Mixing
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Determine point of intersection or find the value of $z$ [closed]
$begingroup$
Let $L_1$ be the line passing through the points $Q_1=(4, −2, −4)$ and $Q_2=(5, −1, −5)$ and let $L_2$ be the line passing through the point $P_1=(−13, −12, 6)$ with direction vector $underline{d}=[6, 3, −6]^T$. Determine whether $L_1$ and $L_2$ intersect. If so, find the point of intersection $Q$. If not, find a value for the $z$-coordinate of $P_1$ so the resulting lines do intersect.
Please tell me what the result is
linear-algebra algebra-precalculus vectors linear-transformations vector-analysis
edited Jan 17 at 20:06
user289143
1,000313
asked Jan 17 at 19:26
DreamVision2017DreamVision2017
318
$endgroup$
closed as off-topic by RRL, Riccardo.Alestra, Adrian Keister, Andrew, Shaun Feb 5 at 15:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Riccardo.Alestra, Adrian Keister, Andrew, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $L_1$ be the line passing through the points $Q_1=(4, −2, −4)$ and $Q_2=(5, −1, −5)$ and let $L_2$ be the line passing through the point $P_1=(−13, −12, 6)$ with direction vector $underline{d}=[6, 3, −6]^T$. Determine whether $L_1$ and $L_2$ intersect. If so, find the point of intersection $Q$. If not, find a value for the $z$-coordinate of $P_1$ so the resulting lines do intersect.
Please tell me what the result is
linear-algebra algebra-precalculus vectors linear-transformations vector-analysis
edited Jan 17 at 20:06
user289143
1,000313
asked Jan 17 at 19:26
DreamVision2017DreamVision2017
318
$endgroup$
closed as off-topic by RRL, Riccardo.Alestra, Adrian Keister, Andrew, Shaun Feb 5 at 15:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Riccardo.Alestra, Adrian Keister, Andrew, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Have you made any effort whatsoever to solve this on your own? Please show your work.
$endgroup$
– amd
Jan 17 at 21:47
$begingroup$
Yes I have, although last time I showed some effort on my question no one even bothered to respond so please don't judge prematurely
$endgroup$
– DreamVision2017
Jan 18 at 18:41
$begingroup$
You haven’t posted a question. You’ve posted a context-free problem statement and then made a demand, however politely. Taken by itself, this is not the sort of “question” that will be well-received on this site. That aside, this one is a minor variation of the exercises that you’ve asked about in previous questions. If you’re having trouble with this subject, you’d be well advised to review and seek help from your instructor/TA.
$endgroup$
– amd
Jan 18 at 21:17
add a comment |
$begingroup$
Let $L_1$ be the line passing through the points $Q_1=(4, −2, −4)$ and $Q_2=(5, −1, −5)$ and let $L_2$ be the line passing through the point $P_1=(−13, −12, 6)$ with direction vector $underline{d}=[6, 3, −6]^T$. Determine whether $L_1$ and $L_2$ intersect. If so, find the point of intersection $Q$. If not, find a value for the $z$-coordinate of $P_1$ so the resulting lines do intersect.
Please tell me what the result is
linear-algebra algebra-precalculus vectors linear-transformations vector-analysis
edited Jan 17 at 20:06
user289143
1,000313
asked Jan 17 at 19:26
DreamVision2017DreamVision2017
318
$endgroup$
Let $L_1$ be the line passing through the points $Q_1=(4, −2, −4)$ and $Q_2=(5, −1, −5)$ and let $L_2$ be the line passing through the point $P_1=(−13, −12, 6)$ with direction vector $underline{d}=[6, 3, −6]^T$. Determine whether $L_1$ and $L_2$ intersect. If so, find the point of intersection $Q$. If not, find a value for the $z$-coordinate of $P_1$ so the resulting lines do intersect.
