every other video on this topic stops when they realize the lines are parallel. thank you for showing what to do for EVERY option, instead of designing a problem that is done at step 1 then stopping at step 1.
Ah, I love how you showed us every possibility, and explained every single step. Even though some steps were not necessary to explain, you did it anyways & that just solidifies our understanding. Thank you so much.
I just came from a patrickJMT video on the same topic and yours helped so much more, not sure if I paid more attention since I still couldn't get it or you just explained all possible outcomes instead of just for that specific problem. Thank you so much you just saved my calc final :)
here's what happens in my maths lectures: Go to class and know what topic to revise (honestly, our teacher's voice is hard to follow up with). Soon, I arrive home, open RUclips and search for your videos and start understanding what the heck was going on... You made my life much easier tbh! Thanks a ton once again Krista, and I am definitely going to continue learning from you, PatrickJMT and Khan Academy :D
Omg thank u so much, i read other articles on how to find perpendicular lines and they said that you have to make sure the lines intersect first. I was so confused until I came across ur video. Thanks
Where was RUclips for high school maths when I needed it? This would have been so valuable... in 1988!! Kids these days don't know how good they have it. :) ETA: With the intersection test, I proceeded by saying that if the lines intersect, the relationship between S and T should be maintained in all three sets of simultaneous equations. Since this relationship is NOT maintained, therefore they must not intersect. It certainly worked in this case, but is this approach generally valid?
Given two lines on the same plane, we can find a normal vector of the plane, by taking the cross product of their two directional vectors. This finds the direction that is mutually perpendicular to both of them, which defines the plane's normal vector. There are an infinite number of possible normal vectors, all of which are scalar multiples of each other, and we simply pick whichever one is most convenient. Then we pick any point on either of the two lines, and shift the plane as required to intersect with it. Consider the following intersecting lines: Line 1: = + t* Line 2: = + u* Take the cross product: cross = , which becomes the plane's normal vector. It doesn't matter which cross product we choose, we'll find a consistent equation for the plane either way. A general equation of any plane, given this normal vector will have the form of: 2*x + 1*y - 2*z = K Plug in our known point, (1, 2, 3), and solve for K: 2*(1) + 1*(2) - 2*(3) = K K = -2 Thus the equation of the plane is: 2*x + y - 2*z = -2
Good video except you missed a step. After finding a line to be parallel test of they intersect. If they are parallel and intersect that means they coincide.
What would my answer be when I find out they are skew would it just be the scalar result of the dot product or do I have to find the magnitude between those two lines?
Skew lines are non-parallel lines that don't intersect. There will be two points where their approach is the closest, but they won't actually hit each other.
Consider the following two lines: Line 1: = + t* Line 2: = + u* where t and u are arbitrary parameters to trace out these lines. Take each component of each line's vector equation, and form an equation. Then equate corresponding equations to each other. x = 1 + 3*t; x = 7 + 1*u y = 2 + 4*t; y = 8 + 2*u z = 3 + 5*t; z = 9 + 3*u Equate x, y, and z across each pair: 1+ 3*t = 7 + u 2 + 4*t = 8 + 2*u 3 + 5*t = 9 + 3*u One of these equations is redundant, so we only need to solve 2 of them together. Multiply first equation by 2, and subtract the second equation to eliminate u: 2*(1+ 3*t = 7 + u) - (2 + 4*t = 8 + 2*u) 2*t = 6 t = 3 Plug back in to the equations that define line 1: x = 10, y = 14, z = 18 Intersection occurs at (10, 14, 18)
Is it possible that when checking for the intersection, one could solve for both s and t , but upon plugging back into the formulas both sides don't agree and that would be skewed, or is it that when its skewed you will always have either s and t cancel out on both sides
If the formulas don't agree, it means you don't have intersecting points. You'll have contradictory equations when setting up the system of equations to solve for the point of intersection. Skew lines will have one pair of contradictory equations, and parallel lines will have two pairs of contradictory equations, when attempting to solve for the point of intersection.
Cant two lines be perpendicular even if they are not intersecting? Like in 3D. At the end you say that it cant possibly be perpendicular because they dont intersect. Maybe the definition you are using for perpendicular calls for the lines to be co-planar?
