Question

If the equation of the plane passing through the points (2,1,2), (1,2,1) and perpendicular to the plane \( 2x-y+2z=1 \) is \( ax+by+cz+d=0 \) then \( \frac{a+b}{c+d} = \)

• A. 0
• B. 1
• C. -1
• D. 2

Hint:

Using the cross product of the normal of the reference plane and the vector formed by the points is the quickest way to find the new normal.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

The normal vector of the required plane is perpendicular to the normal of the given plane and perpendicular to the vector joining the two points. We find this normal vector using the cross product.
Step 2: Key Formula or Approach:

Normal \( \vec{n} = \vec{n_{given}} \times \vec{AB} \).
Step 3: Detailed Explanation:

Points \( A(2,1,2) \) and \( B(1,2,1) \). Vector \( \vec{AB} = (1-2, 2-1, 1-2) = (-1, 1, -1) \). Normal of given plane \( \vec{n_1} = (2, -1, 2) \). Normal of required plane \( \vec{n} = \vec{n_1} \times \vec{AB} \): \( \vec{n} = \begin{vmatrix} i & j & k \\ 2 & -1 & 2 \\ -1 & 1 & -1 \end{vmatrix} \) \( = i(1-2) - j(-2+2) + k(2-1) = -i + 0j + k \). Direction ratios \( (a, b, c) = (-1, 0, 1) \). Let's use \( (1, 0, -1) \) for convenience. Equation of plane through \( (2,1,2) \): \( 1(x-2) + 0(y-1) - 1(z-2) = 0 \) \( x - z = 0 \). Comparing with \( ax+by+cz+d=0 \): \( a=1, b=0, c=-1, d=0 \). Compute expression: \( \frac{a+b}{c+d} = \frac{1+0}{-1+0} = -1 \).
Step 4: Final Answer:

The value is -1.