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Hi, the topic of this video is scalar triple
product, that is a very important topic for

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00:00:04,950 --> 00:00:05,950
JEE.

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00:00:05,950 --> 00:00:09,650
I will probably say this is the most important
topic for JEE, more than cross product more

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00:00:09,650 --> 00:00:14,250
than dot product because this combines cross
product and dot product.

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There are a lot of questions which come in
JEE just based on scalar triple product.

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00:00:18,590 --> 00:00:23,230
It is a very important concept and I hope
that this video will be able to provide you

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a basic understanding of scalar triple product.

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The idea behind video is to first introduce
scalar triple product and then do a few problems

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on this topic.

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There will also be a followup video in which
we will go for more examples of scalar triple

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product.

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So “what is a scalar triple product”?

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Till now we have done dot product which was
a \dot b

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We have done cross product which was a \cross
b.

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Now this is a scalar triple product.

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So as the name suggests — triple means there
are three quantities: vector a, vector b,

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vector c — and it is a scalar product.

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Like dot product was a scalar product, this
is also a scalar product but there will be

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three vector quantities, a b and c.

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And the output

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Hi, the topic of this video is scalar triple
product, that is a very important topic for

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00:01:07,840 --> 00:01:08,840
JEE.

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00:01:08,840 --> 00:01:09,840
I will probably say this is the most important
topic for JEE, more than cross product more

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00:01:09,840 --> 00:01:10,840
than dot product because this combines cross
product and dot product.

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There are a lot of questions which come in
JEE just based on scalar triple product. It

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is a very important concept and I hope that
this video will be able to provide you a basic

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understanding of scalar triple product.

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The idea behind video is to first introduce
scalar triple product and then do a few problems

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on this topic.

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There will also be a followup video in which
we will go for more examples of scalar triple

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product.

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00:01:17,840 --> 00:01:18,840
So “what is a scalar triple product”?

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Till now we have done dot product which was
a \dot b

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We have done cross product which was a \cross
b.

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Now this is a scalar triple product.

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So as the name suggests — triple means there
are three quantities: vector a, vector b,

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vector c — and it is a scalar product.

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Like dot product was a scalar product, this
is also a scalar product but there will be

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three vector quantities, a b and c. And the
output would be a scalar.

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Let us first discuss how we can get a scalar
quantity out of three vectors.

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Let us say we have vector a, and we have vector
b, and we have vector c.

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Now, how can I get a scalar quantity out of
this?

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Can I do a \dot b \dot c?

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This would mean that (let me put brackets
here) a \dot b is a scalar quantity. And I

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cannot dot a scalar with c so. Thus this not
possible.

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Can I do a \cross b \cross c.

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This would mean that this is a vector quantity
as I am taking a vector crossed with another

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vector - and that will give a vector output.

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However, we want a scalar output so this is
also not possible. We will also be discussing

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what kind of product is this but that is the
part of the next videos.

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However, can we do something like this - a
\cross b \dot c. This is vector dotted with

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another vector, and thus gives us scalar output.
This is a correct one. This is what we want

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to know.

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Scalar tripe product is defined as

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a \cross b \dot c. So what does this mean?
How can we then get the answer for the the

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scalar tripe product. The expansion would
be something like this.

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You can imagine that basically you have \i
component and you multiply all of that with

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cx. So rather than doing \i \j \k separately
you can just put cx cy cz in the determinant

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at the top.

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This is an important property which i have
written down. You are doing a \cross b \dot

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c.

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Now can you see something else also which
might or might not be that obvious? We know

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that (a \cross b) \dot c = c \dot (a \cross
b). We know that because this is a dot product.

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Let us call this vector d. Then d \dot c = c
\dot d. So all you are saying is (a \cross

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b) dot c = c \dot (a \cross b)

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Also, there is a property of determinants
that basically means that you can flip two

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rows, the answer would become negative of
the actual answer, and thus if you flip again

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(the second time), then the answer would remain
the same.

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So if I flip (row) a and (row) c, then (row)
a will come to the top and row (c) will come

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to the middle.

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Then when I flip (row) b and (row) c, it will
become a b c. In other words (a \cross b)

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\dot c can also be written as | ax ay az;
bx by bz; cx cy cz|.

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I have done this by flipping rows twice — flipping
c row and a row, and then b row and c row.

