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Find the Cartesian equation of the line passing through $(3,-2,-5)$ and $(3,-2,6)$ in $3$D.

The equation of the line through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by $$ \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) $$ where $\vec{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$ and $\vec{b}-\vec{a}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$ $$ \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} $$

For the line through the points $(3,-2,-5)$ and $(3,-2,6)$ is $$\lambda= \boxed{\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}} $$

Is it the correct solution and how do I make sense of the final equation of the line through the given points ?

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  • $\begingroup$ There can't be a single equation for a line in dimension $3$. $\endgroup$ – Bernard Jul 29 '18 at 9:03
  • $\begingroup$ @Bernard Its 2 equations really. $\endgroup$ – Sorfosh Jul 29 '18 at 9:28
  • $\begingroup$ @Sorfosh: Yes, of course, but the O.P. was about the equation… $\endgroup$ – Bernard Jul 29 '18 at 9:30
  • $\begingroup$ ofcause its two equations. but my doubt is in the equation there is 0 in the denominator, and how do I make sense of it ? $\endgroup$ – ss1729 Jul 29 '18 at 9:31
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I don't think we should divide by zero.

Observe the first two coordinates of the points.

They satisfy $x=3$ and $y=-2$.

The line is the intersection of $x=3$ and $y=-2$.

The formula is only used when $x_1 \ne x_2$, $y_1 \ne y_2$, and $z_1 \ne z_2$.

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  • $\begingroup$ thats whr my doubt is. how do we justify dividing by 0 in the equation ? $\endgroup$ – ss1729 Jul 29 '18 at 9:25
  • $\begingroup$ hmm.. we dont' divide by zero. If the values are the same in $x$ coordiante, we know the lines falls on $x=x_0$. $\endgroup$ – Siong Thye Goh Jul 29 '18 at 9:29
  • $\begingroup$ I think the problem is you do not understand where the equation of the line comes from. Your goal is to eliminate the parameter $t$, so you really just do nothing if the constant in front of it is $0$ $\endgroup$ – Sorfosh Jul 29 '18 at 9:30
  • $\begingroup$ This article might be of your interest, notice that they impose the condition that denominator is non-zero. Also, the article later describe what to do when the values are equal in a particular dimension. $\endgroup$ – Siong Thye Goh Jul 29 '18 at 9:49
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    $\begingroup$ Yes, $x=3, y=-2$. Books are written by humans. Not everything that is written is correct. $\endgroup$ – Siong Thye Goh Jul 30 '18 at 3:33
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Set $A=(3,-2,-5)$, $B=(3,-2,6)$. As any point on the line $(AB)$ is a barycenter $\lambda A+(1-lambda)B$ of $A$ and $B$, the coordinates of a point $M$ of this line are obtained by projections on the axes of this relation: \begin{cases} x_M=\lambda\cdot 2+(1-\lambda)2=2, \\ y_M=\lambda(-3)+(1-\lambda)(-3)=-3,\\ z_M=-5\lambda+6(1-\lambda)=6-11\lambda. \end{cases} This is a parametric representation, and we see that $z_M$ can take any real value, so the cartesian equations are: \begin{cases} x=2, \\ y=-3. \end{cases}

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In $2$-D, how would you find the line passing through $(3,-5)$ and $(3,6)$? Note that the formula $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1} \ \ \text{or} \ \ \frac{x-x_1}{y-y_1}=\frac{x_2-x_1}{y_2-y_1}$$ shows the slope of the line. Thus: $$\frac{x-3}{y+5}=\frac{0}{11} \Rightarrow x-3=0 \Rightarrow x=3.$$ Indeed, it is a vertical line $x=3$ ($y$ can be any value).

Similarly, the line passing through $(3,-2,-5)$ and $(3,-2,6)$ is a vertical line $x=3,y=-2$ ($z$ can be any value).

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By your work $\vec{(0,0,1)}$ is parallel to the line, which gives the answer: $$(x,y,z)=(3,-2,-5)+t(0,0,1).$$

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You have $$x=3, y=-2$$

For $z$ may use parametric form $$z=-5+11t$$

Thus your equation is $$r(t) = (3,-2,-5+11t)$$

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