MHB Vector Proofs: Constant Magnitude and Orthogonality in Finite Dimensional Spaces

[Dustinsfl]

Prove that if $\mathbf{v}(t)$ is any vector that depends on time but which has a constant magnitude, then $\dot{\mathbf{v}}(t)$ is orthogonal to $\mathbf{v}(t)$.
Prove the converse.

We are working with finite dimensional vector spaces.
Let $\mathbf{v}(t) = \sum\limits_{i = 1}^{n}c_iv(t)_i$.
Then
$$
\lVert\mathbf{v}(t)\rVert = \sqrt{\sum\limits_{i = 1}^{n}c_i^2} = \alpha\in\mathbb{C}.
$$
How do I defined $\dot{\mathbf{v}}(t)$?

 

[Fantini]

Hello dwsmith. (Handshake) You don't need to work out the inner product, there is a simpler way.

Since $\mathbf{v}(t)$ has constant magnitude this means $|\mathbf{v}(t)| = c$, for some $c \in \mathbb{R}^+$. Therefore $|\mathbf{v}(t)|^2 = c^2$, but $|\mathbf{v}(t)|^2 = \mathbf{v}(t) \cdot \mathbf{v}(t) = c^2$.

Deriving both sides yields

$$\mathbf{v}' (t) \cdot \mathbf{v}(t) + \mathbf{v}(t) \cdot \mathbf{v}' (t) = 0,$$

but the left side is simply $2 \mathbf{v}'(t) \cdot \mathbf{v}(t)$. We conclude that $\mathbf{v}' (t) \cdot \mathbf{v} (t) = 0$ and $\mathbf{v}'(t)$ is orthogonal to $\mathbf{v}(t)$.

In order to work the converse you simply reverse operations. Hope this has helped! (Yes)

Fantini