Components of Zero Vector Quantity are Zero

Theorem

Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.

Let $\mathbf r$ be expressed in terms of its components:

$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$

Let $\mathbf r$ be the zero vector.

Then:

$x = y = z = 0$

Proof

By definition of the zero vector, the magnitude of $\mathbf r$ is equal to zero.

By Magnitude of Vector Quantity in terms of Components:

$\size {\mathbf r} = \sqrt {x^2 + y^2 + z^2} = 0$

where $\size {\mathbf r}$ denotes the magnitude of $\mathbf r$.

As each of $x$, $y$ and $z$ are real numbers, each of $x^2$, $y^2$ and $z^2$ is non-negative.

so in order for $\sqrt {x^2 + y^2 + z^2} = 0$, it must follow that each of $x$, $y$ and $z$ is zero.

Hence the result.

$\blacksquare$

Sources

• 1970: George Arfken: Mathematical Methods for Physicists (2nd ed.) ... (previous) ... (next): Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach