Combined Operations
Chaining grad, div, and curl - fundamental identities and the Laplacian (Kaplan §3.3/§3.6)
Prereq: §3.5 The Curl
Meet the Gaussian Hill
Before we talk about chaining operators, let's put a concrete object on the table. Take the smooth radial hill $f(x,y) = e^{-(x^2+y^2)}$. It peaks at the origin at height $1$ and decays in every direction. Simple, symmetric, bounded - a good sparring partner.
The surface of $f$. The peak sits at the origin and decays radially. Toggle "+ gradient" to see the gradient vector field, then "+ curl of gradient" to see what happens when we try to curl it.
Spin it around. Notice: the gradient arrows (indigo) always point inward toward the peak - they lead uphill. Could those arrows ever circulate around a loop? If they did, following them would take you in a closed path that climbed the whole way. Impossible. That hints at a deeper fact we'll prove below: the curl of any gradient field vanishes identically.
Take any smooth hill - say $f(x,y) = e^{-(x^2+y^2)}$. Compute its gradient, then take the curl of that gradient. You'll get zero. Always zero, no matter what hill you choose. Now take any spinning flow and compute its curl, then take the divergence. Zero again - always. These aren't coincidences. They're built into the structure of how grad, div, and curl chain together.
Each of our three operators has a specific input/output type:
The types tell us which chains are even legal. We can only feed an operator's output to the next one if the types match. Here are the five valid two-step chains:
Two of these five chains are always zero - no matter what function you feed them. Why? What makes those two special?
curl(grad $f$) = 0 - Always
Let's start with a prediction. The gradient $\nabla f$ points in the direction of steepest ascent of $f$. Could such a field ever form closed loops? If $f$ is a hill, the steepest-ascent arrows always lead uphill - if you followed them in a loop, you'd arrive back where you started at a higher elevation. That's impossible. So intuitively, a gradient field should have zero curl.
Let's verify this algebraically. Take any smooth scalar field $f(x,y,z)$. Its gradient is:
$\nabla f = \left(\dfrac{\partial f}{\partial x},\; \dfrac{\partial f}{\partial y},\; \dfrac{\partial f}{\partial z}\right)$
Now compute the $\mathbf{k}$-component of $\nabla \times (\nabla f)$:
By Clairaut's theorem (equality of mixed partials for smooth functions), $f_{xy} = f_{yx}$. So this is exactly $0$. The same cancellation happens for the $\mathbf{i}$-component ($f_{zy} - f_{yz} = 0$) and the $\mathbf{j}$-component ($f_{xz} - f_{zx} = 0$).
This is arguably the most important identity in vector calculus. It says: gradient fields are always irrotational. If $f$ is elevation, the steepest-ascent directions can't form closed loops - exactly as we predicted.
The converse is equally powerful. If $\nabla \times \mathbf{v} = \mathbf{0}$ everywhere (on a simply connected domain), then $\mathbf{v} = \nabla f$ for some scalar potential $f$. This is the test for conservative fields: compute the curl; if it vanishes, a potential exists.
Try it: verify curl(grad $f$) = 0
Enter any scalar function $f(x,y,z)$. We'll compute $\nabla \times (\nabla f)$ numerically at $(1,1,1)$.
Worked Example: Verifying curl(grad $f$) = 0
The identity says $\operatorname{curl}(\operatorname{grad} f) = \mathbf{0}$ for any smooth $f$. Let's verify with a specific function - if even one component is nonzero, we've disproved a theorem.
Take $f = x^2 y + yz^2$. First, compute the gradient:
Now we take the curl of this gradient field. We need all three components:
$\mathbf{i}$-component: $\;\dfrac{\partial}{\partial y}(2yz) - \dfrac{\partial}{\partial z}(x^2 + z^2) = 2z - 2z = 0$ ✓
$\mathbf{j}$-component: $\;\dfrac{\partial}{\partial z}(2xy) - \dfrac{\partial}{\partial x}(2yz) = 0 - 0 = 0$ ✓
$\mathbf{k}$-component: $\;\dfrac{\partial}{\partial x}(x^2 + z^2) - \dfrac{\partial}{\partial y}(2xy) = 2x - 2x = 0$ ✓
Every component cancels. The identity $\operatorname{curl}(\operatorname{grad} f) = \mathbf{0}$ holds for any smooth scalar field - not because of anything special about $f = x^2 y + yz^2$, but because mixed partials always commute (Clairaut's theorem). Each component of the curl pits $f_{ij}$ against $f_{ji}$, and those are always equal.
