<p>The equation of the plane containing the line \(\dfrac{x - x_1}{l} = \dfrac{y - y_1}{m} = \dfrac{z - z_1}{n}\) is \(a(x - x_1) + b(y - y_1) + c(z - z_1) = 0\), where</p>
<p>\(ax_1 + by_1 + cz_1 = 0\)</p>
<p>\(al + bm + cn = 0\)</p>
<p>\(\dfrac{a}{l} = \dfrac{b}{m} = \dfrac{c}{n}\)</p>
<p>\(lx_1 + my_1 + nz_1 = 0\)</p>
Step-by-Step Solution
Key Concept: A plane containing a given line must have its normal vector perpendicular to the direction vector of the line. This means the dot product of the normal (a, b, c) with the direction vector (l, m, n) must equal zero: al + bm + cn = 0.
Step 1: For a plane to contain a line, every point on the line must satisfy the plane's equation. Step 2: The line passes through (x_1, y_1, z_1) with direction ratios (l, m, n). Any point on the line can be written as (x_1 + lt, y_1 + mt, z_1 + nt) for parameter t. Step 3: Substituting into the plane equation a(x - x_1) + b(y - y_1) + c(z - z_1) = 0: a(lt) + b(mt) + c(nt) = 0 for all values of t Step 4: This is only possible if the coefficient of t equals zero: al + bm + cn = 0 This is the necessary and sufficient condition for the plane to contain the entire line. ∴ Answer: B
Correct Answer: B