Question

If the planes $ 2x + 3y + 4z + 7 = 0 $ and $ 4x + ky + 8z + 1 = 0 $ are parallel, then the equation of the plane passing through the point $ (k, k, k) $ and having the direction ratios of its normal as $ (k-1, k, k+1) $ is

• A. \( x + 2y + 3z = 36 \)
• B. \( 3x + 4y + 5z = 72 \)
• C. \( 4x + 5y + 6z = 90 \)
• D. \( 5x + 6y + 7z = 108 \)

Hint:

For parallel planes, set the ratio of the components of their normal vectors equal. To find the equation of the desired plane, use the normal vector and the point it passes through.

Answer 1

The Correct Option is D

Step 1: The normal vector of the first plane \( 2x + 3y + 4z + 7 = 0 \) is \( \langle 2, 3, 4 \rangle \), and the normal vector of the second plane \( 4x + ky + 8z + 1 = 0 \) is \( \langle 4, k, 8 \rangle \). For the planes to be parallel, the normal vectors must be proportional, meaning the ratio of the corresponding components of the normal vectors must be equal: \[ \frac{2}{4} = \frac{3}{k} = \frac{4}{8} \] Solving \( \frac{2}{4} = \frac{4}{8} \), we find that \( k = 6 \).
Step 2: Now, we need to find the equation of the plane passing through the point \( (k, k, k) = (6, 6, 6) \) with the normal vector \( (k-1, k, k+1) = (5, 6, 7) \). The general equation of a plane is: \[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \] Substitute \( a = 5 \), \( b = 6 \), \( c = 7 \), and the point \( (6, 6, 6) \) into the equation: \[ 5(x - 6) + 6(y - 6) + 7(z - 6) = 0 \] Step 3: Simplify the equation: \[ 5x - 30 + 6y - 36 + 7z - 42 = 0 \] \[ 5x + 6y + 7z = 108 \] Thus, the equation of the plane is \( 5x + 6y + 7z = 108 \).