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Chapter 14: Problem 1227

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

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

Step by step solution

01

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.
02

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., \((A, B)\): \(\begin{bmatrix}1 \\ \sqrt{3}\end{bmatrix}\)
03

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: \(\cos(\alpha) = \frac{1}{2}\) \(\alpha = \frac{\pi}{3}\)
04

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 \(y_0 = 0\). \(p = \frac{|(1)(0) + (\sqrt{3})(0) - 4|}{2} = \frac{4}{2} = 2\)
05

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): \(x \cos(\pi / 3) + y \sin(\pi / 3) = 2\)

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Most popular questions from this chapter

If the length of perpendicular drawn from \((5,0)\) on \(\mathrm{kx}+4 \mathrm{y}=20\) is 1, then \(\mathrm{k}=\ldots \ldots \ldots\) (a) \(3,(16 / 3)\) (b) \(3,-(16 / 3)\) (c) \(-3,(16 / 3)\) (d) \(-3,-(16 / 3)\)

In triangle \(\mathrm{ABC}\), equation of right bisectors of the sides \(\underline{A B}\) and \(\underline{A C}\) are \(x+y=0\) and \(y-x=0\) respectively. If \(\mathrm{A}=(5,7)\) then equation of side \(\mathrm{BC}\) is (a) \(7 \mathrm{y}=5 \mathrm{x}\) (b) \(5 x=y\) (c) \(5 \mathrm{y}=7 \mathrm{x}\) (d) \(5 \mathrm{y}=\mathrm{x}\)

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If the point \([1+(t / \sqrt{2}), 2+(t / \sqrt{2})]\) lies between the two parallel lines \(\mathrm{x}+2 \mathrm{y}=1\) and \(2 \mathrm{x}+4 \mathrm{y}=15\), then the range of \(\mathrm{t}\) is \(\ldots \ldots\) (a) \(0<\mathrm{t}<[5 /(6 \sqrt{2})]\) (b) \(-[(4 \sqrt{2}) / 3]<\mathrm{t}<0\) (c) \(-[(4 \sqrt{2}) / 3]<\mathrm{t}<[(5 \sqrt{2}) / 6]\) (d) None of these
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