Question

If two planes \( 2x - 4y + 3z = 5 \) and \( x + 2y + \lambda z = 12 \) are mutually perpendicular to each other, then \(\lambda =\):

• A. \(-2\)
• B. \(2\)
• C. \(3\)
• D. none of these

Hint:

The planes are perpendicular if the dot product of their normal vectors is zero. \[ \vec{N_1} \cdot \vec{N_2} = 0 \]

Answer 1

The Correct Option is B

Two planes are perpendicular if and only if their normal vectors are perpendicular. Let the normal vectors be: \[ \vec{N_1} = (2, -4, 3), \quad \vec{N_2} = (1, 2, \lambda) \] The condition for perpendicularity is: \[ \vec{N_1} \cdot \vec{N_2} = 0 \] Calculate the dot product: \[ 2 \times 1 + (-4) \times 2 + 3 \times \lambda = 0 \] \[ 2 - 8 + 3\lambda = 0 \] \[ -6 + 3\lambda = 0 \implies 3\lambda = 6 \implies \lambda = 2 \] Thus, the value of \( \lambda \) is: \[ \boxed{2} \]