Question

The \(\mathrm{p}-\alpha\) form of the line \(\mathrm{x}+\sqrt{(3) \mathrm{y}-4}=0 \mathrm{is}\) (a) \(x \cos (\pi / 6)+\operatorname{ysin}(\pi / 6)=2\) (b) \(x \cos (\pi / 3)+y \sin (\pi / 3)=2\) (c) \(x \cos [-(\pi / 3)]+y \sin [-(\pi / 3)]=2\) (d) \(x \cos [-(\pi / 6)]+y \sin [-(\pi / 6)]=2\)

Short answer

The short answer is: The correct choice is (b): \(x \cos(\frac{\pi}{3}) + y \sin(\frac{\pi}{3}) = 2\).

Step-by-step solution

Rewrite the given line equation

We shall first rewrite the given line equation into a more standard form: \(x+\sqrt{3}y - 4=0\) The line equation now is in the form of \(Ax + By + C = 0\), where A = 1, B = \(\sqrt{3}\), and C = -4.

Calculate the normal vector to the line

The normal vector to the line can be found by using the coefficients of x and y, i.e., \((A, B)\): \(\begin{bmatrix}1 \\ \sqrt{3}\end{bmatrix}\)

Calculate the angle α

Now we have to calculate the angle between the normal vector and the x-axis. To do that, we'll use the dot product formula: \(\cos(\alpha) = \frac{\begin{bmatrix}1 & \sqrt{3}\end{bmatrix} \cdot \begin{bmatrix}1 \\ 0\end{bmatrix}}{\left\lVert\begin{bmatrix}1 & \sqrt{3}\end{bmatrix}\right\rVert \left\lVert\begin{bmatrix}1 \\ 0\end{bmatrix}\right\rVert},\) where \(\left\lVert\begin{bmatrix}1 & \sqrt{3}\end{bmatrix}\right\rVert = \sqrt{1^2 + (\sqrt{3})^2} = 2\), and \(\left\lVert\begin{bmatrix}1 \\ 0\end{bmatrix}\right\rVert = 1\). Substitute these values and find the angle: \(\cos(\alpha) = \frac{1}{2}\) \(\alpha = \frac{\pi}{3}\)

Calculate the perpendicular distance p

Now let's find the perpendicular distance p from the line to the origin: \(p = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\), where \((x_0, y_0)\) is the origin. In this case, \(x_0 = 0\) and \(y_0 = 0\). \(p = \frac{|(1)(0) + (\sqrt{3})(0) - 4|}{2} = \frac{4}{2} = 2\)

Write the line equation in p-α form

Now we have all the required information to write the line equation in p-α form: \(x \cos(\frac{\pi}{3}) + y \sin(\frac{\pi}{3}) = 2\) Compare the result to the given options: The correct choice is (b): \(x \cos(\pi / 3) + y \sin(\pi / 3) = 2\)

Key concepts

Straight Line Equation

A straight line equation is a way to represent a line in a plane. The most common form is the standard form, which is given by \(Ax + By + C = 0\). Here, \(A\), \(B\), and \(C\) are constants, and \(x\) and \(y\) are variables representing any point on the line.

The equation in this form helps us understand the properties of the line, such as its slope and intercepts. This form is also flexible enough to be converted into other forms like slope-intercept or point-slope forms depending on what's needed for solving problems.

• The coefficient \(A\) determines the horizontal component, while \(B\) is tied to the vertical component of the line.
• \(C\) shifts the line either upward or downward by a constant amount.
• When analyzing problems, we often convert weird equations to this form to make other calculations easier.

Normal Vector Calculation

The normal vector is a fundamental concept when dealing with straight lines in 2D geometry. Simply put, it is a vector that is perpendicular (or 'normal') to the line.

For a line in the form \(Ax + By + C = 0\), the normal vector is indicated by the pair \((A, B)\). This pair represents both the horizontal and vertical components of the normal. Determining this vector is crucial because:

• It enables us to determine the orientation of the line.
• It is used to find perpendicular distances from points to the line.
• The normal vector is used in finding angles between lines.

Understanding the normal vector is significant when rewriting equations or analyzing their properties.

Perpendicular Distance

The perpendicular distance from a point to a line is a measure of how far off the point is from the line in a straight-line, shortest-path manner. In the context of the origin \((0,0)\) and a line \(Ax + By + C = 0\), this distance is calculated as:

\[p = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\\] where \(x_0\) and \(y_0\) are the coordinates of the point, in this case, the origin.

This formula is especially useful when we want to check how close a point is to our line without having to construct perpendicular lines manually. Understanding this concept helps:

• To gauge the relative positioning of lines and points.
• To verify or solve more complex geometric problems involving distances.
• To convert equations into forms like p-α, which relate to real-world applications such as physics simulations.

Angle with X-axis

Calculating the angle a line makes with the x-axis is an important part of understanding the line's orientation in a plane. In mathematics, this angle \(\alpha\) can be found using the properties of the normal vector and the x-axis itself.

When dealing with a line represented in the standard form \(Ax + By + C = 0\), the angle is calculated using:

\[\cos(\alpha) = \frac{A}{\sqrt{A^2 + B^2}}.\\]This equation is derived from the dot product technique, a method that relates vectors and their angles. The result will range from 0 to 180 degrees, describing how the line is angled in a plane. When calculating this angle:

• It helps identify orientation and slope indirectly.
• It's useful for transforming between different equation forms (like converting to p-α form).
• It aids in visualizing geometric arrangements, crucial in fields like computer graphics or design.