div(curl $\mathbf{v}$) = 0 - Always
Picture the curl of a vector field - those swirling vortex lines we visualized in §3.5. Could a vortex line just start at a point, like water gushing from a faucet? If it did, that point would be a source of the curl field, meaning $\nabla \cdot (\nabla \times \mathbf{v}) > 0$ there. Think about magnetic field lines: they always form closed loops - you can never find a magnetic monopole where field lines begin or end. That's because $\mathbf{B} = \nabla \times \mathbf{A}$, and curl fields have no sources or sinks. Their field lines must close on themselves.
Let's prove this is always the case. Take any smooth vector field $\mathbf{v} = (v_1, v_2, v_3)$. The curl $\nabla \times \mathbf{v}$ has three components, and we take the divergence of that:
Expand this out. We get six second-order mixed partials:
$= \underbrace{v_{3,xy}}_{} - \underbrace{v_{2,xz}}_{} + \underbrace{v_{1,yz}}_{} - \underbrace{v_{3,yx}}_{} + \underbrace{v_{2,zx}}_{} - \underbrace{v_{1,zy}}_{}$
By Clairaut, $v_{3,xy} = v_{3,yx}$, so the first and fourth terms cancel. Similarly $v_{2,xz} = v_{2,zx}$ (second and fifth cancel) and $v_{1,yz} = v_{1,zy}$ (third and sixth cancel). Everything vanishes:
Exactly as our intuition predicted - the algebra confirms that curl fields are always source-free.
The converse: if $\nabla \cdot \mathbf{u} = 0$ everywhere, then $\mathbf{u} = \nabla \times \mathbf{v}$ for some vector field $\mathbf{v}$. Divergence-free (solenoidal) fields are always curls of something.
Try it: verify div(curl $\mathbf{v}$) = 0
Enter any vector field $(F_1, F_2, F_3)$. We'll compute $\nabla \cdot (\nabla \times \mathbf{F})$ numerically at $(1,1,1)$.
Worked Example: Computing Divergence
Before we move on to the Laplacian, let's practice the divergence computation itself - since it appears in several of the identities above, you want this to be automatic.
Let $\mathbf{u} = (x^2 + y^2)\,\mathbf{i} + 2xy\,\mathbf{j} + (z - x)\,\mathbf{k}$. Compute $\operatorname{div}\,\mathbf{u}$.
The divergence is the sum of the three partial derivatives, each taken with respect to the matching variable:
At the point $(1, 2, 3)$: $\operatorname{div}\,\mathbf{u} = 4(1) + 1 = 5$.
What does this mean? Divergence is a scalar. It tells you the local expansion rate of the field at each point - positive means source (fluid is being created there), negative means sink (fluid is being absorbed). Here $\operatorname{div}\,\mathbf{u} = 4x + 1$, so the field acts as a source everywhere $x > -1/4$ and as a sink where $x < -1/4$. At our test point $(1,2,3)$, the positive divergence of $5$ means the field is spreading outward.
Notice the $y$ and $z$ coordinates don't affect the divergence at all - the expansion rate depends only on $x$. That's because the $y$-dependence in $x^2 + y^2$ vanishes under $\partial/\partial x$, and similarly for the other terms. Don't be fooled by the complexity of the original field - the divergence can be much simpler.
The Laplacian
Two of our five chains give zero. But the third one - $\operatorname{div}(\operatorname{grad} f)$ - is genuinely interesting and non-trivial. Let's expand it:
$\nabla \cdot (\nabla f) = \nabla \cdot \!\left(\dfrac{\partial f}{\partial x},\; \dfrac{\partial f}{\partial y},\; \dfrac{\partial f}{\partial z}\right) = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} + \dfrac{\partial^2 f}{\partial z^2}$
This sum of pure second derivatives appears so often that it earns its own symbol:
This is the Laplacian of $f$, also written $\Delta f$. What does it measure? Here's the beautiful interpretation: the Laplacian tells you how much $f$ at a point differs from its average on nearby spheres.
