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

To find the equation of a line passing through the point \( (2, 3, 1) \) and parallel to the line of intersection of two planes, we need to follow these steps: 

1. First, identify the normal vectors of the given planes:
   • For the plane \( x - 2y - z + 5 = 0 \), the normal vector is \(\langle 1, -2, -1 \rangle\).
   • For the plane \( x + y + 3z = 6 \), the normal vector is \(\langle 1, 1, 3 \rangle\).
2. To find the direction vector of the line of intersection, compute the cross product of the normals of the two planes:

\[ \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -1 \\ 1 & 1 & 3 \\ \end{vmatrix} = \mathbf{i}( (-2)(3) - (-1)(1) ) - \mathbf{j}( (1)(3) - (-1)(1) ) + \mathbf{k}( (1)(1) - (-2)(1) ) \] \[ = \mathbf{i}( -6 + 1) - \mathbf{j}( 3 + 1 ) + \mathbf{k}( 1 + 2 ) \] \[ = \langle -5, -4, 3 \rangle \]

1. The direction vector of the line through the point \((2, 3, 1)\) that is parallel to the line of intersection is therefore \(\langle -5, -4, 3 \rangle\).
2. The equation of a line in 3D passing through a point \((x_0, y_0, z_0)\) and having direction vector \(\langle a, b, c \rangle\) is represented parametrically as:

\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \]

1. Therefore, the equation of the line we are seeking is:

\[ \frac{x - 2}{-5} = \frac{y - 3}{-4} = \frac{z - 1}{3} \]

Thus, the correct option is: 
\(\frac{x - 2}{-5} = \frac{y - 3}{-4} = \frac{z - 1}{3}\)