Please tell me what the result is
linear-algebra algebra-precalculus vectors linear-transformations vector-analysis
linear-algebra algebra-precalculus vectors linear-transformations vector-analysis
edited Jan 17 at 20:06
user289143
1,000313
asked Jan 17 at 19:26
DreamVision2017DreamVision2017
318
edited Jan 17 at 20:06
user289143
1,000313
asked Jan 17 at 19:26
DreamVision2017DreamVision2017
318
edited Jan 17 at 20:06
user289143
1,000313
edited Jan 17 at 20:06
user289143
1,000313
edited Jan 17 at 20:06
user289143
1,000313
1,000313
asked Jan 17 at 19:26
DreamVision2017DreamVision2017
318
asked Jan 17 at 19:26
DreamVision2017DreamVision2017
318
asked Jan 17 at 19:26
DreamVision2017DreamVision2017
318
318
closed as off-topic by RRL, Riccardo.Alestra, Adrian Keister, Andrew, Shaun Feb 5 at 15:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Riccardo.Alestra, Adrian Keister, Andrew, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Riccardo.Alestra, Adrian Keister, Andrew, Shaun Feb 5 at 15:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Riccardo.Alestra, Adrian Keister, Andrew, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Have you made any effort whatsoever to solve this on your own? Please show your work.
$endgroup$
– amd
Jan 17 at 21:47
$begingroup$
Yes I have, although last time I showed some effort on my question no one even bothered to respond so please don't judge prematurely
$endgroup$
– DreamVision2017
Jan 18 at 18:41
$begingroup$
You haven’t posted a question. You’ve posted a context-free problem statement and then made a demand, however politely. Taken by itself, this is not the sort of “question” that will be well-received on this site. That aside, this one is a minor variation of the exercises that you’ve asked about in previous questions. If you’re having trouble with this subject, you’d be well advised to review and seek help from your instructor/TA.
$endgroup$
– amd
Jan 18 at 21:17
add a comment |
$begingroup$
Have you made any effort whatsoever to solve this on your own? Please show your work.
$endgroup$
– amd
Jan 17 at 21:47
$begingroup$
Yes I have, although last time I showed some effort on my question no one even bothered to respond so please don't judge prematurely
$endgroup$
– DreamVision2017
Jan 18 at 18:41
$begingroup$
You haven’t posted a question. You’ve posted a context-free problem statement and then made a demand, however politely. Taken by itself, this is not the sort of “question” that will be well-received on this site. That aside, this one is a minor variation of the exercises that you’ve asked about in previous questions. If you’re having trouble with this subject, you’d be well advised to review and seek help from your instructor/TA.
$endgroup$
– amd
Jan 18 at 21:17
$begingroup$
Have you made any effort whatsoever to solve this on your own? Please show your work.
$endgroup$
– amd
Jan 17 at 21:47
$begingroup$
Have you made any effort whatsoever to solve this on your own? Please show your work.
$endgroup$
– amd
Jan 17 at 21:47
$begingroup$
Yes I have, although last time I showed some effort on my question no one even bothered to respond so please don't judge prematurely
$endgroup$
– DreamVision2017
Jan 18 at 18:41
$begingroup$
Yes I have, although last time I showed some effort on my question no one even bothered to respond so please don't judge prematurely
$endgroup$
– DreamVision2017
Jan 18 at 18:41
$begingroup$
You haven’t posted a question. You’ve posted a context-free problem statement and then made a demand, however politely. Taken by itself, this is not the sort of “question” that will be well-received on this site. That aside, this one is a minor variation of the exercises that you’ve asked about in previous questions. If you’re having trouble with this subject, you’d be well advised to review and seek help from your instructor/TA.
$endgroup$
– amd
Jan 18 at 21:17
$begingroup$
You haven’t posted a question. You’ve posted a context-free problem statement and then made a demand, however politely. Taken by itself, this is not the sort of “question” that will be well-received on this site. That aside, this one is a minor variation of the exercises that you’ve asked about in previous questions. If you’re having trouble with this subject, you’d be well advised to review and seek help from your instructor/TA.
