PYQs / GATE CE / 2026 / Set 2 / Q37 GATE CE 2026 Set 2 — Question 37 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Gradient, Divergence & Curl Vector Calculus Engineering Mathematics
Engineering Mathematics → Vector Calculus → Gradient, Divergence & Curl
Last updated 5 September 2026
Question Vector field
V ⃗ \vec{V} V is defined as
V ⃗ = 3 x 2 y z i ^ − 5 x y j ^ + 6 y z 2 k ^ \vec{V} = 3x^2yz \hat{i} - 5xy \hat{j} + 6yz^2 \hat{k} V = 3 x 2 y z i ^ − 5 x y j ^ + 6 y z 2 k ^ The curl of
V ⃗ \vec{V} V at point
( 2 , − 1 , 1 ) (2, -1, 1) ( 2 , − 1 , 1 ) is
Correct answer (A) 6i - 12j - 7k
Solution The curl of a vector field
V ⃗ \vec{V} V is given by
∇ × V ⃗ = ∣ i ^ j ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z V x V y V z ∣ \nabla \times \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ V_x & V_y & V_z \end{vmatrix} ∇ × V = i ^ ∂ x ∂ V x j ^ ∂ y ∂ V y k ^ ∂ z ∂ V z .
Given
V ⃗ = 3 x 2 y z i ^ − 5 x y j ^ + 6 y z 2 k ^ \vec{V} = 3x^2yz \hat{i} - 5xy \hat{j} + 6yz^2 \hat{k} V = 3 x 2 y z i ^ − 5 x y j ^ + 6 y z 2 k ^ :
∇ × V ⃗ = i ^ ( ∂ ∂ y ( 6 y z 2 ) − ∂ ∂ z ( − 5 x y ) ) − j ^ ( ∂ ∂ x ( 6 y z 2 ) − ∂ ∂ z ( 3 x 2 y z ) ) + k ^ ( ∂ ∂ x ( − 5 x y ) − ∂ ∂ y ( 3 x 2 y z ) ) \nabla \times \vec{V} = \hat{i} \left( \frac{\partial}{\partial y}(6yz^2) - \frac{\partial}{\partial z}(-5xy) \right) - \hat{j} \left( \frac{\partial}{\partial x}(6yz^2) - \frac{\partial}{\partial z}(3x^2yz) \right) + \hat{k} \left( \frac{\partial}{\partial x}(-5xy) - \frac{\partial}{\partial y}(3x^2yz) \right) ∇ × V = i ^ ( ∂ y ∂ ( 6 y z 2 ) − ∂ z ∂ ( − 5 x y ) ) − j ^ ( ∂ x ∂ ( 6 y z 2 ) − ∂ z ∂ ( 3 x 2 y z ) ) + k ^ ( ∂ x ∂ ( − 5 x y ) − ∂ y ∂ ( 3 x 2 y z ) ) = i ^ ( 6 z 2 − 0 ) − j ^ ( 0 − 3 x 2 y ) + k ^ ( − 5 y − 3 x 2 z ) = \hat{i}(6z^2 - 0) - \hat{j}(0 - 3x^2y) + \hat{k}(-5y - 3x^2z) = i ^ ( 6 z 2 − 0 ) − j ^ ( 0 − 3 x 2 y ) + k ^ ( − 5 y − 3 x 2 z ) = 6 z 2 i ^ + 3 x 2 y j ^ − ( 5 y + 3 x 2 z ) k ^ = 6z^2 \hat{i} + 3x^2y \hat{j} - (5y + 3x^2z) \hat{k} = 6 z 2 i ^ + 3 x 2 y j ^ − ( 5 y + 3 x 2 z ) k ^ At point
( 2 , − 1 , 1 ) (2, -1, 1) ( 2 , − 1 , 1 ) :
curl V ⃗ = 6 ( 1 ) 2 i ^ + 3 ( 2 ) 2 ( − 1 ) j ^ − ( 5 ( − 1 ) + 3 ( 2 ) 2 ( 1 ) ) k ^ \text{curl } \vec{V} = 6(1)^2 \hat{i} + 3(2)^2(-1) \hat{j} - (5(-1) + 3(2)^2(1)) \hat{k} curl V = 6 ( 1 ) 2 i ^ + 3 ( 2 ) 2 ( − 1 ) j ^ − ( 5 ( − 1 ) + 3 ( 2 ) 2 ( 1 )) k ^ = 6 i ^ − 12 j ^ − ( − 5 + 12 ) k ^ = 6 i ^ − 12 j ^ − 7 k ^ = 6 \hat{i} - 12 \hat{j} - (-5 + 12) \hat{k} = 6 \hat{i} - 12 \hat{j} - 7 \hat{k} = 6 i ^ − 12 j ^ − ( − 5 + 12 ) k ^ = 6 i ^ − 12 j ^ − 7 k ^ .
