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

\(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={1}\)

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

\(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={1}\)