$endgroup$
– amd
Jan 18 at 21:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Find the equation for both lines:
$$L_1 = begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix};L_2 = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Set $L_1=L_2$, to form a set of linear equations in s and t:
$$begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix} = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Rearranging:
$$begin{pmatrix}17 \ 10 \ -10end{pmatrix}=sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}-tbegin{pmatrix}1 \ 1 \ -1end{pmatrix}$$
Now $$17=6s- t, 10=3s-t\implies t=-3,s=frac{7}{3}$$
Verify with the final equation: $$-10neq-6(frac{7}{3})+1(-3)$$
Since this does not hold, we must change either $L_1$ or $L_2$, e.g. $$-4-(t)=x-6(s)\ implies -4-(-3)=x-6(frac{7}{3})\ implies x = 13.$$ So $$L_2 = begin{pmatrix}-13 \ -12 \ 13end{pmatrix} + tbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
answered Jan 17 at 20:00
cluelessatthiscluelessatthis
402313
$endgroup$
$begingroup$
The other person who answered got $1$. At least one of you is wrong.
$endgroup$
– amd
Jan 17 at 21:46
add a comment |
$begingroup$
(Check my arithmetic). $L_1=Q_1+(Q_2-Q_1)s=(4,-2,-4)+(1,1,-1)s$ and $L_2=P_1+d=(-13,-12,6)+(6,3,-6)T$. Match $x$ and $y$ coordinates to get simultaneous equations. $-13+6T=4+s,-12+3T=-2+s$. Solve and get $T=7/3, s=-3$. The $z$ coordinate for $L_1$ is$-1$ while the coordinate for $L_2$ is $4$. To get them to intersect change $z$ coordinate of $P_1$ to $13$.
$Q=(1,-5,-1)$.
answered Jan 17 at 19:58
herb steinbergherb steinberg
2,8032310
$endgroup$
$begingroup$
@amd You are right. I made arith. error at end. Rest of calculation, including point of intersection, is correct. $P_1(z)$ corrected.
$endgroup$
– herb steinberg
Jan 17 at 22:25
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Find the equation for both lines:
$$L_1 = begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix};L_2 = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Set $L_1=L_2$, to form a set of linear equations in s and t:
$$begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix} = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Rearranging:
$$begin{pmatrix}17 \ 10 \ -10end{pmatrix}=sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}-tbegin{pmatrix}1 \ 1 \ -1end{pmatrix}$$
Now $$17=6s- t, 10=3s-t\implies t=-3,s=frac{7}{3}$$
Verify with the final equation: $$-10neq-6(frac{7}{3})+1(-3)$$
Since this does not hold, we must change either $L_1$ or $L_2$, e.g. $$-4-(t)=x-6(s)\ implies -4-(-3)=x-6(frac{7}{3})\ implies x = 13.$$ So $$L_2 = begin{pmatrix}-13 \ -12 \ 13end{pmatrix} + tbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
answered Jan 17 at 20:00
cluelessatthiscluelessatthis
402313
$endgroup$
$begingroup$
The other person who answered got $1$. At least one of you is wrong.
$endgroup$
– amd
Jan 17 at 21:46
add a comment |
$begingroup$
Find the equation for both lines:
$$L_1 = begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix};L_2 = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Set $L_1=L_2$, to form a set of linear equations in s and t:
$$begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix} = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Rearranging:
$$begin{pmatrix}17 \ 10 \ -10end{pmatrix}=sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}-tbegin{pmatrix}1 \ 1 \ -1end{pmatrix}$$
Now $$17=6s- t, 10=3s-t\implies t=-3,s=frac{7}{3}$$
Verify with the final equation: $$-10neq-6(frac{7}{3})+1(-3)$$
Since this does not hold, we must change either $L_1$ or $L_2$, e.g. $$-4-(t)=x-6(s)\ implies -4-(-3)=x-6(frac{7}{3})\ implies x = 13.$$ So $$L_2 = begin{pmatrix}-13 \ -12 \ 13end{pmatrix} + tbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
answered Jan 17 at 20:00
cluelessatthiscluelessatthis
402313
$endgroup$
$begingroup$
The other person who answered got $1$. At least one of you is wrong.
