Problem Statement
Find the vector and Cartesian equations of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines:
$$\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7} \quad \text{and} \quad \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$$
Verified Solution & Marking Scheme
Direction vectors of the two given lines
$$\vec{b}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}, \quad \vec{b}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k}$$
Compute cross product b₁ × b₂ for perpendicular direction
$$\vec{b} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$$
$$= \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 - (-48))$$
$$= 24\hat{i} + 36\hat{j} + 72\hat{k} = 12(2\hat{i} + 3\hat{j} + 6\hat{k})$$
Direction ratios can be taken as $(2, 3, 6)$.
Write Vector and Cartesian Equations
Point $A = \hat{i} + 2\hat{j} - 4\hat{k}$.
Vector Equation:
$$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$$
Cartesian Equation:
$$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$$