Examples : Vector Product Video Lecture | Mathematics (Maths) Class 12 - JEE

hello friends this video on vector
algebra par 27 is brought you by exam
viacom no more fear from exam before
watching this video please make sure
that you have got part 1 to 426 let's
take some examples of vector cross
product now so here we have this two
vectors a and B even here to find a
cross B please note you define magnitude
of a cross B here you don't have to find
a cross B we have to find magnitude of a
cross B and this will be scaler
obviously right I'll show you how so to
start we have another formula of a cross
B was nothing but i JK even a 2 a 3 and
B 1 B 2 is correct this formula video
where I have vector a is even I cap plus
a 2 J cap plus if we keep cap and vector
B is B 1 I cap plus V 2 J cap plus B 3
Giga now if we compare these two
equations I can say that this guy is a 1
this guy's a - this guy's a 3 this guy
is B 1 B 2 and B 3 so in this case my 8
cross B be nothing but i JK and here
have 1 - 7 7 1 - 7 7 here I have 3 - 2 -
3 - 2 and correct this is nothing but I
into - 7 - 2 is minus 14 and minus 77
into minus 2 is again minus 14 minus -
alright - J into 1 into 2 is 2 + 3 into
7 is 21 minus 21 plus K into 1 into
minus 2 is minus 2 and minus 7 into - -7
industry is minus 21 C minus general so
if you're not understanding this
matrices
please wash my mattress videos you
to understand how I am doing this so
this becomes minus 14 plus 14 res 0 I is
gone and this is minus j2 minus 19 that
is 19 G Plus this guy's again 19 K so a
cross B is 19 J plus 19 K cap but I am
asked to find magnitude of a cross B
magnitude of a cross B will be nothing
but root of this square plus this square
that is 19 square plus a square then
that would be 19 to do i do 19 plus 19
is 19 let's take an example here we
define the per unit vector we define the
unit vector which is perpendicular to
this vector and this vector where is the
service so to do this what can we do to
find a perpendicular a vector that is
perpendicular to vector let's suppose P
and Q right what we have to do is we to
find P cross Q correct this will be some
vector which is perpendicular to P cross
Q now I have to find the unit vector so
I will say if we cross Q divided by
magnitude of P customer so this will be
nothing but my unit vector that is
perpendicular to P cross P orbitals can
see if I have period tablets up for some
vector T Phi a stomach that Q here my P
cross Q will be some vector and suppose
this guy this guy will be P cross P
right P cross Q will be some vector
perpendicular to PNG now I have to find
a unit vector which is per which is in
this direction
fine unit vector I had this vector you
divide this vector the magnitude of this
vector you getting it and that's what we
do what is P and P is a tell us B and Q
is immensely so let mine be first P
rector is nothing but a vector plus B
actor that does nothing but three plus
one there is a for I 2 plus 2 for g u-
to 0 that is p Repton Q vector is
nothing but a vector minus reactant a
vector minus V that is 3 minus 1 is 2 I
will take J bar to J minus 2 j 0 j to k
minus -2 that is plus virginia so this
is my Hewitt correct what I am
interested in 0 K what I am interested
in is P cross Q by magnitude of P cross
P so first ad to find P cross P this is
the every be nothing but ijk 4 4 0 4 0
right
- 0 4
hope this is clear deal because we had
this formula P cross Q if you are not
not clear with this watch the previous
video where we explain this so this
becomes I into 4 into 4 is 16 minus 0
into 0 is 0 minus J into 4 into 4 16
minus 2 into 0 is 0 plus J cap into all
this cap here for into 0 is 0 minus 4
into 2 is 8 0 minus this is nothing but
16 I minus 16 minus 8
keaghlan this is my P cross Q what I'm
looking for I am looking for P cross Q
by magnitude of P cross P so let's find
magnitude of P cross 1 this becomes root
of 16 square 16 square H square that is
16 square plus 16 square + H so this is
all you get 24 so my unit vector lid
this BM vector is nothing but my P cross
Q this vector divided by magnitude of P
cross Q because I'm looking for the unit
vector in this direction right so you
inductor is nothing but this vector
divided the manage of this vector so
this becomes 16 I minus 16 J minus HK
all cap divided by 20 this is nothing
