Right answer is (c) \(a = \frac{3}{2} \, \& \, b = 1 \)
For explanation: Let \(f_1 = ax^2 – byz – (a+2)x \, \& \, f_2= 4x^2y+z^3 – 4 \)
\(∴ ∇ f_1 = [2ax – (a+2)]\hat{i} – (-bz)\hat{j} + (-bz)\hat{k} \)
also \( ∇ f_1|_{(1,-1,2)} = (a-2) \hat{i} + b\hat{j} + b\hat{k} \)
And \( ∇ f_2 = (8xy) \hat{i} + (4x^2)\hat{j} + (3z^2)\hat{k}, also \, ∇ f_2|_{(1,-1,2)} = -8\hat{i} + 4\hat{j} + 12\hat{k} \)
Now, if the two surfaces cut each other orthogonally at point P(1,-1,2)
then,
\(∇ f_1 . ∇ f_2 = 0 \)
Or \( [(a-2)\hat{i} + b\hat{j} + b\hat{k}] . (-8\hat{i} + 4\hat{j} + 12\hat{k}) = 0
= -2a + b = -4 …..(i) \)
Also if point P lies on both the curve hence it would also satisfy
Both the curves, i.e substituting P(1,-1,2) in equation
\(= ax^2-byz = (a+2) \)
\(= a+2b = a+2 ⇨ b=1 \)
From eqn (i) \(-2a+1 = -4 => 2a = 3 => a = \frac{3}{2}. \)