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I have a question

Find the distance between the origin and the line x = 3t-1, y = 2-t, z = t.

I know: You find a line perpendicular to the line, and passing through the origin. Check. You then find the intersection between the given line and your line, but I cannot find one. I came up with 2 = 0. Can somebody help me?

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  • $\begingroup$ The line I made was x = 2t, y = 3t, z = -3t. No intersection was found. $\endgroup$
    – Human
    Dec 7 '19 at 15:02
  • $\begingroup$ I rejected the edit. It was not a relevant tag $\endgroup$
    – Human
    Dec 7 '19 at 15:10
  • $\begingroup$ Where’s the plane that you mention in the title? $\endgroup$
    – amd
    Dec 7 '19 at 20:43
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You will need a vector $$\vec{OS}\cdot \vec{a}=0$$ where $\vec{a}$ is the direction vector of the given line. We have $$[x_s,y_s,z_s]\cdot[3,-1,1]=3x_s-y_s+z_s=0$$, the point $(x_s,y_s,z_s)$ is situated on the line so $$3(-1+3t)-2+t+t=0$$ from here you will get $t$

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Consider a vector that is parallel to this line and a vector to a point on the line $(x_0,y_0,z_0)=(2t,3t,-3t)$. Now consider the dot product of these vectors.

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In case you know that the area of the parallelogram defined by two vectors $a$ and $b$ is given by $$\sqrt{\langle a,a\rangle\langle b,b\rangle-\langle a,b\rangle^2}$$ try the following: Define $a=(-1,2,0)$ and $b=(3,-1,1)$. The distance of the line to the origin is the area of the parallelogram given by $a$ and unit vector in direction $b$. Make a picture to see why.

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By definition, the lines must intersect for them to be perpendicular. Just as is the case in two dimensions, there’s only one perpendicular to a line through a given point. On the other hand, there’s an infinite number of vectors perpendicular to the line’s direction, so you can’t simply pick a convenient one (which is what it appears that you did) and expect to get that unique perpendicular line. Most choices, including the one you made, will result in skew lines—no intersection at all.

You have to ensure that the lines intersect. A straightforward way to do that is to find a point on the original line such that its position vector is perpendicular to the line’s direction. In other words, solve $(3,-1,1)\cdot(3t-1,2-t,t)=0$ for $t$. This also happens to be the intersection point that you’re looking for in the first place, so you’re done.

That equation has another geometric interpretation, by the way. The plane $3x-y+z=0$ is perpendicular to the given line. Their intersection is exactly the point that you’re looking for. If you substitute the coordinates of a generic point on your line into this equation, you get exactly the same equation in $t$ that we have in the previous paragraph.

Incidentally, a direction vector for the perpendicular can be computed directly, without solving any equations. Intersecting lines are coplanar, so the perpendicular we’re looking for must also be perpendicular to the normal of the plane defined by the origin and the original line. Take any two points $p_1$ and $p_2$ on the line, say, for $t=0$ and $t=1$; a direction vector of the perpendicular is then $(3,-1,1)\times(p_1\times p_2)$. This still leaves you with having to compute the lines’ intersection, so for this problem, the solution described earlier is better.

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