Equation of the line passing through $(3,-2,-5)$ and $(3,-2,6)$
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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$\begingroup$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 $z_1 \ne z_2$.
$\endgroup$ 6 $\begingroup$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}
$\endgroup$ $\begingroup$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).
$\endgroup$ $\begingroup$By your work $\vec{(0,0,1)}$ is parallel to the line, which gives the answer: $$(x,y,z)=(3,-2,-5)+t(0,0,1).$$
$\endgroup$ $\begingroup$You have $$x=3, y=-2$$
For $z$ may use parametric form $$z=-5+11t$$
Thus your equation is $$r(t) = (3,-2,-5+11t)$$
$\endgroup$ $\begingroup$To make sense of the Division by 0, we have to first see the derivation of this formula to understand what the denominator actually means in this formula.
Lets say, we need to write the function for the line $l$ and we have been given that the line passes through a fixed point $A$ and is parallel to vector $\vec B$ .
Now, let $\vec a = \vec {OA}$
Put the tail of $\vec b$ on $A$
We have $\vec r = \vec a + \lambda\vec b$ such that the tip of $\vec r$ always lies on line $l$.
If we write our results in cartesian form, where
$\vec r = x \hat{i}+y \hat{j} + z\hat{k}$
$\vec a = x_1 \hat{i}+ y_1 \hat{j} + z_1\hat{k}$
$\vec b = a \hat{i}+b \hat{j} + c\hat{k}$
Here, $(x_1, y_1, z_1)$ is the given point through which $l$ passes and $(a, b, c)$ are direction ratio of $\vec b$.
Putting these in the previously derived equation,
$$ \lambda=\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} $$
This formula can be modified if two points are given since the direction ratio of the line passing through $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ would be just the difference of respective coordinates i.e $(x_2-x_1, y_2-y_1, z_2- z_1)$. So you just replace $(a, b ,c)$ with these coordinates to get your formula.
As you can see here, the denominators are actually the direction ratio of $\vec b$. So when you get
$$\lambda= \boxed{\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}} $$
what it actually means is that direction ratio(or cosine) of $\vec b$ along $X$ and $Y$ axis is $0$.
Remember direction cosine is actually the cosine of the angle made by a vector with the 3 axes.
So in this case, your
$$\boxed{cos(\alpha) = cos(\beta) = 0}$$
i.e$$\boxed{\alpha = \beta = \frac{\pi}{2}}$$
So that would mean your line $l$ would make $90°$ with $X$ and $Y$ axis respectively.
Note that this does not mean that your line would be the $Z$ axis. Your line $l$ could be any line which which is perpendicular to the $XY$ plane.
From the given points you can try to visualise that $l$ will intersect $XY$ plane perpendicularly at point $(3, -2, 0)$
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