Mixing
Find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d
$begingroup$
Let L1 be the line passing through the point P1 = (−2, 3, 1) with direction vector →d = <−2, −1, −2>, and let L2 be the line passing through the point P2 = (4, 4, −4) with the same direction vector.
Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d.
What I've solved for so far:
L1: <-2,3,1> + t<-2,-1,2>
L2: <4,4-4> + u<-2,-1,2>
Consider a point (-2,3,1) and a point on L2 which is <4-2u, 4-u, -4-2u>
Distance between the points is:
d$^2$ = (-2u+6)$^2$ + (-u+1)$^2$ + (-2u-5)$^2$
I then differentiated both sides to get:
(4u-12) + (u-1) + (4u+10) = 0
u = $frac13$
Shortest distance between the lines is then:
d$^2$ = (-2($frac13$)+6)$^2$ + (-($frac13$)+1)$^2$ + (-2($frac13$)-5)$^2$
d = $sqrt{61}$
So I've gotten d to equal $sqrt{61}$. I don't know where to go from here and solve for Q1 and Q2.
linear-algebra vectors
asked Nov 12 '18 at 3:03
qbufferqbuffer
63
$endgroup$
add a comment |
$begingroup$
Let L1 be the line passing through the point P1 = (−2, 3, 1) with direction vector →d = <−2, −1, −2>, and let L2 be the line passing through the point P2 = (4, 4, −4) with the same direction vector.
Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d.
What I've solved for so far:
L1: <-2,3,1> + t<-2,-1,2>
L2: <4,4-4> + u<-2,-1,2>
Consider a point (-2,3,1) and a point on L2 which is <4-2u, 4-u, -4-2u>
Distance between the points is:
d$^2$ = (-2u+6)$^2$ + (-u+1)$^2$ + (-2u-5)$^2$
I then differentiated both sides to get:
(4u-12) + (u-1) + (4u+10) = 0
u = $frac13$
Shortest distance between the lines is then:
d$^2$ = (-2($frac13$)+6)$^2$ + (-($frac13$)+1)$^2$ + (-2($frac13$)-5)$^2$
d = $sqrt{61}$
So I've gotten d to equal $sqrt{61}$. I don't know where to go from here and solve for Q1 and Q2.
linear-algebra vectors
asked Nov 12 '18 at 3:03
qbufferqbuffer
63
$endgroup$
add a comment |
$begingroup$
Let L1 be the line passing through the point P1 = (−2, 3, 1) with direction vector →d = <−2, −1, −2>, and let L2 be the line passing through the point P2 = (4, 4, −4) with the same direction vector.
Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d.
What I've solved for so far:
L1: <-2,3,1> + t<-2,-1,2>
L2: <4,4-4> + u<-2,-1,2>
Consider a point (-2,3,1) and a point on L2 which is <4-2u, 4-u, -4-2u>
Distance between the points is:
d$^2$ = (-2u+6)$^2$ + (-u+1)$^2$ + (-2u-5)$^2$
I then differentiated both sides to get:
(4u-12) + (u-1) + (4u+10) = 0
u = $frac13$
Shortest distance between the lines is then:
d$^2$ = (-2($frac13$)+6)$^2$ + (-($frac13$)+1)$^2$ + (-2($frac13$)-5)$^2$
d = $sqrt{61}$
So I've gotten d to equal $sqrt{61}$. I don't know where to go from here and solve for Q1 and Q2.
linear-algebra vectors
asked Nov 12 '18 at 3:03
qbufferqbuffer
63
$endgroup$
Let L1 be the line passing through the point P1 = (−2, 3, 1) with direction vector →d = <−2, −1, −2>, and let L2 be the line passing through the point P2 = (4, 4, −4) with the same direction vector.
Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d.
What I've solved for so far:
L1: <-2,3,1> + t<-2,-1,2>
L2: <4,4-4> + u<-2,-1,2>
Consider a point (-2,3,1) and a point on L2 which is <4-2u, 4-u, -4-2u>
Distance between the points is:
d$^2$ = (-2u+6)$^2$ + (-u+1)$^2$ + (-2u-5)$^2$
I then differentiated both sides to get:
(4u-12) + (u-1) + (4u+10) = 0
u = $frac13$
Shortest distance between the lines is then:
d$^2$ = (-2($frac13$)+6)$^2$ + (-($frac13$)+1)$^2$ + (-2($frac13$)-5)$^2$
d = $sqrt{61}$
So I've gotten d to equal $sqrt{61}$. I don't know where to go from here and solve for Q1 and Q2.
linear-algebra vectors
linear-algebra vectors
asked Nov 12 '18 at 3:03
qbufferqbuffer
63
asked Nov 12 '18 at 3:03
qbufferqbuffer
63
asked Nov 12 '18 at 3:03
qbufferqbuffer
63
asked Nov 12 '18 at 3:03
qbufferqbuffer
63
asked Nov 12 '18 at 3:03
qbufferqbuffer
63
63
add a comment |
add a comment |
1 Answer
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$begingroup$
If the lines have the same direction vector, then they must be parallel.
