00:00Hi guys, so in this
00:02lesson we're going to find
00:03the distance between a point
00:05and a line that might
00:06sound straightforward but it's actually
00:08kind of tricky. So these
00:11abs are finding the shortest
00:12distance between the point, this
00:13and the line, this. Okay,
00:15firstly, let's draw a line.
00:17If in any question, especially
00:19in vectors, if you can
00:20draw a diagram, it's going
00:23to help you. So here's
00:24a line, I don't have
00:25to draw it accurately in
00:27three dimensions or anything like
00:28that.
00:28That's just, I'm just gonna
00:29draw a line. And I'm
00:30gonna draw a point, there's
00:31a point here, P. So
00:34I need to find the
00:35distance, pre the point of
00:36the line. Now obviously, the
00:38shortest distance, it's not like,
00:40it's not like I'm gonna
00:41go, oh, that's the shortest
00:42distance. Clearly the shortest distance
00:44is, we need to go
00:45straight over, well let me
00:48start in the middle here.
00:49We need to go straight
00:50over to the line. Now
00:52what have I done there?
00:53Hopefully you can see that.
00:56This is perpendicular. These two
01:01lines are perpendicular because that
01:03would be the shortest distance.
01:05Imagine if you were the
01:07shortest distance between you and
01:09a wall, you just walk
01:10straight over to the wall,
01:11you would walk in a
01:12perpendicular line. Okay, so that's
01:15B. What's this point here
01:18that is on the line?
01:20Well, we don't know. Let's
01:23call it
01:24point A to begin with.
01:29Okay, now I set the
01:31word perpendicular there and I
01:33said before when you hear
01:35perpendicular in a vectors question,
01:38you should kind of immediately
01:39go, okay, the dot product
01:42must equal zero because that
01:44is what we have for
01:47perpendicular vectors, the dot product
01:49equals zero, but the dot
01:51product of what?
01:52So it would be the
01:53dot product of this vector
01:54either PA or AP. I'm
01:59actually going to go, I
02:00am going to go PA,
02:04will be a little bit
02:05easier to do. So we'll
02:06do PA dot the direction
02:09vector, this one, one, three,
02:12negative one. So this vector
02:13dot this vector must equal
02:16zero. So let's write this
02:17down here a second. So
02:18I'm going to write PA.
02:20pA, that vector dot the
02:24direction vector, 1, 3, negative
02:281. This dot, this has
02:30to equal 0. Okay, but
02:35what is pA? Let's find
02:37pA. So pA is A
02:42minus p, remember, I told
02:43you that. I always said
02:44that's very, very important. If
02:47not, go back and revise
02:48these are vectors, but PA
02:49is A minus P, where
02:50A and P are the
02:51position vectors of capital A
02:54and capital P respectively. So
02:56I have little P, I
02:58have the, because P, remember
03:01is two negative one, three,
03:03that's the point. So I
03:04have the position vector, the
03:05position vector is just two
03:07negative three, negative one, three
03:08as a vector. But what's
03:10the position vector of A?
03:13Well, the position vector of
03:14A is this,
03:16because it's some point, I
03:18don't know what it is
03:18yet, but it's some point
03:20on the line and all
03:21the points on the line
03:22have this position vector. So
03:24it's actually one plus one
03:27lambda, one plus lambda, zero
03:29plus three lambda. So that's
03:31just three lambda and two
03:33minus lambda, two minus lambda.
03:35That is the position vector
03:38of point A. And I'm
03:40going to do that minus
03:41P, which is two
03:44negative 1, 3. So this
03:48is PA, and this is
03:50going to be 1 plus
03:52lambda minus 2, which is
03:54minus 1 plus lambda, 3
03:57lambda plus 1, which is
03:593 lambda plus 1, and
04:012 minus lambda minus 3
04:03is minus 1 minus lambda.
04:09This is A minus P.