but 16 by 24 I cap minus 16 by 24 J cap
minus a by 24 so you want to solve this
further you can solve this 8 is a common
here as it comes to by 3 I can minus 2
by 3 J cap minus 1 by and that is my
unit vector that is perpendicular to my
P and Q vector where P vector is a plus
B and Q vector is a - I have this is
clear so here find any vector that is
perpendicular to right there whatever do
you do a cross product of the two
vectors example in this case I want to
find the vector that's perpendicular to
P and Q this will be P cross P but now
I'm supposed to find the unit vector
that is perpendicular to P and Q in that
case I had this vector to make it a unit
vector the only thing I will do is I
will divide this vector by the magnitude
of this vector so this vector P cross Q
I divide this by the magnitude of P
cross Q and what you get is some n
vector this guy is perpendicular
pthen children so this
this clearing can solve more examples on
this the next we should see is that if a
unit vector the unit vector e where
magnitude of a is 1 obviously makes
angle PI by 3 with I PI by 4 the G and
some acute angle became fine angle theta
so if you know that I write a as a 1 I
plus a 2 G Plus a 3 K right I write a
vector s something like this I can also
write as magnitude of a into cos alpha I
plus cos theta a plus cos gamma XI cap
when alpha is the angle this vector
makes with I can I direction this x-axis
the beta is the angle with this vector
with mix with y-axis and gamma is the
angle which makes it the next I think we
have done this in the scalar part where
you know that we can write both are same
actually
so now we had told that manager of a is
one anyway so 1 into alpha is PI by 3
correct so cos PI by 3 I plus this is
beta is PI by 4 cos Phi by 2 G cap and
gamma is some angle theta chicken this
is my vector right
not since the magnitude of this vector
is one my a cap I can also write as
nothing bad
this guy is wanting to cause pi by three
ah let me factor is further cos PI by
three is what 1 by 2 mu becomes 1 by 2 I
cos PI by 4 is 1 by root 2 1 by J and
this is cos theta he correct so this is
my vector actually this is my victim
thank you is 1 so the magnitude of this
I can also write as if this is my vector
I found this is a vector I can write
that as a nothing but magnitude of this
I component square which is 1 by 2
square plus magnitude of J component
square that is 1 by root 2 square was
key component square cos theta school
and this is given one already given this
one correct so this becomes 1 by 4 plus
1 by 2 plus cos theta cos square theta
is 1 because if you square also this
come once or you get thing is aligned
cos square theta is 1 minus 1 by 4 plus
1 by 2 is 3 by 4 1 minus 3 by 4 that is
one better so theta is cos theta is plus
minus 1 by 2 because cos square theta is
1 by 4 so cos theta is root of this guy
right 1 by 4 so the plus plus 1 by 2 so
I say that theta is either PI by 3 you
can take minus 1 because it's told that
is a cute angle so we'll take theta is
PI by 3 only and we'll take plus 1 by 2
because the stone theta is acute angle
correct
had tea table in an obtuse angle we
could have taken pi minus pi by three
also but this since told Zack you tangle
with a t-test by wedges so what I am
asked I am asked to find Conference of a
rector that is I have to find this guy
so here I have T tag input this value so
my a vector will be 1 by 2 I plus 1 by
rule j-cap plus cos theta is how much
plus 1 by 2 1 by 2 and this is my unit I
hope this is clear see I was told that
my vector makes alpha beta and gamma
angle with ijk I was told the value
I'll find me dumb I wrote this a vector
in this man's fashion a vector is
nothing but magnitude of a do cos alpha
i cos beta J and cos gamma K cap this
gave us one I put the value of alpha
beta and theta is Theta I got my a
vector s just I have to find another how
to find I know once again that magnitude
of a is 1 correct
since magnitude of a is 1 I use this
formula that the square with the square
of this value is 1 and that I got cos
theta is plus 1 by 2 r minus 1 by D so
what it is told the theta is a cute
angle so I'll take lost 1 by 2 pi I have
got it
minus 1 by T and this cos theta is plus
1 by 2 it just causes the plus for
matrix addition and regardless even
thank you
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