If $\nabla^2 f > 0$ at a point $P$, then $f(P)$ is less than the average of $f$ on small spheres around $P$ - the point is a local "valley." If $\nabla^2 f \lt 0$, the point is a local "hill."
Harmonic functions are the special case where $\nabla^2 f = 0$ everywhere. At every point, the value exactly equals the average over surrounding spheres - no hills, no valleys, just perfect balance. These are the solutions to Laplace's equation, one of the most important PDEs in physics.
Example: Take $f = 1/r$ where $r = \sqrt{x^2 + y^2 + z^2}$ - the potential of a point charge. Let's compute each piece:
$\dfrac{\partial f}{\partial x} = -\dfrac{x}{r^3}, \quad \dfrac{\partial^2 f}{\partial x^2} = -\dfrac{1}{r^3} + \dfrac{3x^2}{r^5} = \dfrac{3x^2 - r^2}{r^5}$
By symmetry, $f_{yy} = (3y^2 - r^2)/r^5$ and $f_{zz} = (3z^2 - r^2)/r^5$. Adding:
$\nabla^2(1/r) = \dfrac{3(x^2 + y^2 + z^2) - 3r^2}{r^5} = \dfrac{3r^2 - 3r^2}{r^5} = 0$
So $1/r$ is harmonic everywhere except the origin. The gravitational/electric potential satisfies Laplace's equation outside the source - a fundamental fact of physics.
curl(curl) and the Vector Laplacian
The last chain worth studying is $\nabla \times (\nabla \times \mathbf{u})$. Computing this directly from the curl formula is a slog - you'd expand the outer curl of the inner curl and drown in partial derivatives. But there's a remarkable shortcut:
Here $\nabla^2 \mathbf{u}$ is the vector Laplacian - we just apply the scalar Laplacian to each component:
$\nabla^2 \mathbf{u} = (\nabla^2 u_1)\,\mathbf{i} + (\nabla^2 u_2)\,\mathbf{j} + (\nabla^2 u_3)\,\mathbf{k}$
This identity is sometimes called the "BAC-CAB" identity for $\nabla$ (by analogy with $\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \mathbf{B}(\mathbf{A}\cdot\mathbf{C}) - \mathbf{C}(\mathbf{A}\cdot\mathbf{B})$). It rewrites a complicated double-curl in terms of two simpler operations: a gradient of the divergence, and a vector Laplacian.
Why it matters: In electromagnetism, the wave equation for the electric field comes directly from applying this identity to $\nabla \times (\nabla \times \mathbf{E})$. In regions with no charges ($\nabla \cdot \mathbf{E} = 0$), it simplifies to $\nabla^2 \mathbf{E} = \mu_0 \epsilon_0 \,\partial^2\mathbf{E}/\partial t^2$ - the equation that predicts electromagnetic waves.
Finding Potential Functions
We know that if $\nabla \times \mathbf{F} = \mathbf{0}$, then $\mathbf{F} = \nabla \varphi$ for some scalar potential $\varphi$. But how do we actually find $\varphi$? Here's the systematic method. We'll work through it with a concrete example.
Example: $\mathbf{F} = (2xy + z)\,\mathbf{i} + x^2\,\mathbf{j} + x\,\mathbf{k}$. Is this conservative? If so, find $\varphi$.
We need all three components of $\nabla \times \mathbf{F}$ to vanish:
$\mathbf{i}$: $\;\partial(x)/\partial y - \partial(x^2)/\partial z = 0 - 0 = 0$ ✓
$\mathbf{j}$: $\;-[\partial(x)/\partial x - \partial(2xy+z)/\partial z] = -(1 - 1) = 0$ ✓
$\mathbf{k}$: $\;\partial(x^2)/\partial x - \partial(2xy+z)/\partial y = 2x - 2x = 0$ ✓
Good - $\mathbf{F}$ is conservative. A potential exists.
We need $\partial \varphi / \partial x = 2xy + z$. Integrate in $x$ (treating $y,z$ as constants):
$\varphi = x^2 y + xz + g(y,z)$
The "constant" of integration $g(y,z)$ can depend on the other variables - since we only integrated in $x$, anything that doesn't involve $x$ could be lurking.
