Question

All lines are in the \((x, y)\) plane. Write, in parametric form, the equation of the straight line that is perpendicular to \(\mathbf{r}=(2 \mathbf{i}+4 \mathbf{j})+(\mathbf{i}-2 \mathbf{j}) t\) and goes through (1,0).

Short answer

The parametric equations are \(x = 1 + 2t\) and \(y = 0 + t\).

Step-by-step solution

- Identify the Direction Vector

The given line is described by the vector equation \(\mathbf{r} = (2\mathbf{i} + 4\mathbf{j}) + (\mathbf{i} - 2\mathbf{j})t.\) The direction vector of this line is given by the term multiplied by \(t\), which is \((\mathbf{i} - 2\mathbf{j}).\)

- Find the Perpendicular Direction Vector

A vector perpendicular to \((1, -2)\) can be found by swapping the components and changing the sign of one of them. Therefore, a perpendicular direction vector is \((2, 1)\).

- Write the Parametric Equation

Since the perpendicular line passes through the point (1,0) and has a direction vector \((2, 1)\), the parametric equations can be written as follows: \(x = 1 + 2t, \) and \(y = 0 + t.\)

Key concepts

direction vector

In vector calculus, the direction vector plays a crucial role in describing the orientation of a line. For any given line, the direction vector gives the direction in which the line extends.

The direction vector can be found in the vector equation of a line. For example, consider the line described by \(\textbf{r} = (2\textbf{i} + 4\textbf{j}) + (\textbf{i} - 2\textbf{j})t\).

Here:

• The term multiplied by \t\ is the direction vector. This is \(\textbf{i} - 2\textbf{j}\).
• In component form, this vector is \(1\textbf{i} - 2\textbf{j}\) or, more simply, \(1, -2\).

This vector dictates the direction in which the line moves as \t\ varies. Recognizing and correctly using the direction vector is a key step in many vector calculus problems.

parametric equations

Parametric equations are a powerful way to express the coordinates of the points on a line or curve. They are particularly useful in vector calculus.

To establish parametric equations for a line, we need:

• A direction vector
• A point through which the line passes.

Let's write the parametric equations for a line passing through \((1,0)\) with a direction vector \((2,1)\).

The parametric equations will be:

• \(x = 1 + 2t\)
• \(y = 0 + t\)

Here, \t\ is a parameter that can take any real number value, allowing us to locate any point on the line by plugging in different values of \t\.

perpendicular vectors

Understanding perpendicular vectors is fundamental in vector calculus. Two vectors are perpendicular if their dot product equals zero.

To find a vector perpendicular to a given vector, we can use a simple method:

• Swap the components.
• Change the sign of one component.

For a vector \(1, -2\), a perpendicular vector can be found by:

• Swapping the components to get \(-2, 1\).
• Changing the sign of one component to get \(2, 1\).

This perpendicular vector \((2, 1)\) can now be used in writing parametric equations for new lines, ensuring they are perpendicular to the original line.