I’m so happy that I found your channel, to have someone like you doing all these various and tricky vector equations are pure gold! I subscribed and I will stay 💯
@@GlassofNumbers Definitely, I got exam in mathematics on Monday so your channel is quite helpful! Here is a challenge for you if you’re looking for vector content: This was taken from 2019 Exam in Math for engineering at all R1/R2/X+ Given three points in space A(0, -1, 2), B(-3, 2, -1) and C(1, 2, -3). a) Decide the vectors AB and AC b) Show that the plane Beta which crosses through these points A, B and C is given as the solution to the equation: x + 3y + 2z - 1 = 0 Find also the area to the triangle ABC. c) Decide the coordinates to the point Q in plane Beta which holds the shortest distance to the point E(1, 1, 2) Answer: a) AB=[-3,3,-3] AC=[1,3,-5] b) Area 3sqrt14 c) (1/2, -1/2, 1) I find Q: c) to be the difficult one, and will try this one today :)
Your channel deserves more views and subscribers since you don't solve a+b level questions and you solve the questions all STEM students face on the exams. Thanks for the video!
@@abhinavrao8698 I actually did plan to make a video on finding the distance of a point to the plane. It's on my agenda! 🤔Finding an image of a point on a plane, that's a good one!
It actually doesn't matter. You can switch them. The normal that you obtain (after taking cross product) will be opposite of the one in this video, but it can still be used as a normal for the plane.
Yes, switching the order in the cross product we will get the vector in the opposite direction, which can still be used as a normal vector for the plane.
I do understand how to calculate the normal vector but i dont know how those 2 vectors result in the normal vector. The 2 vectors must lie on the desired plane in order to get its vector? if not then how would we know what specific vectors to use in order to calculate the normal if there is a question with more vectors to it. Amazing video tho!
@@GlassofNumbers But given a situation where we have a line on the desired plane and we have the parametric equation of that line. We also have equations of 2 other planes who's line of intersection is parallel to the disired plane. When i try to find the normal vector of the desired plane, i start off with finding 2 vectors that lie on the plane. One vector we can easily determine is from the parametric equation of the line which i did. I determined the second vector by finding 2 points on the line of intersection of the other plane 2 planes and through those 2 points, i find the direction vector of the line of intersection. So if the line of intersection is parallel to the desired plane, shouldn't the direction vector of the line of intersection assumably lie on the desired plane too? because even though the vector is not on the plane but it is parallel to it which means we can assume that direction vector lies on the plane too. But everytime i use this method my answer is wrong and i dont understand why. Thank you for the response, i have an exam in 8 hoursand you have helped a lot.
wait why doesn't it matter if we sub in the first point or the second point? And if we only sub in the XYZ values of the second point, then does that mean the plane will only pass thru the second point but not the first?
If we sub in the second point, we will still have the plane passing through the first point because we obtained v1 using the two points, and v1 is perpendicular to the normal vector n. Try substituting both points, we will get the same plane!
Good question! I chose the second point because the coordinates are easier. It doesn't matter which point we choose. We could choose the first point and still get the same answer.
Good question! It doesn't matter whether we use the point (4, -2, 1) or (1, 1, 0). We will get the same equation (after simplifying). I used (1, 1, 0) because the numbers are easier to deal with if we need to simplify.
Thanks for being illustrative on every video you make. Love from Hong Kong
Thank you! I am happy to hear that you like that style! Are you taking Calculus 3?
I’m so happy that I found your channel, to have someone like you doing all these various and tricky vector equations are pure gold! I subscribed and I will stay 💯
Thank you! Agree! These problems, while not long, but are often tricky! Glad you like these videos 😁
@@GlassofNumbers Definitely, I got exam in mathematics on Monday so your channel is quite helpful!
Here is a challenge for you if you’re looking for vector content:
This was taken from 2019 Exam in Math for engineering at all R1/R2/X+
Given three points in space A(0, -1, 2), B(-3, 2, -1) and C(1, 2, -3).
a) Decide the vectors AB and AC
b) Show that the plane Beta which crosses through these points A, B and C is given as the solution to the equation:
x + 3y + 2z - 1 = 0
Find also the area to the triangle ABC.
c) Decide the coordinates to the point Q in plane Beta which holds the shortest distance to the point E(1, 1, 2)
Answer: a) AB=[-3,3,-3] AC=[1,3,-5]
b) Area 3sqrt14
c) (1/2, -1/2, 1)
I find Q: c) to be the difficult one, and will try this one today :)
Your channel deserves more views and subscribers since you don't solve a+b level questions and you solve the questions all STEM students face on the exams.