Depends on who's asking. A layman will say yes, but a mathematician will say no. Skew lines can have orthogonal directional vectors, and most people would consider that perpendicular. Even in Engineering, when you read the definition of a worm gear, it calls the two shafts perpendicular, even though their axes are skew lines. But a mathematician will only call two lines perpendicular, if they intersect at right angles. I would call them orthogonal skew lines, if they have orthogonal directional vectors, but don't intersect.
You can form a parametric equation for a 3-D line, given the two points through which it passes. You let one of the points be the initial point when the parameter t=0, and the distance vector between the two points, becomes the directional vector. It's completely arbitrary which point you choose as the initial point, because it's arbitrary where the parameter t=0, and arbitrary how fast the line moves as t increases, so you usually choose whichever is more convenient. Consider point (1, 1, 1) and point (3, 4, 5). The distance vector between these two points is - = . This tells us that the parametric vector equation of this line is: = + t*
You define the lines as parametric vector equations in the form of L1 = p1 + t*d1, and L2 = p2 + u*d2, where p1 & p2 are vectors for the initial points, and d1 & d2 are the directional vectors for the two lines, and L1 and L2 are vectors that trace out the paths of each line. The parameters t and u are the parameters that trace out the three position coordinates to draw the lines. Think of them as time. They are not necessarily both equal, hence I assigned another letter for line L2. This will create a system of 3 equations and 2 unknowns (t and u). If you have intersecting lines, one of these equations will be redundant. If you have skew lines or parallel lines, you'll have contradictory equations. You then solve for either t or u, to find the corresponding intersecting point, and then substitute it in to the corresponding line's definition, to find the point of intersection.
Given two lines that intersect each other (and aren't the same line), take the dot product between their two directional vectors. Then, divide by the product of the magnitudes. Dot product divided by product of magnitudes, tells you the cosine of the angle between two vectors. As an example: Given Line L1 = *t + And Line L2 = *t + Right away, you can see they both intersect point (1, 1, 1). Not that this is relevant, but this allows you to verify that they aren't skew lines. Take the dot product: dot = 108 + 192 + 300 = 600 Take the product of magnitudes: sqrt(12^2 + 16^2 + 15^2) = 25 sqrt(9^2 + 12^2 +20^2) = 25 Product of magnitudes = 625 Cos of angle = 600/625 = 0.96
If I found that they are neither parallel or intersecting can I found the equation of plane containing these lines? I have exam after tomorrow about these things but geometry is always a nightmare 🌚💔😂I hope you're still here and help me 😂
No wonder y they say some people died, went 2 haven and came back 2 earth with good missions, i think your mission is to simply solve calculus problems
every other video on this topic stops when they realize the lines are parallel. thank you for showing what to do for EVERY option, instead of designing a problem that is done at step 1 then stopping at step 1.
+ryan ellis You're welcome! :D
Amen!! Excellent video :)
exactly!!!!
Right!
Thanks for your tutorial!
This woman is saving lives here. I'm so grateful for your videos, they are so clear and thorough.
Thanks, Cecilia! I'm so glad the videos are helping! :D
Ah, I love how you showed us every possibility, and explained every single step. Even though some steps were not necessary to explain, you did it anyways & that just solidifies our understanding. Thank you so much.
You’re literally a goddess, THANK YOU. I’m straight up boutta fail calc 3 but maybe a little less because of you
You're welcome, Zishawn! I hope the videos can continue to be helpful! :D
Just discovered you 3 hours before my exam, wow I wouldn't have written anything if it wasn't for you
This video is soooooo organized!!!! I loved it. Thank you very much. I loved the "if not, then" strusture you used
I LOVE U KRISTA KING
If I could like this video a thousand times I would! Thanks Krista!
Aww, thanks Daniel! So glad it helped! :D
I just came from a patrickJMT video on the same topic and yours helped so much more, not sure if I paid more attention since I still couldn't get it or you just explained all possible outcomes instead of just for that specific problem. Thank you so much you just saved my calc final :)
I'm so glad the video helped, and I hope the final went great!! :D
here's what happens in my maths lectures: Go to class and know what topic to revise (honestly, our teacher's voice is hard to follow up with). Soon, I arrive home, open RUclips and search for your videos and start understanding what the heck was going on... You made my life much easier tbh! Thanks a ton once again Krista, and I am definitely going to continue learning from you, PatrickJMT and Khan Academy :D
I'm glad I can help along the way! 😊
Nicely covered and thanks for going over examples on each case.
