Question

Find the angle between x² + yz = 2 and x + 2y - z = 2 at point 1,1,1.

Answer 1

\(\vec ▽ \)S₁ = (\(\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial z}\)) (x² + yz)

 = (2x\(\hat i\) + z\(\hat j\) + y\(\hat k\))

( \(\vec ▽ \)S₁) _{(1, 1, 1)} = \(2\hat i+\hat j+\hat k\)

 \(\vec ▽ \)S ₂ = \((\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial z})(x + 2y-z)\)

 = \((\hat i+2\hat j-\hat k)\)

Let angle between them is θ

 \(\vec ▽ \)S₁ =  \(\vec ▽ \)S ₂ = (\(2\hat i+\hat j+\hat k\)).(\(\hat i+2\hat j-\hat k\))

⇒ \(\sqrt6.\sqrt6 cos\theta=2 + 2- 1\)

⇒ cos  θ = 3/6 = 1/2 = cos 60°

⇒ θ = 60°

Hence, angle between them is 60°.