# Find the curl of a given vector at the point (1, 2, 3)

1. $\dpi{80}&space;\fn_jvn&space;V&space;=&space;(2x&space;-&space;y^{2})\hat{i}&space;+&space;(3z&space;+&space;x^{2})\hat{j}&space;+&space;(4y&space;-&space;z^{2})\hat{k}$
1. i + 6k
2. 6K
3. 0
4. i
5. i – 6k

The curl of a given vector at the point (1, 2, 3) is i + 6k

### Helping Concept:

1. The curl of vector V is
2. $\dpi{100}&space;\fn_jvn&space;\overrightarrow{\triangledown&space;}\times&space;\overrightarrow{V}&space;=&space;\begin{vmatrix}&space;\hat{i}&space;&&space;\hat{j}&space;&&space;\hat{k}\\&space;\frac{\partial&space;}{\partial&space;x}&space;&&space;\frac{\partial&space;}{\partial&space;y}&space;&&space;\frac{\partial&space;}{\partial&space;z}&space;\\&space;2x&space;-&space;y^{2}&space;&&space;3z&space;+&space;x^{2}&space;&&space;4y&space;-&space;z^{2}&space;\end{vmatrix}$
3. $\dpi{80}&space;\fn_jvn&space;\overrightarrow{\triangledown&space;}\times&space;\overrightarrow{V}&space;=&space;\hat{i}(4-3)&space;-\hat{j}(0-0)+\hat{k}(2x+2y)$
4. $\dpi{80}&space;\fn_jvn&space;\overrightarrow{\triangledown&space;}\times&space;\overrightarrow{V}&space;=&space;\hat{i}&space;+(2x+2y)\hat{k}$
5. put x =1, y = 2 and z = 3 in above
6. $\dpi{80}&space;\fn_jvn&space;\overrightarrow{\triangledown&space;}\times&space;\overrightarrow{V}&space;=&space;\hat{i}&space;+[2(1)+2(2)]\hat{k}$
7. $\dpi{80}&space;\fn_jvn&space;\overrightarrow{\triangledown&space;}\times&space;\overrightarrow{V}&space;=&space;\hat{i}&space;+6\hat{k}$