91Ó°ÊÓ

Parametric equations are a powerful mathematical tool used to define lines, curves, and surfaces. They express a set of quantities as explicit functions of one or more independent parameters. In the context of our exercise, parametric equations of a line provide a systematic way to describe points on that line.

Consider the line equation \(\vec{r}=\vec{a} + s\vec{b} \), where \(\vec{r}\) is the position vector, \(\vec{a}\) is a fixed point on the line, \(\vec{b}\) is the direction vector, and \(s\) is a scalar parameter.

• \(\vec{a}\) gives a starting position on the line.
• \(\vec{b}\) defines the direction in which the line extends.

The scalar \(s\) stretches or shrinks the direction vector \(\vec{b}\). This allows any point on the line to be expressed in terms of \(s\). In the exercise, as \(s\) varies, the parametric form helps identify position vectors \(\vec{a}, \vec{b}, \vec{c},\) and \(\vec{d}\) associated with specific values of \(s\).

Position vectors are fundamental in understanding the location of a point in a space relative to an origin. They are vectors pointing from one reference location, commonly the origin \((0, 0)\), to a specific point.

In the exercise, position vectors for points \(A, B, C,\) and \(D\) are derived using the equation \(\vec{r} = (1,2) + s(-2,3)\). Let’s see how:

• Point \(A\) when \(s = 0\): Position vector \(\vec{a} = (1, 2)\).
• Point \(B\) when \(s = 1\): Position vector \(\vec{b} = (-1, 5)\).
• Point \(C\) when \(s = 2\): Position vector \(\vec{c} = (-3, 8)\).
• Point \(D\) when \(s = 3\): Position vector \(\vec{d} = (-5, 11)\).

Each position vector directs us to a specific point on the line, showing how different values of \(s\) translate into different points in the space.

Scalar multiplication in vector calculations involves multiplying a vector by a scalar to adjust its magnitude (length), without changing its direction. This operation is performed component-wise.

The rule for scalar multiplication is straightforward:

• If \(\vec{v} = (x, y)\) and \(k\) is a scalar, \(k\vec{v} = (kx, ky)\).

In our exercise, this concept is used to verify relationships between vectors:

• To show \(\overrightarrow{AC} = 2 \overrightarrow{AB}\), we multiply \(\overrightarrow{AB} = (-2, 3)\) by 2.
• For \(\overrightarrow{AD} = 3 \overrightarrow{AB}\), we multiply the same vector by 3.

By scaling \(\overrightarrow{AB}\), we determine if it fits the transformations required to represent \(\overrightarrow{AC}\) and \(\overrightarrow{AD}\).

Vector subtraction is technique used to find the difference between two vectors. It is an essential operation for calculating vectors between points, which tell us about magnitude and direction shift from one point to another.

The procedure for vector subtraction is:

• Given vectors \(\vec{u} = (x_1, y_1)\) and \(\vec{v} = (x_2, y_2)\), the subtraction \(\vec{u} - \vec{v} = (x_1 - x_2, y_1 - y_2)\).

In our exercise, we used vector subtraction to compute vectors \(\overrightarrow{AB}\), \(\overrightarrow{AC}\), and \(\overrightarrow{AD}\):

• \(\overrightarrow{AB} = \vec{b} - \vec{a} = (-1, 5) - (1, 2) = (-2, 3)\)
• \(\overrightarrow{AC} = \vec{c} - \vec{a} = (-3, 8) - (1, 2) = (-4, 6)\)
• \(\overrightarrow{AD} = \vec{d} - \vec{a} = (-5, 11) - (1, 2) = (-6, 9)\)

This allows us to understand how far and in what direction each point is from another, facilitating the verification of vector relationships mentioned in the problem.