$endgroup$
– amd
Jan 17 at 21:46
add a comment |
$begingroup$
Find the equation for both lines:
$$L_1 = begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix};L_2 = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Set $L_1=L_2$, to form a set of linear equations in s and t:
$$begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix} = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Rearranging:
$$begin{pmatrix}17 \ 10 \ -10end{pmatrix}=sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}-tbegin{pmatrix}1 \ 1 \ -1end{pmatrix}$$
Now $$17=6s- t, 10=3s-t\implies t=-3,s=frac{7}{3}$$
Verify with the final equation: $$-10neq-6(frac{7}{3})+1(-3)$$
Since this does not hold, we must change either $L_1$ or $L_2$, e.g. $$-4-(t)=x-6(s)\ implies -4-(-3)=x-6(frac{7}{3})\ implies x = 13.$$ So $$L_2 = begin{pmatrix}-13 \ -12 \ 13end{pmatrix} + tbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
answered Jan 17 at 20:00
cluelessatthiscluelessatthis
402313
$endgroup$
Find the equation for both lines:
$$L_1 = begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix};L_2 = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Set $L_1=L_2$, to form a set of linear equations in s and t:
$$begin{pmatrix}4 \ -2 \ -4end{pmatrix} + tbegin{pmatrix}1 \ 1 \ -1end{pmatrix} = begin{pmatrix}-13 \ -12 \ 6end{pmatrix} + sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
Rearranging:
$$begin{pmatrix}17 \ 10 \ -10end{pmatrix}=sbegin{pmatrix}6 \ 3 \ -6end{pmatrix}-tbegin{pmatrix}1 \ 1 \ -1end{pmatrix}$$
Now $$17=6s- t, 10=3s-t\implies t=-3,s=frac{7}{3}$$
Verify with the final equation: $$-10neq-6(frac{7}{3})+1(-3)$$
Since this does not hold, we must change either $L_1$ or $L_2$, e.g. $$-4-(t)=x-6(s)\ implies -4-(-3)=x-6(frac{7}{3})\ implies x = 13.$$ So $$L_2 = begin{pmatrix}-13 \ -12 \ 13end{pmatrix} + tbegin{pmatrix}6 \ 3 \ -6end{pmatrix}$$
answered Jan 17 at 20:00
cluelessatthiscluelessatthis
402313
edited Jan 17 at 20:45
answered Jan 17 at 20:00
cluelessatthiscluelessatthis
402313
answered Jan 17 at 20:00
cluelessatthiscluelessatthis
402313
answered Jan 17 at 20:00
cluelessatthiscluelessatthis
402313
402313
$begingroup$
The other person who answered got $1$. At least one of you is wrong.
$endgroup$
– amd
Jan 17 at 21:46
add a comment |
$begingroup$
The other person who answered got $1$. At least one of you is wrong.
$endgroup$
– amd
Jan 17 at 21:46
$begingroup$
The other person who answered got $1$. At least one of you is wrong.
$endgroup$
– amd
Jan 17 at 21:46
$begingroup$
The other person who answered got $1$. At least one of you is wrong.
$endgroup$
– amd
Jan 17 at 21:46
add a comment |
$begingroup$
(Check my arithmetic). $L_1=Q_1+(Q_2-Q_1)s=(4,-2,-4)+(1,1,-1)s$ and $L_2=P_1+d=(-13,-12,6)+(6,3,-6)T$. Match $x$ and $y$ coordinates to get simultaneous equations. $-13+6T=4+s,-12+3T=-2+s$. Solve and get $T=7/3, s=-3$. The $z$ coordinate for $L_1$ is$-1$ while the coordinate for $L_2$ is $4$. To get them to intersect change $z$ coordinate of $P_1$ to $13$.
$Q=(1,-5,-1)$.
answered Jan 17 at 19:58
herb steinbergherb steinberg
2,8032310
$endgroup$
$begingroup$
@amd You are right. I made arith. error at end. Rest of calculation, including point of intersection, is correct. $P_1(z)$ corrected.
