Distance between two skew lines Vectors 2016 Test

Test papers: • Vectors Lines Points a...
Important Related Video: • Linearly Dependent Vec...
Strategy for Coplanar Vectors and Linear Combination of Vectors in R3.
Coplanar vectors problems can be solved using two different strategies as shown in the video. Consider linear combination of scalar triple product of vectors.
#linearcombination #scalartripleproduct #coplanar #coplanarvectors #anilkumar #globalmathinstitute #edexcel #vectors #vectors_3d #vectors_MCV4U
/ @mathematicstutor Learn From Anil Kumar: www.globalmathinstitute.com

[@huyhuy9848]

I think there is a mistake in Subtraction of vector P1P2 , it should be P2 - P1

[@MathematicsTutor]

Thanks. It worked out since we are taking the absolute value. I should have written P2P1 for those values.
Appreciate your observation.
Thanks

[@perseusgeorgiadis7821]

The order doesn't really matter since the vector is in absolute value but you're right

[@yadpreetcheema4806]

great explanation as always sir.

[@jubilee4]

Is rejection of a vector in this case mistaken for a projection of a vector? The distance is actually the magnitude of the rejection vector from the vector that is calculated using two points to the projection of that same vector onto the second line l2.

[@dinahizwati8486]

it's really helpful. great job!

[@MathematicsTutor]

Thanks for appreciation. Feel free to post questions

[@netanelmad]

Great video, thank you!

[@MathematicsTutor]

Thanks. Here are Related video: ua-cam.com/play/PLJ-ma5dJyAqoRm1pbdY4odhtS-tVLfOl4.html

[@acies9408]

Very good thank you

[@ratlavatgopal7780]

Thanks

[@Youngballer739]

👌

[@phil7645]

nice job

[@samkelomthokozisi2041]

i tend to differ with u sir how u happan to say dis distance is a projection of p1p2 on n while a vector "Projection" formular is not the same as of a distance between skew lines , yes ur calculations are right , but projection part there has been a misconception on that Sir

[@AdityaT03]

@@samkelomthokozisi2041 I think he took the distance of the vector you get by doing the projection.

[@gregoriobiagi2025]

if n is the common perpendicular, why isn't ||n|| the shortest distance?

[@mathieulebot7032]

I assume it's because the normal isn't by nature a vector so it cannot have a magnitude (at least not in the same way as a vector). Yet, I'm still looking for a better explanation but can't manage to find one

[@mathieulebot7032]

tell me if u figured it out since ur comment

[@lonelylord6581]

The normal vector is actually a vector whose direction is normal to those two lines in space. But it is not scaled properly. So we divide by its magnitude to only get the direction. The actual distance between those lines is something else.