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

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


Answer:

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

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