Question

Find the equation of the line passing through the point of intersection of the lines: \[ \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 2}{3}, \quad \frac{x - 1}{0} = \frac{y - 3}{-3} = \frac{z - 7}{2}, \] and perpendicular to these given lines.

Hint:

For lines perpendicular to two given lines, use the cross product of their direction vectors to find the required direction.

Answer 1

1. Point of intersection of the two lines: Let: \[ \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 2}{3} = t_1 \quad \text{and} \quad \frac{x - 1}{0} = \frac{y - 3}{-3} = \frac{z - 7}{2} = t_2. \] From the first line: \[ x = 1 + t_1, \quad y = 2 + 2t_1, \quad z = 2 + 3t_1. \] From the second line: \[ x = 1, \quad y = 3 - 3t_2, \quad z = 7 + 2t_2. \] Equate \( x, y, z \) for consistency: - From \( x \): \( 1 + t_1 = 1 \quad \Rightarrow \quad t_1 = 0. \) - From \( y \): \( 2 + 2t_1 = 3 - 3t_2 \quad \Rightarrow \quad 2 = 3 - 3t_2 \quad \Rightarrow \quad t_2 = \frac{1}{3}. \) Substituting \( t_1 = 0 \) into the first line: \[ x = 1, \quad y = 2, \quad z = 2. \] Thus, the point of intersection is \( (1, 2, 2) \). 
2. Direction vectors of the given lines: - First line: \( \vec{d_1} = \langle 1, 2, 3 \rangle \). - Second line: \( \vec{d_2} = \langle 0, -3, 2 \rangle \). 
3. Direction of the required line: The required line is perpendicular to both \( \vec{d_1} \) and \( \vec{d_2} \). Use the cross product: \[ \vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & -3 & 2 \end{vmatrix}. \] Expanding: \[ \vec{d} = \hat{i} \begin{vmatrix} 2 & 3 \\ -3 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 0 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix}. \] \[ \vec{d} = \hat{i}(4 + 9) - \hat{j}(2 - 0) + \hat{k}(-3 - 0) = \langle 13, -2, -3 \rangle. \] 
4. Equation of the required line: The line passing through \( (1, 2, 2) \) with direction vector \( \langle 13, -2, -3 \rangle \) is: \[ \frac{x - 1}{13} = \frac{y - 2}{-2} = \frac{z - 2}{-3}. \] 
Final Answer: \[ \boxed{\frac{x - 1}{13} = \frac{y - 2}{-2} = \frac{z - 2}{-3}.} \]