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Find the vector and parametric equation of the line through point $P(1,0,1)$ that meets the line with vector equation $$p = (1,2,0) + (2,-1,2)t$$ at points a distance $3$ from $P_0(1,2,0)$

How am I supposed to find a point a distance $3$ from $P_0(1,2,0)$? I get the equation of a sphere: $9 = (x-1)^2 + (y-2)^2 + z^2$, where $P(x,y,z)$ is any point 3 units away. This gives infinite points, but to get the one on the line $p$ I did $$\|P_0P\| = t\|v\|,\\ 3 = t\sqrt{9},\\ t = 1$$ This gives $P(3,1,2)$ then the other line passing through those two points is $$(3,1,2) = (1,0,1) + (a,b,c)s , \\ (2,1,1) = (a,b,c)s$$ this gives three equations $2 = as$, $1 = bs$, $1 = cs$. What do I do now? How can I find a,b,c?

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  • $\begingroup$ There are two such lines. Step one: find a point on $p$ a distance of $3$ from $P_0(1,2,0)$ (use the distance formula). Step two: find the equation of the line through that point and $P(1,0,1)$. $\endgroup$ – user137731 Oct 18 '15 at 14:55
  • $\begingroup$ How am I supposed to find a point a distance 3 from P_0(1,2,0) ? I get the equation of a sphere: 9 = (x-1)^2 + (y-2)^2 + z^2 , where P(x,y,z) is any point 3 units away. This gives infinite points, but to get the one on the line p I did ||P_0P|| = t||v|| 3 = tsqrt(9) t = 1 This gives P(3,1,2) then the other line passing through those two points is (3,1,2) = (1,0,1) + (a,b,c)s (2,1,1) = (a,b,c)s this gives three equations 2 = as 1 = bs 1 = cs what do i do now? How can I find a,b,c... $\endgroup$ – Denis Kartachov Oct 18 '15 at 15:31
  • $\begingroup$ Distance between $(1,2,0)$ and $(1+2t,2-t,2t)$ $= \sqrt{(1-(1+2t))^2+(2-(2-t))^2 +(0-(2t))^2}=3$. Solve for $t$ then plug back into your line to get the point. $\endgroup$ – user137731 Oct 18 '15 at 15:33
  • $\begingroup$ I edited the question. For the future... things to include: what you know and what you don't understand. Things not to include: "need help", "hope someone can help", "thanks", and so on. $\endgroup$ – user147263 Oct 18 '15 at 16:44
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For a parametric equation of a line $(x,y,z) = (x_0, y_0, z_0) + (a,b,c)s$, the line itself is the set of all points $P(x,y,z)$ such that $x = x_0 + as$, $y = y_0 + bs$, and $z = z_0 + cs$. When $s = 1$, this gives you one point $A$; when $s = 2$, this gives you another point $B$; and when you take all values of $s \in \mathbb{R}$, you get the entire line.

Notice that if instead you had $(x,y,z) = (x_0, y_0, z_0) + (a,b,c)s/2$, you'd still have the same line, except this time, $s$ has to be $2$ to give you point $A$, $s = 4$ gives you point $B$, etc. So when you're trying to find the value of $s$ for any one point, you can just choose any $s$ that you want!

So if $(2,1,1) = (a,b,c)s$, choose any $s$ you want and solve it for $a$, $b$, and $c$.

(By the way, there are two points on the line $(1,2,0) + (2,-1,2)t$ that are $3$ units away from $(1,2,0)$. You found one when you set $t = 1$; what if you set $t = -1$?)

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First find the point(s) of the line $r : p=(1,2,0)+(2,-1,2)t$ whose distance to $P_0$ equals $3$. Now find the equation of the line that goes through $P$ and the point you found in the first part.

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  • $\begingroup$ I know what you mean in the first part, I have no trouble doing that. I get the point (3,1,2) Now I don't know what to do. The other line that intersects with p is q = (1,0,1) + (a,b,c)s i just get three equations 2 = as 1 = bs 1 = cs $\endgroup$ – Denis Kartachov Oct 18 '15 at 15:45
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    $\begingroup$ @DenisKartachov That's what you get when you don't say what you already know about the problem: an answer that only says what you don't need. Answerers are not mind-readers. $\endgroup$ – user147263 Oct 18 '15 at 16:36

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