Let F––(t)=t2i–+tj–+k––𝐹_(𝑡)=𝑡2𝑖_+𝑡𝑗_+𝑘_ and G––(t)=i–+tj–+t2k––𝐺_(𝑡)=𝑖_+𝑡𝑗_+𝑡2𝑘_ be two vector functions defined for each t∈R𝑡∈𝑅.

To analyze the vector functions $\mathbf{F}(t)$ and $\mathbf{G}(t)$, we’ll break down their components and explore their properties, such as derivatives, lengths, and potential intersections.

1. Break Down the Problem

• Identify the vector functions:
  1. $\mathbf{F}(t) = t^2 \mathbf{i} + t \mathbf{j} + \mathbf{k}$
  2. $\mathbf{G}(t) = \mathbf{i} + t \mathbf{j} + t^2 \mathbf{k}$

2. Relevant Concepts

• Each vector function has three components in the Cartesian coordinates: $x, y, z$.
• We can analyze the geometric representation and possibly find the intersection or other relevant properties.

3. Analysis and Detail

1. Components of $\mathbf{F}(t)$:

   • $x_F = t^2$
   • $y_F = t$
   • $z_F = 1$

2. Components of $\mathbf{G}(t)$:

   • $x_G = 1$
   • $y_G = t$
   • $z_G = t^2$

3. Finding Intersection Points: Set $\mathbf{F}(t_1) = \mathbf{G}(t_2)$: $t_1^2 = 1 \quad (1)$ $t_1 = t_2 \quad (2)$ $1 = t_2^2 \quad (3)$

From equation (1), we solve $t_1^2 = 1$:

• $t_1 = 1$ or $t_1 = -1$.

Using equation (2):

• If $t_1 = 1$, then $t_2 = 1$.
• If $t_1 = -1$, then $t_2 = -1$.

From equation (3), both cases check out:

• $t_2 = 1$ gives $z = 1$ and $t_2 = -1$ gives $z = 1$.

4. Verify and Summarize

Thus, the intersections or potential coinciding points occur at $(1, 1, 1)$ and $(-1, -1, 1)$.

Final Answer

The intersection points of the vector functions $\mathbf{F}(t)$ and $\mathbf{G}(t)$ are:

• $(1, 1, 1)$
• $(-1, -1, 1)$