PYQs / GATE EE / 2019 / Set 1 / Q49 GATE EE 2019 Set 1 — Question 49 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +2 / -0 Medium Gradient, Divergence & Curl Vector Calculus Engineering Mathematics
Engineering Mathematics → Vector Calculus → Gradient, Divergence & Curl
Last updated 5 September 2026
Question If
A = 2 x i + 3 y j + 4 z k \mathbf{A} = 2x\mathbf{i} + 3y\mathbf{j} + 4z\mathbf{k} A = 2 x i + 3 y j + 4 z k and
u = x 2 + y 2 + z 2 u = x^2 + y^2 + z^2 u = x 2 + y 2 + z 2 , then
div ( u A ) \text{div}(u\mathbf{A}) div ( u A ) at
( 1 , 1 , 1 ) (1, 1, 1) ( 1 , 1 , 1 ) is
______ .
Solution We use the vector identity:
div ( u A ) = ∇ ⋅ ( u A ) = u ( ∇ ⋅ A ) + A ⋅ ( ∇ u ) \text{div}(u\mathbf{A}) = \nabla \cdot (u\mathbf{A}) = u(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla u) div ( u A ) = ∇ ⋅ ( u A ) = u ( ∇ ⋅ A ) + A ⋅ ( ∇ u ) Step 1: Calculate
∇ ⋅ A \nabla \cdot \mathbf{A} ∇ ⋅ A :
∇ ⋅ A = ∂ ∂ x ( 2 x ) + ∂ ∂ y ( 3 y ) + ∂ ∂ z ( 4 z ) = 2 + 3 + 4 = 9 \nabla \cdot \mathbf{A} = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial z}(4z) = 2 + 3 + 4 = 9 ∇ ⋅ A = ∂ x ∂ ( 2 x ) + ∂ y ∂ ( 3 y ) + ∂ z ∂ ( 4 z ) = 2 + 3 + 4 = 9 Step 2: Calculate
∇ u \nabla u ∇ u :
∇ u = ∂ u ∂ x i + ∂ u ∂ y j + ∂ u ∂ z k = 2 x i + 2 y j + 2 z k \nabla u = \frac{\partial u}{\partial x}\mathbf{i} + \frac{\partial u}{\partial y}\mathbf{j} + \frac{\partial u}{\partial z}\mathbf{k} = 2x\mathbf{i} + 2y\mathbf{j} + 2z\mathbf{k} ∇ u = ∂ x ∂ u i + ∂ y ∂ u j + ∂ z ∂ u k = 2 x i + 2 y j + 2 z k Step 3: Evaluate at
( 1 , 1 , 1 ) (1, 1, 1) ( 1 , 1 , 1 ) :
u = 1 2 + 1 2 + 1 2 = 3 u = 1^2 + 1^2 + 1^2 = 3 u = 1 2 + 1 2 + 1 2 = 3 ∇ ⋅ A = 9 \nabla \cdot \mathbf{A} = 9 ∇ ⋅ A = 9 A = 2 ( 1 ) i + 3 ( 1 ) j + 4 ( 1 ) k = 2 i + 3 j + 4 k \mathbf{A} = 2(1)\mathbf{i} + 3(1)\mathbf{j} + 4(1)\mathbf{k} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} A = 2 ( 1 ) i + 3 ( 1 ) j + 4 ( 1 ) k = 2 i + 3 j + 4 k ∇ u = 2 ( 1 ) i + 2 ( 1 ) j + 2 ( 1 ) k = 2 i + 2 j + 2 k \nabla u = 2(1)\mathbf{i} + 2(1)\mathbf{j} + 2(1)\mathbf{k} = 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} ∇ u = 2 ( 1 ) i + 2 ( 1 ) j + 2 ( 1 ) k = 2 i + 2 j + 2 k Step 4: Substitute into the identity:
div ( u A ) = 3 ( 9 ) + ( 2 i + 3 j + 4 k ) ⋅ ( 2 i + 2 j + 2 k ) \text{div}(u\mathbf{A}) = 3(9) + (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) \cdot (2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) div ( u A ) = 3 ( 9 ) + ( 2 i + 3 j + 4 k ) ⋅ ( 2 i + 2 j + 2 k ) = 27 + ( 2 × 2 + 3 × 2 + 4 × 2 ) = 27 + (2 \times 2 + 3 \times 2 + 4 \times 2) = 27 + ( 2 × 2 + 3 × 2 + 4 × 2 ) = 27 + ( 4 + 6 + 8 ) = 27 + (4 + 6 + 8) = 27 + ( 4 + 6 + 8 ) = 27 + 18 = 45 = 27 + 18 = 45 = 27 + 18 = 45 The final answer is 45.
