Question

Equation of the line through the point (2, 3, 1) and parallel to the line of intersection of the planes \( x - 2y - z + 5 = 0 \) and \( x + y + 3z = 6 \) is:

• A. \( \frac{x - 2}{-5} = \frac{y - 3}{-4} = \frac{z - 1}{3} \)
• B. \( \frac{x - 2}{5} = \frac{y - 3}{-4} = \frac{z - 1}{3} \)
• C. \( \frac{x - 2}{5} = \frac{y - 3}{4} = \frac{z - 1}{3} \)
• D. \( \frac{x - 2}{4} = \frac{y - 3}{3} = \frac{z - 1}{2} \)
• E. \( \frac{x - 2}{-4} = \frac{y - 3}{-3} = \frac{z - 1}{2} \)

Hint:

The line of intersection of two planes is always parallel to $\vec{n_1} \times \vec{n_2}$. This vector represents the direction where both planes are "advancing" together.

Answer 1

The Correct Option is A

Concept: The direction of the line of intersection of two planes is perpendicular to the normal vectors of both planes. Thus, the direction vector \( \vec{b} \) is the cross product of the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \).

Step 1: Identify the normal vectors.
Plane 1: \( x - 2y - z + 5 = 0 \Rightarrow \vec{n_1} = (1, -2, -1) \). Plane 2: \( x + y + 3z = 6 \Rightarrow \vec{n_2} = (1, 1, 3) \).

Step 2: Calculate the cross product \( \vec{n_1} \times \vec{n_2} \).
\[ \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -1 \\ 1 & 1 & 3 \end{vmatrix} \] \[ \vec{b} = \hat{i}(-6 - (-1)) - \hat{j}(3 - (-1)) + \hat{k}(1 - (-2)) \] \[ \vec{b} = -5\hat{i} - 4\hat{j} + 3\hat{k} \] Direction ratios are \( (-5, -4, 3) \).

Step 3: Form the line equation through (2, 3, 1).
\[ \frac{x - 2}{-5} = \frac{y - 3}{-4} = \frac{z - 1}{3} \]