Question

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

Hint:

To find the line perpendicular to two given lines, use the cross product of their direction vectors to obtain the direction ratios.

Answer 1

Step 1: Parameterize the two given lines
The parametric equations of the lines are: \[ l_1: \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda, \] so any point on \( l_1 \) is: \[ (1 + \lambda, 1 + 2\lambda, 2 + 3\lambda). \] For the second line: \[ l_2: \frac{x - 1}{0} = \frac{y}{-3} = \frac{z - 7}{2} = \mu, \] so any point on \( l_2 \) is: \[ (1, -3\mu, 7 + 2\mu). \] Step 2: Find the point of intersection of the two lines
Equating the coordinates of \( l_1 \) and \( l_2 \): \[ 1 + \lambda = 1, \quad 1 + 2\lambda = -3\mu, \quad 2 + 3\lambda = 7 + 2\mu. \] From \( 1 + \lambda = 1 \), \( \lambda = 0 \). Substitute \( \lambda = 0 \) into the other equations: \[ 1 = -3\mu, \quad 2 = 7 + 2\mu \implies \mu = -1. \] Thus, the point of intersection is: \[ (1, 1, 5). \] Step 3: Find the direction ratios of the required line
The direction ratios of the given lines are: \[ \vec{d_1} = \langle 1, 2, 3 \rangle, \quad \vec{d_2} = \langle 0, -3, 2 \rangle. \] The direction ratios of the line perpendicular to both \( l_1 \) and \( l_2 \) are given by: \[ \vec{d} = \vec{d_1} \times \vec{d_2}. \] Calculate the cross product: \[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & -3 & 2 \end{vmatrix} = \hat{i}(4 - (-9)) - \hat{j}(2 - 0) + \hat{k}(-3 - 0). \] \[ \vec{d} = \langle 13, -2, -3 \rangle. \] Step 4: Write the equation of the required line
The equation of the required line passing through \( (1, 1, 5) \) with direction ratios \( \langle 13, -2, -3 \rangle \) is: \[ \frac{x - 1}{13} = \frac{y - 1}{-2} = \frac{z - 5}{-3}. \]