Question

Let plane $ \pi $ pass through $ (1, 0, 1) $ and be perpendicular to both: - $ 2x + 3y - z = 2 $ - $ x - y + 2z = 1 $ Let another plane pass through point $ (11, 7, 5) $, be parallel to $ \pi $, and have the form: $$ ax + by - z + d = 0 $$ Then evaluate: $$ \frac{a}{b} + \frac{b}{d} $$

• A. 3
• B. 0
• C. 2
• D. -2

Hint:

When a plane is perpendicular to two given planes, use the cross product of their normals to get the required plane's normal.

Answer 1

The Correct Option is D

The normal vector \( \vec{n}_{\pi} \) of plane \( \pi \) is perpendicular to normals of: - \( \vec{n}_1 = (2, 3, -1) \) - \( \vec{n}_2 = (1, -1, 2) \) Hence, \( \vec{n}_{\pi} \) is parallel to cross product: \[ \vec{n}_{\pi} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix} = \mathbf{i}(3 \cdot 2 - (-1)(-1)) - \mathbf{j}(2 \cdot 2 - (-1)(1)) + \mathbf{k}(2 \cdot (-1) - 3 \cdot 1) = \langle 6 - 1,\ - (4 + 1),\ -2 - 3 \rangle = \langle 5,\ -5,\ -5 \rangle \] So, direction ratios of required plane: \( a = 5,\ b = -5,\ c = -5 \) Equation becomes: \[ 5x - 5y - 5z + d = 0 \Rightarrow \text{Passes through } (11, 7, 5): \Rightarrow 5(11) - 5(7) - 5(5) + d = 0 \Rightarrow 55 - 35 - 25 + d = 0 \Rightarrow -5 + d = 0 \Rightarrow d = 5 \] Now: \[ \frac{a}{b} + \frac{b}{d} = \frac{5}{-5} + \frac{-5}{5} = -1 -1 = \boxed{-2} \]