Question

Create a vector-valued function whose graph matches the given description. The line through points (1,2,3) and \((4,5,6),\) where \(\vec{r}(0)=\langle 1,2,3\rangle\) and \(\vec{r}(1)=\langle 4,5,6\rangle\)

Short answer

\(\vec{r}(t) = \langle 1 + 3t, 2 + 3t, 3 + 3t \rangle\).

Step-by-step solution

Find the Direction Vector

To create a vector-valued function for a line, we need a direction vector. Given the points \((1,2,3)\) and \((4,5,6)\), the direction vector \(\vec{d}\) can be found by subtracting the first point from the second point: \(\vec{d} = \langle 4 - 1, 5 - 2, 6 - 3 \rangle = \langle 3, 3, 3 \rangle\).

Write the Vector-Valued Function

A line can be expressed as a vector-valued function in the form \(\vec{r}(t) = \vec{r}_0 + t \vec{d}\), where \(\vec{r}_0\) is the position vector of a point on the line and \(\vec{d}\) is the direction vector. Given that \(\vec{r}(0) = \langle 1, 2, 3 \rangle\), the function becomes \(\vec{r}(t) = \langle 1, 2, 3 \rangle + t \langle 3, 3, 3 \rangle\).

Simplify the Function

Expand the vector-valued function by incorporating the direction vector: \(\vec{r}(t) = \langle 1 + 3t, 2 + 3t, 3 + 3t \rangle\). This is the simplified form of the vector-valued function representing the line through the given points.

Key concepts

Direction Vector

When dealing with vector-valued functions that describe lines, a key element is the direction vector. To find the direction vector, we take two points which the line passes through—in this case, the points are (1, 2, 3) and (4, 5, 6). We subtract the coordinates of the first point from the second point to find this direction vector. It involves simple subtraction:

• For the x-component: 4 - 1 = 3
• For the y-component: 5 - 2 = 3
• For the z-component: 6 - 3 = 3

Thus, the direction vector is \(\langle 3, 3, 3 \rangle\).
This direction vector essentially dictates the direction in which the line extends in 3D space. It remains constant throughout any point along the line and plays a crucial role in forming the vector-valued function.

Line Through Points

A line through points in 3D can be represented by using a vector-valued function.
Think of it as a path connecting these points through a continuous line in space. In our case, the line starts at point (1, 2, 3) and travels in the direction of (4, 5, 6).
The formula for a line through two points can be written as:\[\vec{r}(t) = \vec{r}_0 + t \vec{d}\]here, \( \vec{r}_0 \) is the position vector of the initial point and \( \vec{d} \) is the direction vector.
This function allows us to easily describe the line graphically and analytically, as it ties together the starting point and the ongoing direction.

Position Vector

A position vector is your starting spot in a vector-valued function for a line. It's crucial because it acts as your 'launch pad' from which everything else is measured along the line.

• For our exercise, the first point given is (1, 2, 3).
• This position can be written as a vector \( \vec{r}_0 = \langle 1, 2, 3 \rangle \).

The position vector \( \vec{r}_0\) is plugged directly into the vector-valued function formula, which is used to calculate any point on the line. This foundation allows for creating an equation that tells us about every other point along the way, based on the shift dictated by the direction vector.

Parametric Representation

Parametric representation is a neat way to express a line using parameters, which in this case is often represented by \(t\). This makes it quite flexible and useful.
In our line equation:\[\vec{r}(t) = \langle 1, 2, 3 \rangle + t \langle 3, 3, 3 \rangle\]Each component of the final vector \(\vec{r}(t)\) is expressed as a function of \(t\), which can vary:

• The x-component is: \(1 + 3t\)
• The y-component is: \(2 + 3t\)
• The z-component is: \(3 + 3t\)

By substituting any value of \(t\), you can find corresponding points on the line.
This format provides an easy way to graph the line, display its direction, and find specific points, making it a powerful tool in calculus and vector analysis.