Let  F––(t)=2ti–−5j–+t2k––𝐹_(𝑡)=2𝑡𝑖_−5𝑗_+𝑡2𝑘_ and G––(t)=(1−t)i–+1tk––𝐺_(𝑡)=(1−𝑡)𝑖_+1𝑡𝑘_ where t∈R+𝑡∈𝑅+.

Break Down the Problem

1. We have two vector functions $\mathbf{F}(t)$ and $\mathbf{G}(t)$.
2. The functions are defined as follows:
   • $\mathbf{F}(t) = 2t \mathbf{i} - 5 \mathbf{j} + t^2 \mathbf{k}$
   • $\mathbf{G}(t) = (1-t) \mathbf{i} + 1t \mathbf{k}$
3. The goal is to analyze these vector functions, their relationship, or other properties as required.

Relevant Concepts

1. Vector Functions: Each function describes a vector in terms of a parameter $t$.
2. Addition and Subtraction of Vectors: Understanding how to manipulate vector functions is crucial.
3. Limits and Continuity: We need to consider the implications of $t$ being in $\mathbb{R}^+$ (positive real numbers).

Analysis and Detail

1. Expressing the Vectors:

   • $\mathbf{F}(t) = (2t, -5, t^2)$ in coordinate form.
   • $\mathbf{G}(t) = (1-t, 0, t)$ in coordinate form.

2. Finding Relationships:

   • We can set $\mathbf{F}(t)$ equal to $\mathbf{G}(t)$ to find common points, if any: $(2t, -5, t^2) = (1-t, 0, t)$
   • This gives us three equations, one for each component:
     1. $2t = 1 - t$
     2. $-5 = 0$ (which has no solution)
     3. $t^2 = t$

Verify and Summarize

1. From the Equations:
   • The first equation $2t + t = 1$ leads to $3t = 1$ or $t = \frac{1}{3}$.
   • The second equation is a contradiction, indicating that there are no points where these vectors are equal.
   • The third equation $t^2 = t$ gives $t(t-1) = 0$, implying $t = 0$ or $t = 1$. Since $t$ is restricted to positive reals, we take $t = 1$.

Final Answer

Since $\mathbf{F}(t)$ and $\mathbf{G}(t)$ have no common points for $t \in \mathbb{R}^+$, there are no intersections or equal points in the defined range. The first component yields one valid solution, but the second component shows a contradiction, indicating no equality exists for positive $t$.