The line $L_1$ that goes through the point $A(4,3,-2)$ and its parallel to the line $(x=1+3t, y=2-4t, z= 3-t)$, if $P(m,n,-5)$ belongs to $L_1$, determine the values for $m$ and $n$
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2026-04-07 09:18:31.1775553511
Finding point in two parallel lines in 3d?
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A line can be specified by two things:
-a direction
-a point that the line passes through
we want to separate the line you've been given into these parts, and then pick out the directional part. the line is given by
$$(1+3t,2-4t,3-t)$$
and we get various $(x,y,z)$ points one the line by substituting various values for $t$. But since vectors add we can separate this into two parts:
$$(1+3t,2-4t,3-t)=(1,2,3)+(3t,-4t,-t)=(1,2,3)+(3,-4,-1)t$$
Now look at what this last description says: to construct the line, start out at $(1,2,3)$ and then go various distances (i.e., $t$ values) in the direction $(3,-4,-1)$. So the line hits $(1,2,3)$ and then travels in the direction $(3,-4,-1)$. This is the part we want. For our line to be parallel to this one, it must also travel in this direction. So our new line is of the form
$$(4,3,-2)+(3,-4,-1)t$$
It has the same direction (i.e. is parallel), but passes through a different starting point. What you want to find out is at what point this line hits a $z$ value of $-5$. The equation for the $z$ coordinates is simply
$$z=-2-t$$
For some $t$ value, $z=-5$. From the equation above we see that this happens when $t=3$. When $t=3$ the equation of the line becomes
$$(4,3,-2)+3*(3,-4,-1)$$ $$(4,3,-2)+(9,-12,-3)$$ $$(13,-9,-5)$$
And so we see that $m=13$ and $n=-9$.
The overall method here is:
-find the equation of the line we're being asked about. Here we used the fact that it was parallel to some other line and passed through the given point, which is all we need to know.
-we now have the line in terms of a parameter $t$, but want information about its coordinates. Use the equations to solve for the appropriate $t$ value.
-find out where the line is at that particular $t$ value and read off the wanted information.