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Correct answer (A) 6i - 12j - 7k
Solution The curl of a vector field
V ⃗ \vec{V} V is given by
∇ × V ⃗ = ∣ i ^ j ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z V x V y V z ∣ \nabla \times \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ V_x & V_y & V_z \end{vmatrix} ∇ × V = i ^ ∂ x ∂ V x j ^ ∂ y ∂ V y k ^ ∂ z ∂ V z .
Given
V ⃗ = 3 x 2 y z i ^ − 5 x y j ^ + 6 y z 2 k ^ \vec{V} = 3x^2yz \hat{i} - 5xy \hat{j} + 6yz^2 \hat{k} V = 3 x 2 y z i ^ − 5 x y j ^ + 6 y z 2 k ^ :
∇ × V ⃗ = i ^ ( ∂ ∂ y ( 6 y z 2 ) − ∂ ∂ z ( − 5 x y ) ) − j ^ ( ∂ ∂ x ( 6 y z 2 ) − ∂ ∂ z ( 3 x 2 y z ) ) + k ^ ( ∂ ∂ x ( − 5 x y ) − ∂ ∂ y ( 3 x 2 y z ) ) \nabla \times \vec{V} = \hat{i} \left( \frac{\partial}{\partial y}(6yz^2) - \frac{\partial}{\partial z}(-5xy) \right) - \hat{j} \left( \frac{\partial}{\partial x}(6yz^2) - \frac{\partial}{\partial z}(3x^2yz) \right) + \hat{k} \left( \frac{\partial}{\partial x}(-5xy) - \frac{\partial}{\partial y}(3x^2yz) \right) ∇ × V = i ^ ( ∂ y ∂ ( 6 y z 2 ) − ∂ z ∂ ( − 5 x y ) ) − j ^ ( ∂ x ∂ ( 6 y z 2 ) − ∂ z ∂ ( 3 x 2 y z ) ) + k ^ ( ∂ x ∂ ( − 5 x y ) − ∂ y ∂ ( 3 x 2 y z ) ) = i ^ ( 6 z 2 − 0 ) − j ^ ( 0 − 3 x 2 y ) + k ^ ( − 5 y − 3 x 2 z ) = \hat{i}(6z^2 - 0) - \hat{j}(0 - 3x^2y) + \hat{k}(-5y - 3x^2z) = i ^ ( 6 z 2 − 0 ) − j ^ ( 0 − 3 x 2 y ) + k ^ ( − 5 y − 3 x 2 z ) = 6 z 2 i ^ + 3 x 2 y j ^ − ( 5 y + 3 x 2 z ) k ^ = 6z^2 \hat{i} + 3x^2y \hat{j} - (5y + 3x^2z) \hat{k} = 6 z 2 i ^ + 3 x 2 y j ^ − ( 5 y + 3 x 2 z ) k ^ At point
( 2 , − 1 , 1 ) (2, -1, 1) ( 2 , − 1 , 1 ) :
curl V ⃗ = 6 ( 1 ) 2 i ^ + 3 ( 2 ) 2 ( − 1 ) j ^ − ( 5 ( − 1 ) + 3 ( 2 ) 2 ( 1 ) ) k ^ \text{curl } \vec{V} = 6(1)^2 \hat{i} + 3(2)^2(-1) \hat{j} - (5(-1) + 3(2)^2(1)) \hat{k} curl V = 6 ( 1 ) 2 i ^ + 3 ( 2 ) 2 ( − 1 ) j ^ − ( 5 ( − 1 ) + 3 ( 2 ) 2 ( 1 )) k ^ = 6 i ^ − 12 j ^ − ( − 5 + 12 ) k ^ = 6 i ^ − 12 j ^ − 7 k ^ = 6 \hat{i} - 12 \hat{j} - (-5 + 12) \hat{k} = 6 \hat{i} - 12 \hat{j} - 7 \hat{k} = 6 i ^ − 12 j ^ − ( − 5 + 12 ) k ^ = 6 i ^ − 12 j ^ − 7 k ^ .
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