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However, if you think about this, whatever
was in the \dot came at the top.

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I can write this also as a \dot ( b \cross
c).

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What I am saying is whatever is in the dot
came at the top row or the single vector came

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at the top row.

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I have written this down as (a \cross b) \dot
c = a \dot (b \cross c). This is a super important

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property.

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a \dot (b \cross c) = (a \cross b) \dot c.
In other words, you can flip the sign of \dot

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and \cross.

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The first property that we have discussed
was this.

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The second property that I want to discuss
is (a \cross b) \dot c = a \dot (b \cross

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c).

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Third property is a \dot (b \cross c) = (a
\cross b) \dot c. And sometimes

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this is written as [a b c], since the position
of \dot and \cross doesn’t matter.

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It doesn’t matter you are taking dot here
or cross here. It will always come out to

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be same.

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I hope that this is making sense: we first
said in a determinant and when we flipped

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the rows twice the value remains the same.
Also because the top row belonged to the alone

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vector, we said a \dot (b \cross c) is same
as (a \cross b) \dot c.

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We also know that (a \cross b) \dot c =
c \dot (a \cross b) = (c \cross a) \dot b.

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Or this can also be [c a b]. Because now we
are taking the associative property of dot.

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Similarly, now I can write this as b \dot
(c \cross a). Am flipping the dots here again,

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and now i am changing this as (b \cross c)
\dot a i.e. changing dot and cross. And this

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is [b c a].

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This is a very very important property.

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What i am trying to say here is (a \cross
b) \dot c. This seems to be a cycling property.

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Cyclic property says that a should go to b,
and then b should go to c, and then c should

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go to a.

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So what is happening is [a b c]. You can always
think it terms of cyclic [a b c]

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The cycle is always [a b c] and that means
as long as you have [a b c] written in the

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correct cycle, then the value remains the
same.

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In other words, [a b c] = [c a b] = [b c a].

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I just want to highlight one other thing after,
[a b c] = - [a c b].

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So if you flip two things, the cycle breaks.

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It is not [a b c], it is [a c b] and so you
have broken the cycle. Or you have reversed

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the cycle.

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So rather than [a b c], you are now going
[a c b].

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Thats why you have a negative sign and you
can check that so if you just flip two rows

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- just b and c - only once you do row exchange,
then there is negative sign in front of the

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determinant.

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Thats why there is a negative sign here.

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Again you can make the cycle that [a b c]
= - [b a c]. The cycle is again [a c b], not

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[a b c].

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This [a b c] = - [c b a]

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This is a very very important property. I
hope that you are able to now understand what

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we are doing.

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We discussed a lot of things very quickly.

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This is a scalar product: (a \cross b) \dot
c — because (a \cross b) is a vector and

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then you are dotting it with (vector) c.

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The second thing we discussed was that you
can flip the position of \dot and \cross.

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We showed through the determinant property
that when you flip rows twice, you can conclude

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that (a \cross b) \dot c = a \dot (b \cross
c).

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The third property is that there is a cyclic
property to the whole system.

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We saw this by exchanging the \dot and \cross.
And by using the property of dot product i.e.

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(a \cross b) \dot c = c \dot (a \cross b).

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You should take some time to digest these
things because these will be used in and out.

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They are very very important for the chapter
of vectors.

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So please please take your time to understand
these things. Slowly go through this by writing

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these equations and then you should be able
to understand this.

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I don’t think this a very difficult concept
to grasp but for once you have to really understand

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it.

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So two things: you can exchange the sign of
\dot and \cross, and there is a cyclic property.

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For evaluating (a \cross b) \dot c, you can
just use the determinant.

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The next thing I want to discuss — like
all other products — is the physical significance.

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What is the physical significance of (a \cross
b) \dot c.

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What basically we have to realize is that
this condition a \dot ( b \cross c) =0, then

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a b c are coplanar.

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This is the first part of the significance.
I want to stress on this quite a lot because

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you will often use this in questions.

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What does this mean?

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Let us say you have two vectors, b and c.

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Two vectors are always coplanar. If you have
two vectors, you can always pass a plane through

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them.

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b and c are two vectors and you have passed
a plane through them.

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(b \cross c) is a vector which is perpendicular
to both b and c. This is something we have

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discussed in the topic of cross product and
if you can’t recall it, please go and check

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back the notes on cross product.

