9.6 Finding Perpendiculars (CORE 1 - Chapter 9: Vectors)

#hindsmaths Using perpendiculars to find the shortest distances between parallel and skew lines, points and lines, points and planes. Also using perpendiculars to solve geometric problems
*** IN EXAMPLE 26 I SHOULD HAVE WRITTEN/SAID *PERPENDICULAR* NOT *PARALLEL* ***
0:00 Intro
3:25 Example 24
13:39 Example 25
23:52 Example 26
33:16 Example 27
39:48 Example 28
52:50 Example 29
1:08:46 End/Recap

[leilatani]

these are the best videos on the world

[lfcmaniac]

at 1:08:35 could you use Y as a position vector instead of (6, 0, -4) as your positional vector?

[labibatahsin]

Hi, in example one how comes you chose vector a as (1, 2, -1) but for vector b, you chose it to be a general point on line 2 ?
Also, in the book they do it differently but they set t=μ-λ. Would you please be able to explain how they define that variable? thank you

[prod_cio]

on Example 28, my first idea was to

consider x+2y+2z-5, which results in -2
Which means I know the coordinate P is 'beneath' the plane

Then I found the normal normalised (unit vector I think?)
which is just n/3

Then I extended this by the scalar of double the length, as this will go through the plane and then extend another length out

So basically I had

P + (n/3)*(4/3)
Which gave me the same coordinates

Would this method miss any sort of important results, or steps in between? How would the MS view this
Thank you

[hamzahwaheed]

Aren’t these the questions that came up in the C4 papers for a level maths

[gordonwalker]

for example 25 can you not just cross product the two direction vectors to end up with AB? much quicker imo