Understanding the Problem: Line Parallel to a Plane

The problem asks us to find the value of 'k' for which a given straight line is parallel to a specific plane. The straight line is defined by the intersection of two planes, and the condition for a line to be parallel to a plane involves the relationship between the line's direction vector and the plane's normal vector.

Key Concepts for Line and Plane Relationships

To solve this problem, we need to understand a few fundamental concepts from 3D analytical geometry:

• Normal Vector of a Plane: For a plane defined by the equation \(Ax + By + Cz + D = 0\), its normal vector is \(\vec{n} = (A, B, C)\). This vector is perpendicular to the plane.
• Direction Vector of a Line: A line in 3D space can be represented by a direction vector \(\vec{d} = (l, m, n)\) that indicates its orientation.
• Line as Intersection of Two Planes: When a line is formed by the intersection of two planes, its direction vector is perpendicular to the normal vectors of both planes. Therefore, the direction vector of such a line can be found by taking the cross product of the normal vectors of the two intersecting planes.
• Condition for Line Parallel to a Plane: A line with direction vector \(\vec{d}\) is parallel to a plane with normal vector \(\vec{n_p}\) if and only if the line's direction vector is perpendicular to the plane's normal vector. Mathematically, this means their dot product is zero: \(\vec{d} \cdot \vec{n_p} = 0\).

Step-by-Step Solution to Find 'k'

1. Determining the Normal Vectors of the Intersecting Planes

The straight line is given by the intersection of the planes:

• Plane 1: \(x + y + 3z - 2 = 0\)
• Plane 2: \(2x + y - z - 3 = 0\)

The normal vector for Plane 1, \(\vec{n_1}\), is the coefficients of x, y, and z:

\[ \vec{n_1} = (1, 1, 3) \]

The normal vector for Plane 2, \(\vec{n_2}\), is the coefficients of x, y, and z:

\[ \vec{n_2} = (2, 1, -1) \]

2. Calculating the Direction Vector of the Straight Line

The direction vector \(\vec{d}\) of the straight line, which is the intersection of Plane 1 and Plane 2, is given by the cross product of their normal vectors, \(\vec{n_1} \times \vec{n_2}\).

\[ \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 3 \\ 2 & 1 & -1 \end{vmatrix} \]

Expanding the determinant:

\[ \vec{d} = \mathbf{i}((1)(-1) - (3)(1)) - \mathbf{j}((1)(-1) - (3)(2)) + \mathbf{k}((1)(1) - (1)(2)) \]

\[ \vec{d} = \mathbf{i}(-1 - 3) - \mathbf{j}(-1 - 6) + \mathbf{k}(1 - 2) \]

\[ \vec{d} = -4\mathbf{i} - (-7)\mathbf{j} + (-1)\mathbf{k} \]

\[ \vec{d} = -4\mathbf{i} + 7\mathbf{j} - \mathbf{k} \]

So, the direction vector of the line is \(\vec{d} = (-4, 7, -1)\).

3. Identifying the Normal Vector of the Parallel Plane

The plane to which the straight line is parallel is given by the equation:

\[ 3x + 2y + kz - 4 = 0 \]

The normal vector for this plane, \(\vec{n_p}\), is:

\[ \vec{n_p} = (3, 2, k) \]

4. Applying the Parallelism Condition to Find 'k'

For the straight line to be parallel to the plane, their direction vector \(\vec{d}\) must be perpendicular to the plane's normal vector \(\vec{n_p}\). This means their dot product must be zero:

\[ \vec{d} \cdot \vec{n_p} = 0 \]

Substituting the components of \(\vec{d}\) and \(\vec{n_p}\):

\[ (-4)(3) + (7)(2) + (-1)(k) = 0 \]

\[ -12 + 14 - k = 0 \]

\[ 2 - k = 0 \]

Solving for \(k\):

\[ k = 2 \]

Final Answer Summary

By finding the direction vector of the line (using the cross product of the normal vectors of the two planes defining it) and then applying the condition that this direction vector is perpendicular to the normal vector of the parallel plane, we determined the value of 'k'.

The calculated value of \(k\) is 2.