488] Since |t|^2 = frac19$|t|^2 = \frac{1}{9}$, squaring components yields [cite: 1488]: lambda^2 = 1$\lambda^2 = 1$ [cite: 1488] ### Step 2: Vector magnitude resolution Expand target expression using standard inner dot product expansions [cite: 1488]: |3lambda hatd + mu vecc|^2 = 9lambda^2|hatd|^2 + mu^2|vecc|^2 + 6lambdamu(hatd cdot vecc)$$|3\lambda \hat{d} + \mu \vec{c}|^2 = 9\lambda^2|\hat{d}|^2 + \mu^2|\vec{c}|^2 + 6\lambda\mu(\hat{d} \cdot \vec{c})$$ [cite: 1488] Substitute values evaluated throughout sections [cite: 1488, 1489]: = 9(1)(1) + (lambda)^2(2lambda^2) + 6lambda(-lambda)(1)$$= 9(1)(1) + (\lambda)^2(2\lambda^2) + 6\lambda(-\lambda)(1)$$ = 9 + 2lambda^4 - 6lambda^2 = 9 + 2(1) - 6(1) = 5$$= 9 + 2\lambda^4 - 6\lambda^2 = 9 + 2(1) - 6(1) = 5$$ [cite: 1489, 1490] ### Pattern Recognition Translating vector cross equalities directly into linear scale parameter multipliers prevents manual determinant expansions, leaving clean system variables behind. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra