\(\vec ▽
\)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°.