We need $\partial \varphi / \partial y = x^2$. Differentiating our expression:
$\partial \varphi / \partial y = x^2 + \partial g / \partial y$
So $\partial g / \partial y = x^2 - x^2 = 0$. This means $g$ doesn't depend on $y$ at all: $g(y,z) = h(z)$.
We need $\partial \varphi / \partial z = x$. With $\varphi = x^2 y + xz + h(z)$:
$\partial \varphi / \partial z = x + h'(z)$
So $h'(z) = 0$, meaning $h(z) = C$ (a true constant).
Common mistake: forgetting that the "constant" of integration in Step 2 can depend on $y$ and $z$. If you write $\varphi = x^2 y + xz + C$ right away, you get lucky here - but in general, $g(y,z)$ might contain terms like $y^2 z$ that you'd miss.
Analyze Any Field
Enter any 3D vector field and we'll compute everything at $(1,1,1)$: divergence, curl, whether it's conservative, and more.
Build a Chain
Time to cement the typing rules. Drag (or click) an input tile into the dropzone to start, then stack operators on it. The chain validates its types live: the badge after each tile shows the output type, or "invalid" if the previous output doesn't match what the next operator eats. When the chain is valid, you'll see what it simplifies to.
Click a tile already in the chain to truncate from that point. Two chains always collapse to zero - can you find them both?
Practice Problems - §3.6
From Kaplan, problems after §3.6
Prove the two fundamental identities: $\operatorname{curl}(\operatorname{grad} f) = \mathbf{0}$ and $\operatorname{div}(\operatorname{curl}\,\mathbf{v}) = 0$.
Step 1: Prove $\operatorname{curl}(\operatorname{grad} f) = \mathbf{0}$.
$\operatorname{grad} f = \dfrac{\partial f}{\partial x}\mathbf{i} + \dfrac{\partial f}{\partial y}\mathbf{j} + \dfrac{\partial f}{\partial z}\mathbf{k}$. The curl is:
$$\operatorname{curl}(\operatorname{grad} f) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f_x & f_y & f_z \end{vmatrix}$$
The $\mathbf{i}$-component: $f_{zy} - f_{yz} = 0$ by equality of mixed partials. Similarly for $\mathbf{j}$ ($f_{xz} - f_{zx} = 0$) and $\mathbf{k}$ ($f_{yx} - f_{xy} = 0$). $\square$
Step 2: Prove $\operatorname{div}(\operatorname{curl}\,\mathbf{v}) = 0$.
Let $\mathbf{v} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$. Then $\operatorname{curl}\,\mathbf{v} = (R_y - Q_z)\mathbf{i} + (P_z - R_x)\mathbf{j} + (Q_x - P_y)\mathbf{k}$.
Taking the divergence:
$$\operatorname{div}(\operatorname{curl}\,\mathbf{v}) = (R_y - Q_z)_x + (P_z - R_x)_y + (Q_x - P_y)_z$$
$$= R_{yx} - Q_{zx} + P_{zy} - R_{xy} + Q_{xz} - P_{yz} = 0$$
by equality of mixed partials ($R_{yx} = R_{xy}$, $Q_{zx} = Q_{xz}$, $P_{zy} = P_{yz}$). $\square$
Verify $\nabla \times \mathbf{v} = \mathbf{0}$ and find $f$ with $\nabla f = \mathbf{v}$ for $\mathbf{v} = e^{xy}[(2y^2+yz^2)\,\mathbf{i} + (2xy+xz^2+2)\,\mathbf{j} + 2z\,\mathbf{k}]$.
$v_1 = e^{xy}(2y^2+yz^2)$, $\;v_2 = e^{xy}(2xy+xz^2+2)$, $\;v_3 = 2ze^{xy}$.
To verify curl = 0, we need to carefully apply the product rule through $e^{xy}$. Let's check the $\mathbf{k}$-component: $\partial v_2/\partial x - \partial v_1/\partial y$.
