Find the equation of a line that satisfies the conditions:

  1. Perpendicular to the $X$-axis
  2. The same $X$-intercept as the plane $x-y+2z-3=0$
  3. $45°$ angle with the $Y$-axis

Attempting the problem:

To be perpendicular to the $X$ axis the Direction vector of the line $d = (a,b,c)$ must be $d = (0,b,c)$

The X-intercept of the plane $x-y+2z-3 = 0$ is $x = 3$, $X = 3$ is a condition of the line.

$l = \left[ \begin{array}{c} x = 3\\ y = ?\\ z = ? \end{array} \right] $

For a $45°$ angle with $(0,1,0):$ $$(0,b,c) \cdot (0,1,0) = \sqrt{1} \cdot\sqrt{b^2 + c^2} \cdot \frac{\sqrt{2}}{2}$$ $$2b = \sqrt{2(b^2+c^2)}$$ $$2b^2 = b^2 + c^2$$ $$b = c$$

$l = (3,?,?) + m(0,b,b) $

Not sure how to complete the question.


1 Answer 1


When you got that the x-intercept of the line is $3$, it already means that when $x=3$, $y=z=0$. So your '?' in $(3, ?,?)$ are both $0$.

For another part, you got the direction ratios to be $(0,b,b)$. You just divide them by $b$, to get DR's to be $(0,1,1)$

  • $\begingroup$ Divide what by b? $\endgroup$
    – lakada
    Jun 16, 2017 at 22:30
  • $\begingroup$ @lakada The direction ratios. If you got vector parallel to line to be $0 i +bj+bk$ ($i,j$ and $k$ represent unit vector around the $x,y$ and $z-$ axis respectively) but since we only need to deal with the direction, we can divide the vector by any non-zero constant. This won't alter the direction of the vector.In other words, you can write the line to be $$L=3i+0j+0k+m(0i+bj+bk)=3i+mb(j+k)$$ Now let $mb=t$, which also is a parameter, so you get your line to be $$L=3i+t(j+k)$$ $\endgroup$ Jun 16, 2017 at 23:23

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