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Whenever we have (b \cross c), let us say
n, it is perpendicular to both b and c.

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What we are saying is that this n vector is
perpendicular to the plane of b and c.

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Because b \cross c is n, a \dot n = 0. What
does a \dot n = 0 mean? This mean a and n

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are perpendicular.

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If a and n are perpendicular, a has to lie
in the plane of b and c. or a, b, c are coplanar.

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What does coplanar mean? It means that if
there are three vectors, you can pass a plane

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through them.

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With two vectors, you can always pass a plane
through them.

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Let us say three vectors are like this (demonstration),
you cannot pass a plane through them.

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Only with two vectors, you can always pass
a plane.

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If there are three vectors, you cannot always
pass a plane through them. When can you pass

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a plane through them? Whenever a \dot (b \cross
c) = 0 or the scalar triple product = 0.

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How are we explaining this?

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b \cross c is a vector perpendicular to both
b and c. Or is perpendicular to the plane

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of b and c. When a \dot n = 0, that implies
a and (b \cross c) are perpendicular. Since

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dot product is zero whenever two vectors are
perpendicular. So this forces a to lie in

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the plane of b and c.

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This is very important and you just have to
remember that whenever three vectors are coplanar,

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the scalar triple product is 0.

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Coplanarity also means
that you can represent the third vectors as

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linear combination of the first two vectors.
This is also a way to understand coplanarity.

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When can you represent a vector as a linear
combination of a and b?

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Let us say there are a and b, and a plane
passes through them. Then basically you are

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saying that you are stretching one vector
or compressing a vector as l and m can take

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any real values.

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Let us say you made one vector half and other
vector double, and then now you are adding

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them. So basically completing some kind of
addition, and addition forces you to not leave

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that plane.

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That is why these three things are coplanar.

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You can also imagine that if I put c=la+mb
here: if I take cross of b with b, it will

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become zero. Then, we will be left with (a
\cross b) \dot a. Then you can flip \dot and

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\cross, and it will become (a \cross a) \dot
b. Cross of a with a will also be zero.

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You can check. You can just put this expression
back and check that this always comes out

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to be zero.

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So coplanarity means that one vector can also
be written as the linear combination of other

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two vectors, and scalar triple product [a
b c] is zero.

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I hope that is giving some insight into the
significance of scalar tripe product.

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Dot product was the projection of a vector
on the other vector.

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00:18:05,070 --> 00:18:07,870
Cross product was the vector perpendicular
to both the vectors.

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Scalar triple product = 0 means a condition
for coplanarity.

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00:18:13,230 --> 00:18:28,039
Second thing that I want you to learn as a
formula, if there is

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a parallelopiped — parallelopiped is like
a 3-D parallelogram. Volume of parallelopiped

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is [a b c] where three sides are a vector,
b vector and c vector.

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00:18:57,540 --> 00:19:02,460
There are a lot of proofs for this but I don’t
think you need to understand the proofs.

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a vector is this entire side, c vector is
this entire side, and b vector is this entire

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00:19:10,350 --> 00:19:17,070
side. Then the volume of parallelopiped is
[a b c].

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00:19:17,070 --> 00:19:59,840
If there is a tetrahedron, with sides a, b
and c, then volume would be [a b c]/6.

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00:19:59,840 --> 00:20:05,309
If this is a little bit unclear to you, don’t
worry too much about this. Just go back and

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00:20:05,309 --> 00:20:07,150
look at some images of tetrahedron and parallelopiped.

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00:20:07,150 --> 00:20:11,980
Just remember that whenever the side are a,b
and c, then the volume of parallelopiped is

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[a b c] and the volume of tetrahederon is
[a b c]/6.

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00:20:18,020 --> 00:20:28,220
Let us recap whatever we have discussed. This
ends the property discussion of scalar triple

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product.

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00:20:29,220 --> 00:20:36,110
We have discussed that (a \cross b) \dot c
can be written as a determinant.

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00:20:36,110 --> 00:20:47,250
We also learned that you can flip \dot and
\cross. I showed this by flipping the rows

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00:20:47,250 --> 00:20:50,210
twice and hence (a \cross b) \dot c = (a \cross
b) \dot c.