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Solution We use the vector identity:
div ( u A ) = ∇ ⋅ ( u A ) = u ( ∇ ⋅ A ) + A ⋅ ( ∇ u ) \text{div}(u\mathbf{A}) = \nabla \cdot (u\mathbf{A}) = u(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla u) div ( u A ) = ∇ ⋅ ( u A ) = u ( ∇ ⋅ A ) + A ⋅ ( ∇ u ) Step 1: Calculate
∇ ⋅ A \nabla \cdot \mathbf{A} ∇ ⋅ A :
∇ ⋅ A = ∂ ∂ x ( 2 x ) + ∂ ∂ y ( 3 y ) + ∂ ∂ z ( 4 z ) = 2 + 3 + 4 = 9 \nabla \cdot \mathbf{A} = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial z}(4z) = 2 + 3 + 4 = 9 ∇ ⋅ A = ∂ x ∂ ( 2 x ) + ∂ y ∂ ( 3 y ) + ∂ z ∂ ( 4 z ) = 2 + 3 + 4 = 9 Step 2: Calculate
∇ u \nabla u ∇ u :
∇ u = ∂ u ∂ x i + ∂ u ∂ y j + ∂ u ∂ z k = 2 x i + 2 y j + 2 z k \nabla u = \frac{\partial u}{\partial x}\mathbf{i} + \frac{\partial u}{\partial y}\mathbf{j} + \frac{\partial u}{\partial z}\mathbf{k} = 2x\mathbf{i} + 2y\mathbf{j} + 2z\mathbf{k} ∇ u = ∂ x ∂ u i + ∂ y ∂ u j + ∂ z ∂ u k = 2 x i + 2 y j + 2 z k Step 3: Evaluate at
( 1 , 1 , 1 ) (1, 1, 1) ( 1 , 1 , 1 ) :
u = 1 2 + 1 2 + 1 2 = 3 u = 1^2 + 1^2 + 1^2 = 3 u = 1 2 + 1 2 + 1 2 = 3 ∇ ⋅ A = 9 \nabla \cdot \mathbf{A} = 9 ∇ ⋅ A = 9 A = 2 ( 1 ) i + 3 ( 1 ) j + 4 ( 1 ) k = 2 i + 3 j + 4 k \mathbf{A} = 2(1)\mathbf{i} + 3(1)\mathbf{j} + 4(1)\mathbf{k} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} A = 2 ( 1 ) i + 3 ( 1 ) j + 4 ( 1 ) k = 2 i + 3 j + 4 k ∇ u = 2 ( 1 ) i + 2 ( 1 ) j + 2 ( 1 ) k = 2 i + 2 j + 2 k \nabla u = 2(1)\mathbf{i} + 2(1)\mathbf{j} + 2(1)\mathbf{k} = 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} ∇ u = 2 ( 1 ) i + 2 ( 1 ) j + 2 ( 1 ) k = 2 i + 2 j + 2 k Step 4: Substitute into the identity:
div ( u A ) = 3 ( 9 ) + ( 2 i + 3 j + 4 k ) ⋅ ( 2 i + 2 j + 2 k ) \text{div}(u\mathbf{A}) = 3(9) + (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) \cdot (2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) div ( u A ) = 3 ( 9 ) + ( 2 i + 3 j + 4 k ) ⋅ ( 2 i + 2 j + 2 k ) = 27 + ( 2 × 2 + 3 × 2 + 4 × 2 ) = 27 + (2 \times 2 + 3 \times 2 + 4 \times 2) = 27 + ( 2 × 2 + 3 × 2 + 4 × 2 ) = 27 + ( 4 + 6 + 8 ) = 27 + (4 + 6 + 8) = 27 + ( 4 + 6 + 8 ) = 27 + 18 = 45 = 27 + 18 = 45 = 27 + 18 = 45 The final answer is 45.
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More questions on Vector Calculus ← Q48 Full paper Q50 → 