Question

Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).

Hint:

Always ensure the given line equation is transformed into the standard form where the variables \(x\), \(y\), and \(z\) have coefficients of \(+1\) before extracting the direction ratios from the denominators. \nobreak

Answer 1

Concept: To find the equation of a line, we need two components: a specific point on the line, and its directional vector. Here, the required point is the intersection point of two given lines, and the directional vector is derived from a third line to which our target line is parallel.

Step 1: Finding the point of intersection of the two given lines.
The equations of the given lines are: \[ \text{Line 1:} \quad \vec{r} = (1 + 3\lambda)\hat{i} + (1 - \lambda)\hat{j} - \hat{k} \] \[ \text{Line 2:} \quad \vec{r} = (4 + 2\mu)\hat{i} + 0\hat{j} + (-1 + 3\mu)\hat{k} \] Equating the corresponding components of the two lines to find their intersection point: From the \(\hat{j}\) components: \[ 1 - \lambda = 0 \quad \Rightarrow \quad \lambda = 1 \] Substitute \(\lambda = 1\) into the expression for Line 1 to get the exact point coordinates: \[ x = 1 + 3(1) = 4, \quad y = 1 - 1 = 0, \quad z = -1 \] Let's check this against Line 2 for verification using the \(\hat{i}\) components: \[ 4 + 2\mu = 4 \quad \Rightarrow \quad \mu = 0 \] Using \(\mu = 0\) for the \(\hat{k}\) component of Line 2 gives \( z = -1 + 3(0) = -1 \), which matches perfectly. Thus, the point of intersection is \( P(4, 0, -1) \). The position vector is \( \vec{a} = 4\hat{i} - \hat{k} \).

Step 2: Determining the direction vector from the parallel line condition.
The target line is parallel to the line: \[ \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \] Convert this reference line into standard symmetric form: \[ \frac{x - 1}{-2} = \frac{y - 7}{3} = \frac{z - 0}{1} \] The direction ratios of this line (and therefore our parallel target line) are: \[ (-2, 3, 1) \] So, the parallel direction vector is \( \vec{b} = -2\hat{i} + 3\hat{j} + \hat{k} \).

Step 3: Formulating the final Vector and Cartesian equations.
Vector Equation: Using the vector equation structure \( \vec{r} = \vec{a} + t\vec{b} \): \[ \vec{r} = (4\hat{i} - \hat{k}) + t(-2\hat{i} + 3\hat{j} + \hat{k}) \] Cartesian Equation: Using the standard Cartesian structure \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \): \[ \frac{x - 4}{-2} = \frac{y - 0}{3} = \frac{z - (-1)}{1} \quad \Rightarrow \quad \frac{x - 4}{-2} = \frac{y}{3} = \frac{z + 1}{1} \] Both forms are correctly shown in options (A) and (C).