prove that $\overrightarrow{F(t)} = (\cos t)\hat{a} + (\sin t)\hat{b}$ is a unit circle

Question

Let $\hat{a},\hat{b} \in \mathbb{R^3}$ be two perpendicular unit vectors. If $\overrightarrow{F(t)} = (\cos t)\hat{a} + (\sin t)\hat{b}$ with $t \in [0 ,2\pi]$. How can I prove that this vector function is a unit circle in $\mathbb{R^3}$? I tried a to define a plane in $U$ such that $\hat{a},\hat{b} \in U$ and then define a basis for every point in that plane as $\beta = \{\hat{a},\hat{b},\hat{a} \times \hat{b} \}$ and writing $\overrightarrow{F}$ in terms of that basis. But this things I learned in Linear Algebra and this Is a Calculus II class, I'm no allowed to use things from other classes.

Answer 1

We have that \begin{align*} \lvert \mathbf{F}(t) \rvert^2 &= \mathbf{F}(t) \cdot \mathbf{F}(t) = \cos^2 t \; \mathbf{a} \cdot \mathbf{a} + 2 \cos t \sin t \; \mathbf{a} \cdot \mathbf{b} + \sin^2t \;\mathbf{b} \cdot \mathbf{b} \\ &= \cos^2t + \sin^2t \\ &= 1 \end{align*} where $\mathbf{a} \cdot \mathbf{a} = 1$ and similarly for $\mathbf{b}$ since they are unit vectors, and $\mathbf{a} \cdot \mathbf{b} = 0$ since they are perpendicular.

So $\mathbf{F}(t)$ always lies on the unit circle. Note also that $$\mathbf{F}(t) \cdot \mathbf{a} = \cos t, \quad \mathbf{F}(t) \cdot \mathbf{b} = \sin t$$ so, we see that the projections of $\mathbf{F}(t)$ onto the perpendicular directions $\mathbf{a}$ and $\mathbf{b}$ vary from $-1$ to $1$, which (perhaps by drawing a diagram with $\mathbf{a}$ and $\mathbf{b}$ as axes in the centre of a circle) suffices to conclude that every point on the circle is visited by $\mathbf{F}(t)$.

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As you've found, really $\mathbf{a}$ and $\mathbf{b}$ are a good basis for your subspace in which this defines a unit circle, since with respect to that basis, your components are $(\cos t, \sin t)$.