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Find the equation of a line $n$ passing through point $T(2,3,1)$ which is also perpendicular to a line given by equation $$p:\frac{x+1}{2}=\frac{y}{-1}=\frac{z+2}{1}$$

I did this using two methods both giving me the same solution and both differing from the solution that's apparently correct, which is $$n:\frac{x-2}{-3}=\frac{y-3}{-3}=\frac{z-1}{1}$$

My solutions:
$1$. method
We can find orthogonal projection of a point $T$ on a line $p$ by finding a intersection point between some plane $\alpha$, such that it contains point $T$ and is perpendicular to line $p$, and line $p$. We know, from the fact that $p$ is perpendicular to $\alpha$, that $\vec{n_\alpha}=(2,-1,1)$. After plugging in point $T$, we get: $$\alpha: 2x-y+z-2=0$$ Writing line $p$ in its parametric form, there is: $$x=2t-1$$ $$y=-t$$ $$z=t-2$$ where $t\in\mathbb{R}$. So we from the the system of equation of $\alpha$ and $p$ in the written form we can find that $t=1$. From whis follows point of intersection and also orthogonal projection $S(1,-1,-1)$ After putting the points $T$ and $S$ in the equation of line between two points, the result I got is: $$n:\frac{x-1}{1}=\frac{y+1}{4}=\frac{z+1}{2}$$ And obviously the vector of direction of aforementioned solution is not colinear to the solution I got.

$2$. method
Let $$n:\frac{x-2}{a}=\frac{y-3}{b}=\frac{z-1}{c}$$ There need to be 2 condition satisfied. First that $\vec{n}\cdot\vec{p}=0$, where $\vec{n}$ and $\vec{p}$ are vector of directions of $n$ and $p$, respectively. And second that these two lines are perpendicular to each other.
From first condition we can find the equation $(a,b,c)\cdot(2,-1,1)=2a-b+c=0$.
For the second condition we can take the point $M(-1,0,-1)$ from the line $p$ and then we get the equation: $$\begin{vmatrix} -3 & -3 & -3\\ a & b & c\\ 2 & -1 & 1 \end{vmatrix}=6a+3b-9c=0 $$

After solving this system of two equation we get $(a,4a,2a)$ from which we can set $\vec{n}=(1,4,2)$. This gives the solution: $$n:\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-1}{2}$$ Which is indeed the same as with $1$. method just at a different point.

If I'm not mistaken this method is fine, so is it possible that my solution is the right one, or did I make some mistake in the calcution?

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  • $\begingroup$ Shouldn't line $p$ in its parametric form have $y=-t$ rather than $y=t+1$? $\endgroup$ Commented May 18, 2023 at 21:43
  • $\begingroup$ Did you do a sanity check on your solution before posting? It seems that your line neither contains $(2,3,1)$ nor is $\perp$ to the line $p.$ $\endgroup$ Commented May 18, 2023 at 21:45
  • $\begingroup$ @AnneBauval I did and looks like I did it wrong, somehow I was sure that $\vec{n}\cdot\vec{p}=0$. But I'll check how the colution changes based on the previously commented $y=-t$. Bunch of mistakes I made $\endgroup$
    – bb_823
    Commented May 18, 2023 at 21:49
  • $\begingroup$ @AnneBauval $n$ would still contain $T$ even if I did make the mistake previously cause $T$ is in the calculation, but then again I also made a mistake in the substitution in the end. $\endgroup$
    – bb_823
    Commented May 18, 2023 at 21:55
  • $\begingroup$ @J.W.Tanner I'll change my question using the correct form, cause it still doesn't yield the result I'm looking for $\endgroup$
    – bb_823
    Commented May 18, 2023 at 21:56

2 Answers 2

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The given line is

$ \dfrac{x+1}{2} = \dfrac{y}{-1} = \dfrac{z+2}{1} = t$

so the parametric equation of the given line is

$(x, y, z) = (-1, 0, -2) + t (2, -1, 1) $

The required line passes through $T = (2,3,1)$ and is perpendicular to the given line. Let the point of intersection between the two lines be

$ P = (-1, 0, -2) + t (2, -1, 1) $

Then we want $TP \perp (2, -1, 1) $ , i.e.

$( (-3, -3, -3) + t (2, -1, 1) ) \cdot (2, -1, 1) = 0 $

This gives us

$ -6 + t (6) = 0 $

so $ t= 1 $ and $P = (1,-1,-1) $

So now the required line is just the line $TP$, the direction vector of which is

$d = P - T = (-1, -4, -2) $

or $d = (1,4, 2) $

Therefore, the equation of the line is

$ \dfrac{ x - 2 }{1} = \dfrac{ y - 3}{4} = \dfrac{z - 1}{2} $

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The direct method is this: A perpendicular dropped from point $M_0(x_0, y_0, z_0)$ , on to a straight line $L_1$ with following equation:

$\frac {x-x_1}{l_1}=\frac {y-y_1}{m_1}=\frac {z-z_1}{n_1}$

is given by:

$\begin{cases}l_1(x-x_0)+m_1(y-y_0)+n_1(z-z_0)=0\\\begin {vmatrix}x-x_0 &y-y_0 & z-z_0\\x_1-x_0 & y_1-y_0 & z_1-z_0\\l_1 & m_1 & n_1 \end{vmatrix}=0\end {cases}$

We have:

$(x_0. y_0, z_0)=(2, 3, 1)$

$(x_1, y_1, z_1)=(-1, 0, -2)$

$(l_1, m_1, n_1)=( 2, -1 ,1)$

Putting values we finally get a system of two equation of two planes which is one method of representing of a line as intersection of these planes. Generally if a line is represented by following equations:

$\begin{cases}A_1x+B_1y+C_1z=-D_1\\A_2x+B_y+C_z=-D_2\end{cases}$

then we have:

$l=\begin{vmatrix}B_1 & C_1\\B_2 & C_2\end{vmatrix}$

$m=\begin{vmatrix}C_1 & A_1\\C_2 & A_2\end{vmatrix}$

$n=\begin{vmatrix}A_1 & B_1\\A_2 & B_2\end{vmatrix}$

and the equation of perpendicular line is:

$\frac{x-x_0}l=\frac{y-y_0}m=\frac{z-z_0} n$

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