Given $\vec{i} + \vec{j} + \vec{k}$. Two given lines intersected, if $7\vec{i} + 10\vec{j} + 13\vec{k} + s(2\vec{i} + 3\vec{j} + 4\vec{k}) = 3\vec{i} + 5\vec{j} + 7\vec{k} + t(\vec{i} + 2\vec{j} + 3\vec{k})$
Step-by-Step Solution
Key Concept: Two lines intersect when there exist scalar parameters satisfying the vector equation simultaneously; solve by comparing coefficients of $\vec{i}$, $\vec{j}$, $\vec{k}$ components.
Expanding the vector equation: $(7 + 2s)\vec{i} + (10 + 3s)\vec{j} + (13 + 4s)\vec{k} = (3 + t)\vec{i} + (5 + 2t)\vec{j} + (7 + 3t)\vec{k}$. Comparing coefficients: $7 + 2s = 3 + t$ gives $2s - t = -4$ (equation i); $10 + 3s = 5 + 2t$ gives $3s - 2t = -5$ (equation ii); $13 + 4s = 7 + 3t$ gives $4s - 3t = -6$ (equation iii). Solving equations (i) and (ii): from (i) $t = 2s + 4$, substituting into (ii) gives $3s - 2(2s + 4) = -5$, so $s = -3$ and $t = -2$. Verification with equation (iii): $4(-3) - 3(-2) = -12 + 6 = -6$ ✓. The required point is $7\vec{i} + 10\vec{j} + 13\vec{k} + (-3)(2\vec{i} + 3\vec{j} + 4\vec{k}) = \vec{i} + \vec{j} + \vec{k}$.
Correct Answer: 4