Given the line \( y = -\frac{2}{3}x + 5 \), find a vector equation and parametric equations describing the line.

Solution

Since the slope is \( -\frac{2}{3} \), we may choose direction vector \( \vec{d} = (3, -2) \).

A point on the line is \( (0, 5) \) (the \( y \)-intercept), and so the vector equation is \( \vec{r} = (0, 5) + t(3, -2),~t \in \mathbb{R}\) and the parametric equations are \(x = 3t\) and \(y = 5 - 2t, ~ t \in \mathbb{R}\).

The above example shows the standard method for finding the vector/parametric equations of a line.

As we saw in the previous module, we may be required to go in the opposite direction to find the Cartesian equation of a line given the vector/parametric equations.

Method 3

Solve each of the parametric equations for \(t\) to get \( t = \dfrac{x}{3} \) and \( t = -\dfrac{y - 5}{2} \).

Setting these equal we get

We can only solve for \( t \) this way if the direction vector, \( \vec{d} \), has non-zero components.

This form of the equation is called the scalar equation of a line in \( \mathbb{R}^2 \) (\(2\)-space).

Given the line \( y = -\frac{2}{3}x + 5 \), find a vector equation and parametric equations describing the line.

Solution

Consider the coefficients of \(x\) and \(y\) in the scalar equation \( \textcolor{Mulberry}{2}x + \textcolor{Violet}{3}y - 15 = 0 \).

Define the vector \( \vec{n} = (\textcolor{Mulberry}{2}, \textcolor{Violet}{3}) \) (formed by taking these highlighted coefficients in the scalar equation).

Graph vector \(\vec{n}=(2,3)\) along with the direction vector for the line, \(\vec{d}=(3,-2)\).

The graph suggests that perhaps the two vectors \( \vec{d} \) and \( \vec{n} \) are orthogonal.

This can be verified by taking the dot product

Hence, the two vectors are indeed orthogonal.

The vector \( \vec{n} \) is known as a normal vector.

Definition

State a normal vector for the line in ${ℝ}^2$ with the following equation :

(Q)*0.0(Q)*1.0(Q)*2.0(Q)*3.0(Q)*4.0(Q)*5.0

Note : Answer in the form $\left(x_0,y_0\right)$

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