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I am given a point $A =(-1, 0, 1)$. A straight line is defined as the following system: $\{ y + 3x = 6$ and $ z-2 = 0\}$ . They are one system. This defines a line in 3D space but I don't understand it visually. Let this line be $T$.

Let line $S$ be a line drawn from point $A$ to that line $T$ defined by the given system before so that the angle between them is 90 degrees. Line $S \perp T$. Find the length of $S$ and straight line equation of $T$.

The length will probably be $|\vec AT|$ (line $S$ basicly) calculated by $\sqrt {x^2 + y ^2 + z ^2}$ or atleast so I think. Maybe this clears up a bit.

Help appreciated. Thanks.

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    $\begingroup$ Nothing is clear in your question :-) You don't define whatis point P. As to why there is two lines, you misinterpret the fact that $z=2$ is the equation of a plan, not a line. The intersection of two plans is a line, so the set of 2 equations is defining a line in a 3D space. The rest of your question, to be honest, is not comprehensible to me. Could state very precisely what is the problem at hand, and what are your issues with it, and also what you have tried and where your failed? $\endgroup$ – Martigan Dec 6 '17 at 12:16
  • $\begingroup$ $z=2$ is not a straight line, it is a plane. $\endgroup$ – 5xum Dec 6 '17 at 12:19
  • $\begingroup$ @Martigan I cleared up the question a bit, maybe you could take a look now. Thanks. $\endgroup$ – student126 Dec 6 '17 at 13:33
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Here's an outline. $T$ is the intersection of two planes. A vector $v$ parallel to $T$ lies in both planes. So $v$ is perpendicular to the normal vectors of both planes. Let $n$ and $m$ be the normal vectors to each plane. Then you can take $v$ to be their cross product. Then find any point on the line. $P=(0,6,2)$ will work. Parametric equations of the line are given by $P+vt.$

The length of $S$ (a segment, really) is the distance from $A$ to $T$. Pick any point on the line. The same $P$ will do. Let $w$ be the vector from $P$ to $A$. This forms the hypotenuse of a right triangle. If you project $w$ onto $v$, the length of this projection is one leg of the right triangle. The distance you want is the other leg.

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First you want to understand what the line is (visually). It's nothing but the line $y+3x=6$ in $2d$ in the plane $z=2$. In standard $2d$ coordinate system we assume $z=0$. In this question we simply shifted the line $2$ units in the direction of $z$.

To find the distance $AT$, first we write the equation of the line $T$:

$$\frac{x-0}{1}=\frac{(y-6)}{-3}=\frac{(z-2)}{0}=r$$

Here $r$ is an arbitrary constant.

So, a general point on this line $T$ is $(r, 6-3r, 2)$.

Let's call this point $P$.

Now the line joining this point and $A$ i.e., line $AP$ (same as what you call line $S$ in the question) should be perpendicular to line $T$.

Direction ratios of $AP$ are $(r+1,6-3r,1)$

Direction ratios of line $T$ are $(1, -3, 0)$

Condition of perpendicularity:

$(r+1)1 +(6-3r)(-3) + 1(0) =0$

Thus, $r = 1.7$

Therefore, point $P$ is $(1.7, 0.9, 2)$

Now just apply distance formula between $A$ and $P$

Therefore required distance $AP = sqrt(9.1)$

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