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A point $(x,y,z)$ moves parallel to $x$-axis. Which of the three variables $x$, $y$, $z$ remains fixed?
A. $x$
B. $y$ and $z$
C. $x$ and $y$
D. $z$ and $x$

Answer
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Hint: A line parallel to the $x$-axis has an equation of the type $y = b$ , $z = c$ , which intersects the $y - z$ axis at the position $(0,b,c)$ . These lines are parallel to the $x$-axis and parallel to it at a distance of $b$, $c$ units. Since it is parallel to the $x$-axis, the slope of the equation for the line is zero.

Formula Used: Direction cosines can be expressed as:
$l = \cos \alpha $ , $m = \cos \beta $ and $n = \cos \gamma $ .

Complete step-by-step solution:
Firstly, we would find the direction cosines of a line parallel to the $x$-axis. As we are familiar with the concept of direction cosines and direction ratios. The cosine of the angles that a line is making with the $x$-axis, $y$-axis and $z$-axis i.e., $\cos \alpha $ , $\cos \beta $ and $\cos \gamma $ , respectively.


As the line is parallel to the $x$-axis, it will make angle $0^\circ $ with the $x$-axis. Similarly, it will form an angle of $90^\circ $ with the other two axes i.e., $y$-axis and $z$-axis.
So, we have $\alpha = 0^\circ $, $\beta = \gamma = 90^\circ $
Hence, direction cosines are
$l = \cos 0^\circ = 1$
$m = n = \cos 90^\circ = 0$
So, we will get the equation of $x$-axis
$y = 0$ and $z = 0$ . Hence, $y$ and $z$ remain fixed.
So, the correct option will be B.

Note: As in this question, where we were required to determine the equation of a line that was parallel to $x$-axis and passes through the point $(x,y,z)$ , it is obvious that the only axes we will be considering are $y$ and $z$ . A straight line that is parallel to the $x$-axis will always have a slope of zero since there is no slope for a line that is parallel to an axis.