+Tim Kidd You're welcome!
the offspring of you and patrickJMT would solve world peace using math
Omg! I was thinking the same thing lol
Lucy Pham 😂
Omg thought of that too 😂
Omg thank u so much, i read other articles on how to find perpendicular lines and they said that you have to make sure the lines intersect first. I was so confused until I came across ur video. Thanks
I feel so confidence when I watch your vids, I know that I will understand the problem at the end.
:D
I am just amazed that how you simplify these topics and make them easy for us👍😀
I'm so glad you like the videos, Hardik! :D
You're not uploading videos, you're saving some people's lives
I'm happy to help!! :D
Thanks a lot! If we had a teacher like you we wouldn't be raging at theory. :(
great videos, awesome quality, clear explenations. love it
Exactly what I wanted... Tks Krista
Thank you for this. We never went over this class or on the assignments and yet, teacher says we have to know this for this test tomorrow :P
i appreciate your videos so much 😭 you explain it so so well
Thank you Karina, I'm so glad the videos are helping! :)
Amazing video. Totally helped me with math class cuz i had to skip a few class due to family business. Now I get to finish homework lolz
I'm so glad it helped! :)
@@kristakingmath ty so much
Your examples are crystal clear. Thank you so much, Krista!
You're so welcome! Glad they can help!
thank you . I was having a trouble with it , but now everything is clear .
I'm glad it could help!
Thank you so much for explaining the whole process. It helped so much. Liked!
Thanks!
OMG THANK YOU, now i wont fail my course! subbed :)
Have a quiz over this in fifteen minutes, thanks for the review!
You're welcome, hope the quiz went great! :)
Krista you are really awesome ! Thanks for your videos .
Glad they're helping! :D
Your content is amazing...keep uploading such material
2014 is 9 years ago this is shocking, good explanation btw you really helped me
THANKS SO MUCH IM STUDYING FOR MA MATH TEST TMR ANS THIS HELPED SO MUCHH ❤❤❤
Very clear explanation. Thank you so much!
I'm glad it could help!
Oh gawd, thank you so much! Really helpful and through.
so good to understand the way how to solve questions ^_^
Thank you, this was insanely helpful
how is it that this channel always has the exact question im looking at, without me even searching it
Thank you soooo much. This video helped me A LOT!❤❤
You're so welcome! I’m so glad it helped!! 🤓
Thank you so much ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️ that's what I need
Great video. I liked it. It's gonna help me help my nephew...
Love that you're helping your nephew!!
Where was RUclips for high school maths when I needed it? This would have been so valuable... in 1988!! Kids these days don't know how good they have it. :)
ETA: With the intersection test, I proceeded by saying that if the lines intersect, the relationship between S and T should be maintained in all three sets of simultaneous equations. Since this relationship is NOT maintained, therefore they must not intersect. It certainly worked in this case, but is this approach generally valid?
Yeah and back then tuition costed a fraction of what it is today even after adjusting for inflation. Kids back then didnt know how good they had it
Awesome Explanation!
:D
Thanks you help me a lot 10^(1000000000....)!
I like your slow and clear explanation!
Thank you so much, Abdalla, I'm so glad it helped! :D
You are a very help full lady :)
Too good explanation mam thanks 😊☺️
You're welcome, Aditya, I'm so glad you liked it! :D
Thank you for the great video! How do I find the equation of the plane given L1 and L2
Given two lines on the same plane, we can find a normal vector of the plane, by taking the cross product of their two directional vectors. This finds the direction that is mutually perpendicular to both of them, which defines the plane's normal vector. There are an infinite number of possible normal vectors, all of which are scalar multiples of each other, and we simply pick whichever one is most convenient. Then we pick any point on either of the two lines, and shift the plane as required to intersect with it.
Consider the following intersecting lines:
Line 1: = + t*
Line 2: = + u*
Take the cross product: cross = , which becomes the plane's normal vector. It doesn't matter which cross product we choose, we'll find a consistent equation for the plane either way.