Thanks for the video!
Thank you!! Yes, those are the problems that STEM students face on exams, so showing details and full explanations are my goals 😁
Excellent explanation 😊
Thank you! 😁
dude thank you so much, didn't understand this from any other lectures. Thanks for the simple explanation!!
Good to hear you like my explanation! 😁Please help me share the video with others!
@@GlassofNumbers sure, can you do a video on finding an image of a point on a plane please? also the distance of a point from a plane??
@@abhinavrao8698 I actually did plan to make a video on finding the distance of a point to the plane. It's on my agenda! 🤔Finding an image of a point on a plane, that's a good one!
@@GlassofNumbers really looking forward to it exam in 5 days🤓
Made the one for distance from a point to a plane: ua-cam.com/video/rmJa8PfMKL8/v-deo.html. How was your exam?
it is the same process when dealing a parallel to the plane?
Almost, but to find the normal, we only need the same normal from that parallel plane
I understand that the cross product is not commutative, but how do we know whether v1 is going to be our 'a vector' or v2 is our 'b vector'?
It actually doesn't matter. You can switch them. The normal that you obtain (after taking cross product) will be opposite of the one in this video, but it can still be used as a normal for the plane.
Thank you very much. It's very detail. Love it.
Thank you! I am happy you love the video!! Please share it with others 👍😁
If I reversed the order you used for the cross product would that still be fine?
Yes, switching the order in the cross product we will get the vector in the opposite direction, which can still be used as a normal vector for the plane.
Thank you!!
Glad you like the video! Please help me share the video with others 😁👍
thank you
I am happy you like my video! Please help me share it!!
@@GlassofNumbers aye
For the equation, why did you use the point (1,1,0) for Xo, Yo, and Zo instead of (4,-2,1)?
We can pick either point. I just picked one with easy numbers.
I do understand how to calculate the normal vector but i dont know how those 2 vectors result in the normal vector. The 2 vectors must lie on the desired plane in order to get its vector? if not then how would we know what specific vectors to use in order to calculate the normal if there is a question with more vectors to it. Amazing video tho!
Yeah, you are correct that the two vectors must lie in the desired plane so that we can take their cross product to get the normal vector.
@@GlassofNumbers But given a situation where we have a line on the desired plane and we have the parametric equation of that line. We also have equations of 2 other planes who's line of intersection is parallel to the disired plane. When i try to find the normal vector of the desired plane, i start off with finding 2 vectors that lie on the plane.
One vector we can easily determine is from the parametric equation of the line which i did.
I determined the second vector by finding 2 points on the line of intersection of the other plane 2 planes and through those 2 points, i find the direction vector of the line of intersection.
So if the line of intersection is parallel to the desired plane, shouldn't the direction vector of the line of intersection assumably lie on the desired plane too? because even though the vector is not on the plane but it is parallel to it which means we can assume that direction vector lies on the plane too.
But everytime i use this method my answer is wrong and i dont understand why.
Thank you for the response, i have an exam in 8 hoursand you have helped a lot.
wait why doesn't it matter if we sub in the first point or the second point? And if we only sub in the XYZ values of the second point, then does that mean the plane will only pass thru the second point but not the first?
If we sub in the second point, we will still have the plane passing through the first point because we obtained v1 using the two points, and v1 is perpendicular to the normal vector n. Try substituting both points, we will get the same plane!
I was wondering why did you choose the second point not the first one to find the equation? Does it matter if which point we choose?
Good question! I chose the second point because the coordinates are easier. It doesn't matter which point we choose. We could choose the first point and still get the same answer.
Can we put any point from given points in equation of plane???
Yes, you can use either given points 😁
@@GlassofNumbers thanks 🤩
thanksssss brooooo
No problem! Please share my video to others 😁
Hi is there any other vector tutorial video?
Yes, there are other ones, and I am working on uploading more 😁
What if we are only given 1 point on the plane?
Then we will not have a unique answer
Does it matter if we write point (4,-2,1) while subtracting from r vector? as you did in 12:22
Good question! It doesn't matter whether we use the point (4, -2, 1) or (1, 1, 0). We will get the same equation (after simplifying). I used (1, 1, 0) because the numbers are easier to deal with if we need to simplify.
Best
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What if we only had one point to passing through
If there is only one point, then there are infinitely many planes perpendicular to the given plane and passing through that point.
Best
Thanks 😁👍 please help me share this video with others!