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: Parametric representation of the lines.
Line \(l_1\): \[l_1: \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda,\] A general point on \(l_1\) is \((1 + \lambda, 1 + 2\lambda, 2 + 3\lambda)\).
Line \(l_2\): \[l_2: \frac{x - 1}{0} = \frac{y}{-3} = \frac{z - 7}{2} = \mu,\] A general point on \(l_2\) is \((1, -3\mu, 7 + 2\mu)\).

Step 2: Determine the intersection point of the two lines.
Equating the coordinates of points on \(l_1\) and \(l_2\): \[1 + \lambda = 1, \quad 1 + 2\lambda = -3\mu, \quad 2 + 3\lambda = 7 + 2\mu.\] From \(1 + \lambda = 1\), we get \(\lambda = 0\). Substituting \(\lambda = 0\) into the other equations yields: \[1 = -3\mu, \quad 2 = 7 + 2\mu \implies \mu = -1.\] The point of intersection is \((1, 1, 5)\).

Step 3: Calculate 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 a line perpendicular to both \(l_1\) and \(l_2\) are given by their cross product: \[\vec{d} = \vec{d_1} \times \vec{d_2}.\] Computing the cross product: \[\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\1 & 2 & 3 \\0 & -3 & 2 \end{vmatrix}= \hat{i}(4 - (-9)) - \hat{j}(2 - 0) + \hat{k}(-3 - 0).\] This results in \[\vec{d} = \langle 13, -2, -3 \rangle.\]

Step 4: Formulate the equation of the required line.
The equation of the line passing through \((1, 1, 5)\) with direction ratios \(\langle 13, -2, -3 \rangle\) is: \[\frac{x - 1}{13} = \frac{y - 1}{-2} = \frac{z - 5}{-3}.\]