$endgroup$
– herb steinberg
Jan 17 at 22:25
add a comment |
$begingroup$
(Check my arithmetic). $L_1=Q_1+(Q_2-Q_1)s=(4,-2,-4)+(1,1,-1)s$ and $L_2=P_1+d=(-13,-12,6)+(6,3,-6)T$. Match $x$ and $y$ coordinates to get simultaneous equations. $-13+6T=4+s,-12+3T=-2+s$. Solve and get $T=7/3, s=-3$. The $z$ coordinate for $L_1$ is$-1$ while the coordinate for $L_2$ is $4$. To get them to intersect change $z$ coordinate of $P_1$ to $13$.
$Q=(1,-5,-1)$.
answered Jan 17 at 19:58
herb steinbergherb steinberg
2,8032310
$endgroup$
$begingroup$
@amd You are right. I made arith. error at end. Rest of calculation, including point of intersection, is correct. $P_1(z)$ corrected.
$endgroup$
– herb steinberg
Jan 17 at 22:25
add a comment |
$begingroup$
(Check my arithmetic). $L_1=Q_1+(Q_2-Q_1)s=(4,-2,-4)+(1,1,-1)s$ and $L_2=P_1+d=(-13,-12,6)+(6,3,-6)T$. Match $x$ and $y$ coordinates to get simultaneous equations. $-13+6T=4+s,-12+3T=-2+s$. Solve and get $T=7/3, s=-3$. The $z$ coordinate for $L_1$ is$-1$ while the coordinate for $L_2$ is $4$. To get them to intersect change $z$ coordinate of $P_1$ to $13$.
$Q=(1,-5,-1)$.
answered Jan 17 at 19:58
herb steinbergherb steinberg
2,8032310
$endgroup$
(Check my arithmetic). $L_1=Q_1+(Q_2-Q_1)s=(4,-2,-4)+(1,1,-1)s$ and $L_2=P_1+d=(-13,-12,6)+(6,3,-6)T$. Match $x$ and $y$ coordinates to get simultaneous equations. $-13+6T=4+s,-12+3T=-2+s$. Solve and get $T=7/3, s=-3$. The $z$ coordinate for $L_1$ is$-1$ while the coordinate for $L_2$ is $4$. To get them to intersect change $z$ coordinate of $P_1$ to $13$.
$Q=(1,-5,-1)$.
answered Jan 17 at 19:58
herb steinbergherb steinberg
2,8032310
edited Jan 17 at 22:28
answered Jan 17 at 19:58
herb steinbergherb steinberg
2,8032310
answered Jan 17 at 19:58
herb steinbergherb steinberg
2,8032310
answered Jan 17 at 19:58
herb steinbergherb steinberg
2,8032310
2,8032310
$begingroup$
@amd You are right. I made arith. error at end. Rest of calculation, including point of intersection, is correct. $P_1(z)$ corrected.
$endgroup$
– herb steinberg
Jan 17 at 22:25
add a comment |
$begingroup$
@amd You are right. I made arith. error at end. Rest of calculation, including point of intersection, is correct. $P_1(z)$ corrected.
$endgroup$
– herb steinberg
Jan 17 at 22:25
$begingroup$
@amd You are right. I made arith. error at end. Rest of calculation, including point of intersection, is correct. $P_1(z)$ corrected.
$endgroup$
– herb steinberg
Jan 17 at 22:25
$begingroup$
@amd You are right. I made arith. error at end. Rest of calculation, including point of intersection, is correct. $P_1(z)$ corrected.
$endgroup$
– herb steinberg
Jan 17 at 22:25
add a comment |
$begingroup$
Have you made any effort whatsoever to solve this on your own? Please show your work.
$endgroup$
– amd
Jan 17 at 21:47
$begingroup$
Yes I have, although last time I showed some effort on my question no one even bothered to respond so please don't judge prematurely
$endgroup$
– DreamVision2017
Jan 18 at 18:41
$begingroup$
You haven’t posted a question. You’ve posted a context-free problem statement and then made a demand, however politely. Taken by itself, this is not the sort of “question” that will be well-received on this site. That aside, this one is a minor variation of the exercises that you’ve asked about in previous questions. If you’re having trouble with this subject, you’d be well advised to review and seek help from your instructor/TA.
$endgroup$
– amd
Jan 18 at 21:17