A general equation of any plane, given this normal vector will have the form of:
2*x + 1*y - 2*z = K
Plug in our known point, (1, 2, 3), and solve for K:
2*(1) + 1*(2) - 2*(3) = K
K = -2
Thus the equation of the plane is:
2*x + y - 2*z = -2
Great Video. Very helpful!
Thanks, Math! I'm glad it helped! :D
thanks Krista!
Good video except you missed a step. After finding a line to be parallel test of they intersect. If they are parallel and intersect that means they coincide.
this is so helpful thank you!!
"I LOVE YOU"...THANK YOU SO MUCH
+Isi Hummetli You're welcome!
just you are welcome((
thumps up for you,, good work I loved it 💯💯
Best vid on this topic fs fs
Thank you so much!! 😊
What would my answer be when I find out they are skew would it just be the scalar result of the dot product or do I have to find the magnitude between those two lines?
If they we're intersecting, where would they intersect using this method? I'm guessing I would plug in the t and s values somewhere. Is this so?
i dont understand when it can be considered skewed. could you please briefly explain one more time here?
Thanks in advance!
Skew lines are non-parallel lines that don't intersect. There will be two points where their approach is the closest, but they won't actually hit each other.
Great lecture (y) thanks alot :)
Technically, L_1 \dot L_2 is undefined as the dot product is not defined for lines, but only for vectors
very helpful, great vid!
Thanks, Andy! :)
Hello, may be a 5th test to know if vectors are colinear as running along a single line.
Thanks :)
2/4 etc do the job, I saw another like x1*y2 = y1*x2, that give the same result :)
Youre very good teacher.. Keep going
Thanks, Aiman! :)
Would it be possible to do an example where the lines intersect and then show the process of finding that point?
Consider the following two lines:
Line 1: = + t*
Line 2: = + u*
where t and u are arbitrary parameters to trace out these lines.
Take each component of each line's vector equation, and form an equation. Then equate corresponding equations to each other.
x = 1 + 3*t; x = 7 + 1*u
y = 2 + 4*t; y = 8 + 2*u
z = 3 + 5*t; z = 9 + 3*u
Equate x, y, and z across each pair:
1+ 3*t = 7 + u
2 + 4*t = 8 + 2*u
3 + 5*t = 9 + 3*u
One of these equations is redundant, so we only need to solve 2 of them together. Multiply first equation by 2, and subtract the second equation to eliminate u:
2*(1+ 3*t = 7 + u) - (2 + 4*t = 8 + 2*u)
2*t = 6
t = 3
Plug back in to the equations that define line 1:
x = 10, y = 14, z = 18
Intersection occurs at (10, 14, 18)
Thanks, Krista for saving my ass even seven years later :)
I'm so glad the video helped, Anouk! :)
Is it possible that when checking for the intersection, one could solve for both s and t , but upon plugging back into the formulas both sides don't agree and that would be skewed, or is it that when its skewed you will always have either s and t cancel out on both sides
If the formulas don't agree, it means you don't have intersecting points. You'll have contradictory equations when setting up the system of equations to solve for the point of intersection. Skew lines will have one pair of contradictory equations, and parallel lines will have two pairs of contradictory equations, when attempting to solve for the point of intersection.
Actually, they can be perpendicular and skew at the same time.
Yeah I was thinking about it
omg I was doing this EXACT problem from my calc book with the same numbers and came here to learn how to do it
Love it when that happens! :D
very helpful!!!!!! thank you so much!
You're welcome, Quang, I'm so glad it helped! :)
Cant two lines be perpendicular even if they are not intersecting? Like in 3D. At the end you say that it cant possibly be perpendicular because they dont intersect. Maybe the definition you are using for perpendicular calls for the lines to be co-planar?
Depends on who's asking. A layman will say yes, but a mathematician will say no.
Skew lines can have orthogonal directional vectors, and most people would consider that perpendicular. Even in Engineering, when you read the definition of a worm gear, it calls the two shafts perpendicular, even though their axes are skew lines. But a mathematician will only call two lines perpendicular, if they intersect at right angles.
