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

1.	Find the angle between x2 + yz = 2 and x + 2y - z = 2 at point 1,1,1.
Answer» \(\vec ▽\)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)} = \(2\hat i+\hat j+\hat k\) \(\vec ▽\)S₂ = \((\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial z})(x + 2y-z)\) = \((\hat i+2\hat j-\hat k)\) Let angle between them is θ \(\vec ▽\)S₁ = \(\vec ▽\)S₂ = (\(2\hat i+\hat j+\hat k\)).(\(\hat i+2\hat j-\hat k\)) ⇒ \(\sqrt6.\sqrt6 cos\theta=2 + 2- 1\) ⇒ cos θ = 3/6 = 1/2 = cos 60° ⇒ θ = 60° Hence, angle between them is 60°.

Answer»

\(\vec ▽\)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)} = \(2\hat i+\hat j+\hat k\)

\(\vec ▽\)S₂ = \((\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial z})(x + 2y-z)\)

= \((\hat i+2\hat j-\hat k)\)

Let angle between them is θ

\(\vec ▽\)S₁ = \(\vec ▽\)S₂ = (\(2\hat i+\hat j+\hat k\)).(\(\hat i+2\hat j-\hat k\))

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

⇒ cos θ = 3/6 = 1/2 = cos 60°

⇒ θ = 60°

Hence, angle between them is 60°.

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