$\dfrac{\partial v_2}{\partial x} = ye^{xy}(2xy+xz^2+2) + e^{xy}(2y+z^2) = e^{xy}[2xy^2+xyz^2+2y+2y+z^2]$
$\dfrac{\partial v_1}{\partial y} = xe^{xy}(2y^2+yz^2) + e^{xy}(4y+z^2) = e^{xy}[2xy^2+xyz^2+4y+z^2]$
These are equal, so the $\mathbf{k}$-component is 0 ✓. The other components cancel by similar (lengthy) calculations.
This time it's easier to start with $v_3$ (the simplest component):
$f = \displaystyle\int 2ze^{xy}\,dz = z^2 e^{xy} + g(x,y)$
$\partial f / \partial x = yz^2 e^{xy} + g_x(x,y)$. We need $e^{xy}(2y^2 + yz^2)$, so $g_x = 2y^2 e^{xy}$.
Integrate: $g(x,y) = \displaystyle\int 2y^2 e^{xy}\,dx = 2y\,e^{xy} + h(y)$.
Now $f = z^2 e^{xy} + 2ye^{xy} + h(y)$, so:
$\partial f/\partial y = xz^2 e^{xy} + 2e^{xy} + 2xye^{xy} + h'(y) = e^{xy}(xz^2 + 2 + 2xy) + h'(y)$
We need $e^{xy}(2xy + xz^2 + 2)$ - that matches exactly, so $h'(y) = 0$ and $h = C$.
Prove that $\nabla \cdot (\nabla \times \mathbf{v}) = 0$ for any smooth vector field $\mathbf{v}$. Verify for $\mathbf{v} = x^2 yz\,\mathbf{i} - x^3 y^3\,\mathbf{j} + xyz^2\,\mathbf{k}$.
Write $\mathbf{v} = (v_1, v_2, v_3)$. Then:
$\nabla \cdot (\nabla \times \mathbf{v}) = \dfrac{\partial}{\partial x}(v_{3,y} - v_{2,z}) + \dfrac{\partial}{\partial y}(v_{1,z} - v_{3,x}) + \dfrac{\partial}{\partial z}(v_{2,x} - v_{1,y})$
$= v_{3,yx} - v_{2,zx} + v_{1,zy} - v_{3,xy} + v_{2,xz} - v_{1,yz}$
By Clairaut: $v_{3,yx} = v_{3,xy}$, $v_{2,zx} = v_{2,xz}$, $v_{1,zy} = v_{1,yz}$. The six terms cancel in pairs, giving $0$.
$\mathbf{i}$: $\;\partial(xyz^2)/\partial y - \partial(-x^3 y^3)/\partial z = xz^2 - 0 = xz^2$
$\mathbf{j}$: $\;-[\partial(xyz^2)/\partial x - \partial(x^2 yz)/\partial z] = -(yz^2 - x^2 y) = x^2 y - yz^2$
$\mathbf{k}$: $\;\partial(-x^3 y^3)/\partial x - \partial(x^2 yz)/\partial y = -3x^2 y^3 - x^2 z$
$\dfrac{\partial}{\partial x}(xz^2) + \dfrac{\partial}{\partial y}(x^2 y - yz^2) + \dfrac{\partial}{\partial z}(-3x^2 y^3 - x^2 z)$
$= z^2 + (x^2 - z^2) + (-x^2)$
$= z^2 + x^2 - z^2 - x^2 = 0$ ✓
Given $\mathbf{v} = 2x\mathbf{i} + y\mathbf{j} - 3z\mathbf{k}$, verify $\operatorname{div}\,\mathbf{v} = 0$ and find all $\mathbf{u}$ such that $\operatorname{curl}\,\mathbf{u} = \mathbf{v}$.
Step 1: Verify divergence-free.
$\operatorname{div}\,\mathbf{v} = \dfrac{\partial(2x)}{\partial x} + \dfrac{\partial(y)}{\partial y} + \dfrac{\partial(-3z)}{\partial z} = 2 + 1 - 3 = 0$ ✓
Step 2: Find a particular $\mathbf{u}_0$ with $\operatorname{curl}\,\mathbf{u}_0 = \mathbf{v}$.
Try $\mathbf{u}_0$ with $u_{0z} = 0$, i.e. $\mathbf{u}_0 = A\mathbf{i} + B\mathbf{j}$. Then $\operatorname{curl}\,\mathbf{u}_0 = (-B_z)\mathbf{i} + (A_z)\mathbf{j} + (B_x - A_y)\mathbf{k}$.
