Question

If the straight line equation \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \) runs completely parallel to the plane surface given by \( A x + 2y + 3z = 5 \), find the value of the coefficient \( A \).

• A. \( 9 \)
• B. \( 0 \)
• C. \( -9 \)
• D. \( -4 \)

Hint:

Remember this geometric inversion rule: when a line is parallel to a plane, you use the perpendicular dot product condition (\( a_1a_2 + b_1b_2 + c_1c_2 = 0 \)) because the plane's normal vector runs perpendicular to its surface.

Answer 1

The Correct Option is C

Concept: If a straight line runs parallel to a flat plane, the line's directional movement vector \( \vec{m} \) must run perpendicular to the plane's surface normal vector \( \vec{n} \). As a result, their scalar dot product must evaluate to exactly zero: \[ \vec{m} \cdot \vec{n} = 0 \]

Step 1: Extract the directional component vectors from the equations.

• Extract the line's direction vector components from the denominators: \( \vec{m} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
• Extract the plane's normal vector components from the variable coefficients: \( \vec{n} = A\hat{i} + 2\hat{j} + 3\hat{k} \)

Step 2: Set up the scalar dot product equation and equate it to zero.
Multiply the corresponding vector components and sum them up: \[ \vec{m} \cdot \vec{n} = (2)(A) + (3)(2) + (4)(3) = 0 \] Simplify the arithmetic multiplication terms: \[ 2A + 6 + 12 = 0 \] \[ 2A + 18 = 0 \]

Step 3: Isolate the unknown coefficient variable \( A \).
Move the constant numbers across the equality sign and divide by the coefficient: \[ 2A = -18 \quad \Rightarrow \quad A = -9 \]