00:01
So in this question, we are looking at two vectors a and b, and we are looking at finding the scalar projection of a on b, the scalar projection of b on a, and the cosine of the angle between a and b.
00:14
Now let's look at these questions one by one.
00:18
The scalar projection, in mathematical sense, the scalar projection of a vector a on b vector, is given by the term it's given by the expression a vector dot b vector divided by modulus of the the vector that is being projected on which is b vector so with respect to that so we'll we'll calculate what b vector model is in the first place the modulus of b vector is given by the modulus of b vector is given by square root of the coefficients of b now if you look at the coefficients of b it's minus two uh two and minus one therefore we can say a square root of minus two the whole squared plus two the whole squared uh minus one the whole squared now that is equal to uh minus two the whole square is four plus two squared is also four plus one therefore you get nine square root of nine is equal to three so we get the value of 3 in this place for b vector and let's look at the dot product of a and b so a vector so let's write it down in the numerator we would get a vector which is 4i as you can see this is 4i 4i vector for a and you do not have a a j vector so i'm writing 0j but you do have a minus 3k in vector and this has to be dotted with the b b vector which is minus 2j i'm sorry minus 2i a vector plus 2 j vector plus minus k vector so if you did the dot product which is nothing but your a dot dot a vector .bctor.
02:47
This would give, so basically how we do it is we multiply the i's with eyes with jays, j s with j s and k's with k's.
02:58
So you get four into minus two, which is minus eight, you get minus eight and then zero into two will be zero, minus three into minus 1 is plus 3.
03:17
Therefore you get minus 8 plus 3 which is equal to minus 5.
03:23
So this is what you get for a .b.
03:31
So that's the a dot b vector.
03:34
Therefore you get so what we need is the projection of a on b which would mean a so rewriting a dot b vector divided by modulus of b vector is going to be so a dot b vector it a vector dot b vector is minus 5 divided by the modulus of b vector is 3 so they get minus 5 by 3 so that's the projection of a on b similarly we are looking at the projection of b on a because they've asked us the scalar projection of b on a so let's look at the scalar projection of b on a the scalar projection of b on a b vector on a vector is given by similar formula that we looked at here.
04:32
It would be b vector.
04:34
Dot a vector divided by a vector.
04:43
The modulus of a vector.
04:46
So the b vector dot, so before we look at the b vector dot a vector in the same fashion as we did in the previous problem, let's look at the modulus of a vector here.
04:58
The model is of a vector.
04:59
So let's look at the a vector once again.
05:01
So this is going to be 4i minus j 0 j minus 3k because there is no j in this case.
05:09
So just for the sake of completeness, we would say this is going to be square root of 4 squared plus 0 squared minus 3 plus minus 3 the whole square.
05:27
So that would be equal to square root of 4 square is 16 plus 3 squared is 9 which is going to be square root of 25 which is equal to 5...