04:12or PA. So this dot
04:17this must equal zero. Let's
04:21bring that down here. Therefore,
04:25all right, therefore, therefore, minus
04:28one plus lambda, three lambda
04:31plus one, minus one, minus
04:35lambda, this dot this
04:40the direction vector 1, 3,
04:43negative 1 equals 0. Okay,
04:46let me re -explain why
04:49I'm doing this. So I
04:51want to find a lot,
04:54I want to find the
04:55point A. And I know
04:58that this vector PA, which
05:02ends up in this, and
05:03the direction vector are perpendicular,
05:05hence their dot product equals
05:070. Okay.
05:08Now, you can see here
05:10once I have a dot
05:11product equals zero and one
05:12variable, I'm going to be
05:14able to find lambda. Hence,
05:16I'm going to be able
05:17to find the position vector
05:19A, or I'll even be
05:22able to find the vector
05:23PA, which is this. This
05:26is PA. Okay, so this,
05:28the dot product of this
05:30times of this and this
05:31is this times one, which
05:33is minus one plus lambda,
05:36plus
05:36This times 3, which is
05:399 lambda plus 3, plus
05:43this times negative 1, which
05:45is 1 plus lambda negatives,
05:50negative 2. Negative times negative
05:52is positive, and this equals
05:530. So I'm left with
05:579 lambda, 10 lambda, 11
06:00lambda,
06:04equals negative one plus one
06:07is gone negative three. So
06:09lambda equals negative three over
06:1311. Fine, nearly there. Therefore,
06:20I have my lambda. This
06:22is PA. So let's do
06:23therefore again, therefore PA is
06:29equal to this with this
06:31for lambda. So I have
06:33minus 1 minus 3 1.
06:383 times 3 times minus
06:433 1. 3 times minus
06:473 1. And minus 1
06:53plus minus minus plus 3
06:571.
07:01And that's obviously simplified. This
07:03minus 1 minus 3 1
07:041 is minus 14 1
07:071 minus 11. Minus 11,
07:11minus 14 1 1. This
07:13will be minus 9 1
07:151 plus 1. Which is
07:192 1 and then finally
07:23minus 1 plus 3 1
07:261 will be minus
07:29is eight 11ths. And let's
07:34finish this off. The magnitude
07:37of PA, which is going
07:38to be the shortest distance
07:40I told you this wasn't
07:41easy. The shortest distance between
07:42the point and the line
07:43is going to be the
07:44magnitude of this. So it
07:50is 14 11th squared plus
07:56two.
07:5711th squared plus eight 11th
08:02squared and this I'm definitely
08:06Not gonna do it on
08:09my own I am I
08:11gonna use my calculator just
08:12while that is opening let's
08:14go back here and look
08:14at the question The shortest
08:17distance between the point and
08:19the line is going to
08:20be the distance of the
08:22vector PA hence why I've
08:24gone to all this trouble
08:25I could have found the
08:29point A using the same
08:30method, subbing it once I
08:32got lambda, I could have
08:32subbed it in here, and
08:34then found the distance between
08:35the two points. But what
08:36I'm doing is essentially exactly
08:38the same thing. So let
08:43me just move this here.
08:46I have a calculator, and
08:51I'm going to do the
08:53square root of the square
08:53root of the square root
08:53of the square root of
08:53the square root of the
08:53square root of 14 11th
08:58squared. So I need a
09:03bracket and I need this
09:06and I need 14 over
09:0811 and I need to
09:12close my bracket squared plus
09:17two 11ths. Let's just do
09:19it like this two 11ths.
09:21squared plus eight. Eight, 11th
09:32squared gives me 1 .4771
09:36equals 1 .4771. That's good
09:44enough for me and that's
09:46it. So yeah, certainly guys,
09:48I've seen that question and
09:49number
09:49times and past, past paper
09:51questions. It's, it, it actually,
09:53it's one of those that
09:54actually looks easier than it
09:56is. It's like shortest distance
09:57pre and point in line.
09:58That can't be hard, but
10:00it is because you have
10:01to go through all this,
10:02all this process. So hopefully
10:04that makes sense. Definitely practice
10:05one or two of those.
10:06Well, no, more than one
10:08or two, practice like seven
10:09or eight of them to
10:11make sure you know what
10:12you're doing. Okay, see you
10:13in the next lesson.