# At what points does the helix r(t)=<\sin t, \cos t, t> intersect the spher

At what points does the helix $$\displaystyle{r}{\left({t}\right)}={<}{\sin{{t}}},{\cos{{t}}},{t}{>}$$intersect the sphere $$\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={5}$$

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Maciej Morrow
Step 1
We have to find points where helix $$\displaystyle{r}{\left({x}\right)}={\left\langle{\sin{{t}}},{\cos{{t}}}\right\rangle}$$ intersects the sphere $$\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={5}$$
Take each vector component from the given vector and plug it into the sphere's equation ($$\displaystyle{x}=\text{vector's }\ {x}-\text{component},{y}=\text{vector's }\ ,{y}-\text{component},{z}=\text{vector's }\ ,{z}-\text{component}$$)
In this way, we get,
$$\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={5}$$
$$\displaystyle{{\sin}^{{2}}{t}}+{{\cos}^{{2}}{t}}+{t}^{{2}}={5}$$
$$\displaystyle{\left({{\sin}^{{2}}{t}}+{{\cos}^{{2}}{t}}\right)}+{t}^{{2}}={5}\ \ \ \ {\left(\text{apply rule }\ {{\sin}^{{2}}\theta}+{{\cos}^{{2}}\theta}={1}\right)}$$
$$\displaystyle{1}+{t}^{{2}}={5}$$
$$\displaystyle{1}+{t}^{{2}}-{1}={5}-{1}$$
$$\displaystyle{t}^{{2}}={4}$$
$$\displaystyle{t}^{{2}}={\left(\pm{2}\right)}^{{2}}$$
$$\displaystyle{t}=\pm{2}$$
Step 2
$$\displaystyle\text{Using the values for t, plug in t for each of the vector's components.}$$
$$\displaystyle\text{Due to the }\ \pm\ \text{ in the solution t, there will be two sets of intersecting points.}$$
For t=2, we have
Set 1:$$\displaystyle{<}{\sin{{\left({2}\right)}}},{\cos{{\left({2}\right)}}},{2}{>}$$
Set 1:$$\displaystyle{\left({0.909},{0.416},{2}\right)}$$
For t=-2, we have,
Set 2:$$\displaystyle{<}{\sin{{\left(-{2}\right)}}},{\cos{{\left(-{2}\right)}}},-{2}{>}$$
Set 2:$$\displaystyle{\left(-{0.909},-{0.416},-{2}\right)}$$
Result
Set 1:$$\displaystyle{\left({0.909},{0.416},{2}\right)}$$
Set 2:$$\displaystyle{\left(-{0.909},-{0.416},-{2}\right)}$$
These two sets represent the intersection points of helix $$\displaystyle{r}{\left({x}\right)}={\left\langle{\sin{{t}}},{\cos{{t}}},{t}\right\rangle}$$ and sphere
$$\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={1}$$