I would call them orthogonal skew lines, if they have orthogonal directional vectors, but don't intersect.
Huh. Your example question is the exact same as my homework question so that worked out xD
That was very helpful thank you very much
You're welcome, Abdullah, I'm so glad it helped! :D
Great video thanks a million
What do we do if we're provided with points instead of parametric equations?
You can form a parametric equation for a 3-D line, given the two points through which it passes. You let one of the points be the initial point when the parameter t=0, and the distance vector between the two points, becomes the directional vector. It's completely arbitrary which point you choose as the initial point, because it's arbitrary where the parameter t=0, and arbitrary how fast the line moves as t increases, so you usually choose whichever is more convenient.
Consider point (1, 1, 1) and point (3, 4, 5). The distance vector between these two points is - = . This tells us that the parametric vector equation of this line is:
= + t*
Thank you so much!!
they can also be coincident tho
is it possible for two lines to be skew and perpendicular, do they have to touch each other to be perpendicular?
If the lines are perpendicular, they'll touch each other at some point. They can't be skew and perpendicular at the same time.
she is the best
Thnx Love your such a nice tutor
You're welcome, Omar! 😊
Thank you very much.
u r a legend
If you find that the lines intersect, how do you find the point where they intersect?
You define the lines as parametric vector equations in the form of L1 = p1 + t*d1, and L2 = p2 + u*d2, where p1 & p2 are vectors for the initial points, and d1 & d2 are the directional vectors for the two lines, and L1 and L2 are vectors that trace out the paths of each line. The parameters t and u are the parameters that trace out the three position coordinates to draw the lines. Think of them as time. They are not necessarily both equal, hence I assigned another letter for line L2.
This will create a system of 3 equations and 2 unknowns (t and u). If you have intersecting lines, one of these equations will be redundant. If you have skew lines or parallel lines, you'll have contradictory equations. You then solve for either t or u, to find the corresponding intersecting point, and then substitute it in to the corresponding line's definition, to find the point of intersection.
Thanks a LOT😍😀
You're welcome, Rider! :)
on my homework problem it is also asking for to find " the cosine of the angle of intersection" How to find that ? .
Given two lines that intersect each other (and aren't the same line), take the dot product between their two directional vectors. Then, divide by the product of the magnitudes. Dot product divided by product of magnitudes, tells you the cosine of the angle between two vectors.
As an example:
Given Line L1 = *t +
And Line L2 = *t +
Right away, you can see they both intersect point (1, 1, 1). Not that this is relevant, but this allows you to verify that they aren't skew lines.
Take the dot product:
dot = 108 + 192 + 300 = 600
Take the product of magnitudes:
sqrt(12^2 + 16^2 + 15^2) = 25
sqrt(9^2 + 12^2 +20^2) = 25
Product of magnitudes = 625
Cos of angle = 600/625 = 0.96
@@carultch Thank you! I already graduated haha.
my friend likes your handwriting
WTF! This is the exact same problem and given with our homework.😂😂😂
2022 , ty
you are amazing
HAHA this is the exact problem. Early Trans. sect 12.5 problem 19 thanks!!
Wow, that makes so much more sense (^^), thank you!
:D
I liked it a lot "" smart lady ""
Omg Your are Amazing :)
Finally I Understood it
like + sub :)
Awesome! I'm so glad it could help!
thank you so much
You're welcome, Sir! :D
Thank you so much,
You are welcome, Muhammad!
THANK YOU!!!!!!!!!
+DaeHan2321 You're so welcome! Glad you liked it.
thank you
You're welcome! :)
thanks!
best explanation, thanks for saving my ass
Good video ty
Thanks, glad you enjoyed it! :)
thanks mam!
You're welcome, TG! :)
If I found that they are neither parallel or intersecting can I found the equation of plane containing these lines?
I have exam after tomorrow about these things but geometry is always a nightmare 🌚💔😂I hope you're still here and help me 😂
No wonder y they say some people died, went 2 haven and came back 2 earth with good missions, i think your mission is to simply solve calculus problems
I found t=11 and s=6 and when I put it in the two z equations I found them to be true.
nevermind, I saw that the two y-equations don't equal each other but the two x-equations and z-equations do when I plug t=11 and s=6 in them.