Question

Let the line \(\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}\) lie in the plane \(x + 3y - \alpha z + \beta = 0\), then the value of \((\beta - \alpha)\) is equal to

• A. \(1\)
• B. \(13\)
• C. \(7\)
• D. \(-6\)

Hint:

For a line lying in a plane, always use both conditions:
• direction vector \(\perp\) plane normal
• point on line satisfies plane equation e}

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
If a line lies entirely in a plane, its direction vector must be perpendicular to the plane's normal, and any point on the line must satisfy the plane equation.
Step 2: Key Formula or Approach:
1. Dot product of direction ratios \((3, -5, 2)\) and plane normal \((1, 3, -\alpha)\) is zero.
2. Point \((2, 1, -2)\) must satisfy \(x + 3y - \alpha z + \beta = 0\).
Step 3: Detailed Explanation:
1. Find \(\alpha\):
\[ 3(1) + (-5)(3) + 2(-\alpha) = 0 \] \[ 3 - 15 - 2\alpha = 0 \implies -12 = 2\alpha \implies \alpha = -6 \] 2. Find \(\beta\): Substitute \(\alpha = -6\) and point \((2, 1, -2)\) into the plane equation:
\[ 2 + 3(1) - (-6)(-2) + \beta = 0 \] \[ 2 + 3 - 12 + \beta = 0 \implies -7 + \beta = 0 \implies \beta = 7 \] 3. Calculate \(\beta - \alpha\):
\[ \beta - \alpha = 7 - (-6) = 13 \] Step 4: Final Answer:
The value is \(13\).