We need: $-B_z = 2x$, $A_z = y$, $B_x - A_y = -3z$.
From $A_z = y$: $A = yz + \alpha(x,y)$.
From $-B_z = 2x$: $B = -2xz + \beta(x,y)$.
From $B_x - A_y = -3z$: $(-2z + \beta_x) - (z + \alpha_y) = -3z + \beta_x - \alpha_y$. Set $\beta_x - \alpha_y = 0$, e.g. $\alpha = \beta = 0$.
So $\mathbf{u}_0 = yz\mathbf{i} - 2xz\mathbf{j}$.
Step 3: General solution.
Since $\operatorname{curl}(\operatorname{grad} f) = \mathbf{0}$ for any scalar $f$, all solutions are:
for any smooth scalar function $f(x,y,z)$.
Show that $\nabla^2 f = \nabla \cdot (\nabla f)$ and verify that $f = 1/r$ (where $r = \sqrt{x^2 + y^2 + z^2}$) satisfies $\nabla^2 f = 0$ for $r \neq 0$.
$\nabla \cdot (\nabla f) = \nabla \cdot \left(\dfrac{\partial f}{\partial x},\; \dfrac{\partial f}{\partial y},\; \dfrac{\partial f}{\partial z}\right) = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} + \dfrac{\partial^2 f}{\partial z^2} = \nabla^2 f$ ✓
This is simply the definition: div eats the gradient vector and sums the three second derivatives.
$\dfrac{\partial}{\partial x}\!\left(\dfrac{1}{r}\right) = -\dfrac{x}{r^3}$
$\dfrac{\partial^2}{\partial x^2}\!\left(\dfrac{1}{r}\right) = -\dfrac{r^3 - x \cdot 3r^2 \cdot (x/r)}{r^6} = -\dfrac{r^3 - 3x^2 r}{r^6} = \dfrac{3x^2 - r^2}{r^5}$
By symmetry: $\dfrac{\partial^2(1/r)}{\partial y^2} = \dfrac{3y^2 - r^2}{r^5}$, $\;\dfrac{\partial^2(1/r)}{\partial z^2} = \dfrac{3z^2 - r^2}{r^5}$.
Adding:
$\nabla^2(1/r) = \dfrac{3x^2 + 3y^2 + 3z^2 - 3r^2}{r^5} = \dfrac{3r^2 - 3r^2}{r^5} = 0$ ✓
So $1/r$ is harmonic for $r \neq 0$.
Prove: $\operatorname{curl}(\operatorname{curl}\,\mathbf{v}) = \operatorname{grad}(\operatorname{div}\,\mathbf{v}) - \nabla^2 \mathbf{v}$.
Step 1: Compute $\operatorname{curl}\,\mathbf{v}$.
Let $\mathbf{v} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$. Then:
$\operatorname{curl}\,\mathbf{v} = (R_y - Q_z)\mathbf{i} + (P_z - R_x)\mathbf{j} + (Q_x - P_y)\mathbf{k}$
Step 2: Check the $\mathbf{i}$-component of $\operatorname{curl}(\operatorname{curl}\,\mathbf{v})$.
The $\mathbf{i}$-component of $\operatorname{curl}(\operatorname{curl}\,\mathbf{v})$ is:
$(Q_x - P_y)_y - (P_z - R_x)_z = Q_{xy} - P_{yy} - P_{zz} + R_{xz}$
Meanwhile, $[\operatorname{grad}(\operatorname{div}\,\mathbf{v})]_x = (P_x + Q_y + R_z)_x = P_{xx} + Q_{yx} + R_{zx}$, and $(\nabla^2\mathbf{v})_x = P_{xx} + P_{yy} + P_{zz}$.
So $[\operatorname{grad}(\operatorname{div}\,\mathbf{v}) - \nabla^2\mathbf{v}]_x = Q_{yx} + R_{zx} - P_{yy} - P_{zz} = Q_{xy} - P_{yy} - P_{zz} + R_{xz}$. ✓
Step 3: Complete by symmetry.
By the same calculation pattern, the $\mathbf{j}$ and $\mathbf{k}$ components match similarly, completing the proof. $\square$