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00:20:50,210 --> 00:20:53,700
Then we learned about the cyclic property
of [a b c]. And that there will be a minus

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sign whenever the cycle is broken or reversed.

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00:20:55,809 --> 00:20:59,780
Then we discussed two things about physical
significance.

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Condition for coplanarity: one vector is a
linear combination of other two vectors and

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that is also expressed in terms of (a \cross
b) \dot c = 0. Since b \cross c is a vector

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perpendicular to the plane of b and c, and
if a \dot n=0, a has to lie in the place of

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b and c.

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And finally, if a,b and c are sides of parallelepiped
and tetrahedron, then the volume is [a b c]

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and [a b c]/6.

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These are the properties we have discussed.
Please remember them and learn them. They

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should be on your fingertips whenever you
see them.

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Let us try to do some very quick problems.

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First problem is: If a, b, c are coplanar,
what is

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the value of [2a - b, 2b - c, 2c -a]?

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We have been told that there are three vectors
a,b,c that are coplanar. Then we have been

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asked to find the scalar triple product of
vectors of the linear combination of a,b and

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c.

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One way to solve this problem is to take a
cross and then take dot, and trying to find

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something in terms of [a b c].

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My point here is to really make you understand
what this means. Think about three vectors

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that are coplanar: you have two vectors like
this and third vector like this — they are

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in a plane.

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What you are doing is that you are just taking
a linear combination of these three vectors.

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Can you ever go out of the plane? Can you
ever create a vector which is out of the plane?

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For instance, if c = la + mb, since a, b,
c are coplanar.

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Then 2a-b is a linear combination of a and
b.

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2b-c, because c = la + mb, is also a linear
combination of a and b.

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2c-a is also a linear combination of a and
c.

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So even all these vectors are just linear
combination of a and b. That means that they

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are in the same plane. So you don’t even
need to do anything. You don’t have to calculate.

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You can just directly write that this is equal
to zero.

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Because a, b, c are coplanar, any linear combination
of these vectors will also lie in the same

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plane. So these three vectors are in the same
plane. So you just have to understand that

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since these vectors a coplanar, their linear
combinations has to be coplanar, so this has

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to be zero since this is condition of scalar
triple product.

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I can bet that a lot of students will try
to solve this problem. They will start with

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[a b c]=0 and then they will expand this by
taking a cross here and dot here, and opening

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up the brackets and trying to solve it somehow.
All you have to do is to understand that coplanarity

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means that that linear combination of the
vectors will lie in the same plane.

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I hope that this gives you some insight about
scalar triple product.

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The second problem that I want to do is: What
is the value of \lambda such that \lambda

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\i + \j + \k, \i + 2 \lambda \j + \k and \i
+ \j + \k are coplanar.

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So what is
the value of \lambda such that these three

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vectors are coplanar.

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The question is begging you to remember the
condition of coplanarity that a \dot (b \cross

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c) = 0. Thats it.

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How can we write a \dot (b \cross c) is as
we have discussed a determinant. Let us call

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this a vector, b vector and c vector.

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Then a \dot (b \cross c) = |\lambda 1 1; 1
2\lamba 1; 1 1 1 | = 0.

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I hope with time you will become quick in
opening up determinants.

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If you open up the determinant, you will find
that this is 2 \lambda ^ 2 - 3 \lambda + 1

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= 0. You can check that this what it comes
out.

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That would give you the answer \lambda = 1,
1/2. In an objective test, this could be a

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multiple-choice problem.

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Can we check what the answer \lambda = 1 mean?
If \lambda = 1, then this becomes \i + \j

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+ \k and then this becomes \i + \j + \k. So
they are same vectors. So if they are the

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same vectors, that means their cross will
always be zero.

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So if they are the same vectors, this basically
become two vectors, which will obviously be

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coplanar. Rather than three vectors, you have
basically two vectors which are always coplanar.

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If \lambda=1/2, then this become \i + \j +
\k, and then these two become same. So again

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they become two vectors and they are always
coplanar.

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I hope that with this you are physically able
to understand the meaning of scale triple

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product.

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In this video, we covered very important topic
of scalar triple product. We discussed its

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properties. We discussed 4 properties. And
then we did two examples to understand the

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meaning of scalar triple product and how to
apply it.

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In the next video, we will discuss vector
triple product. If you enjoyed this video,

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please check out the next video. Thank you!