So the plane perpendicular to line $P_1+td = (-2-2t, 3-t, 1-2t)$ will also be perpendicular to the line $P_2+ud = (4-2u, 4-u, -4-2u)$.
Since that plane passes through the point $$Q_1=(-2,3,1)$$ and is perpendicular to the direction $(2,1,2)$. Its equation is $2(x+2) +1(y-3) + 2(z-1) = 0$. Which simplifies to
$$2x+y+2z = -1$$
It will intersect the other line when $(4-2u)+(4-u)+2(-4-2u) = -1$, which is when
$u = dfrac 13$. This gives us the point $$Q_2 = frac 13(10, 11, -14)$$. Then
$$d = left |(-2,3,1) - frac 13(10, 11, -14) right | = sqrt{61}$$
answered Jan 17 at 21:54
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If the lines have the same direction vector, then they must be parallel.
So the plane perpendicular to line $P_1+td = (-2-2t, 3-t, 1-2t)$ will also be perpendicular to the line $P_2+ud = (4-2u, 4-u, -4-2u)$.
Since that plane passes through the point $$Q_1=(-2,3,1)$$ and is perpendicular to the direction $(2,1,2)$. Its equation is $2(x+2) +1(y-3) + 2(z-1) = 0$. Which simplifies to
$$2x+y+2z = -1$$
It will intersect the other line when $(4-2u)+(4-u)+2(-4-2u) = -1$, which is when
$u = dfrac 13$. This gives us the point $$Q_2 = frac 13(10, 11, -14)$$. Then
$$d = left |(-2,3,1) - frac 13(10, 11, -14) right | = sqrt{61}$$
answered Jan 17 at 21:54
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
If the lines have the same direction vector, then they must be parallel.
So the plane perpendicular to line $P_1+td = (-2-2t, 3-t, 1-2t)$ will also be perpendicular to the line $P_2+ud = (4-2u, 4-u, -4-2u)$.
Since that plane passes through the point $$Q_1=(-2,3,1)$$ and is perpendicular to the direction $(2,1,2)$. Its equation is $2(x+2) +1(y-3) + 2(z-1) = 0$. Which simplifies to
$$2x+y+2z = -1$$
It will intersect the other line when $(4-2u)+(4-u)+2(-4-2u) = -1$, which is when
$u = dfrac 13$. This gives us the point $$Q_2 = frac 13(10, 11, -14)$$. Then
$$d = left |(-2,3,1) - frac 13(10, 11, -14) right | = sqrt{61}$$
answered Jan 17 at 21:54
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
If the lines have the same direction vector, then they must be parallel.
So the plane perpendicular to line $P_1+td = (-2-2t, 3-t, 1-2t)$ will also be perpendicular to the line $P_2+ud = (4-2u, 4-u, -4-2u)$.
Since that plane passes through the point $$Q_1=(-2,3,1)$$ and is perpendicular to the direction $(2,1,2)$. Its equation is $2(x+2) +1(y-3) + 2(z-1) = 0$. Which simplifies to
$$2x+y+2z = -1$$
It will intersect the other line when $(4-2u)+(4-u)+2(-4-2u) = -1$, which is when
$u = dfrac 13$. This gives us the point $$Q_2 = frac 13(10, 11, -14)$$. Then
$$d = left |(-2,3,1) - frac 13(10, 11, -14) right | = sqrt{61}$$
answered Jan 17 at 21:54
steven gregorysteven gregory
18.2k32258
$endgroup$
If the lines have the same direction vector, then they must be parallel.
So the plane perpendicular to line $P_1+td = (-2-2t, 3-t, 1-2t)$ will also be perpendicular to the line $P_2+ud = (4-2u, 4-u, -4-2u)$.
Since that plane passes through the point $$Q_1=(-2,3,1)$$ and is perpendicular to the direction $(2,1,2)$. Its equation is $2(x+2) +1(y-3) + 2(z-1) = 0$. Which simplifies to
$$2x+y+2z = -1$$
It will intersect the other line when $(4-2u)+(4-u)+2(-4-2u) = -1$, which is when
$u = dfrac 13$. This gives us the point $$Q_2 = frac 13(10, 11, -14)$$. Then
$$d = left |(-2,3,1) - frac 13(10, 11, -14) right | = sqrt{61}$$
answered Jan 17 at 21:54
steven gregorysteven gregory
18.2k32258
answered Jan 17 at 21:54
steven gregorysteven gregory
18.2k32258
answered Jan 17 at 21:54
steven gregorysteven gregory
18.2k32258
answered Jan 17 at 21:54
steven gregorysteven gregory
18.2k32258
18.2k32258
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add a comment |
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