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Now in multiplication of vectors, we will discuss the dot product which we also call the scalar product or the inner product. Two vectors can be multiplied by two ways only, either dot product or cross product.
So, let me consider a two coordinate system means a planar coordinate system. Let's say I have a vector \(\vec { a } \) which is completely horizontal and is having only one component, I have another vector \(\vec b\) which is inclined, so it is in two components. \(\theta \) is the angle between vector \(\vec a \) and \(\vec b \). Now I would like to find their interaction or their multiplication. $$ \vec { a } .\vec { b } =ab\cos { \theta } $$ The component which is along \(\vec { a }\) is \(b\cos { \theta }\)
\(\cos { \theta } \) I defined as $$ \frac { base }{ hypotenous } =\cos { \theta } $$ Here hypotenuse is equal to \( \vec { b } \), so $$ base=b\cos { \theta } $$ Now if we say that what we are using \(\cos { \theta }\) in dot products then this \(\cos { \theta }\) is actually coming from the fact that we only want to calculate the component of vector \(\vec b \) which is along vector \(\vec a \). Similarly, $$ \vec { b } .\vec { a } =ab\cos { \theta } $$ The dot product of two vectors is commutative which means you can change their order. $$ \vec { a } .\vec { b } =\vec { b } .\vec { a } $$ I can now say that $$ \vec { a } .\vec { b } =a{ b }_{ x } $$ $$ \vec { b } .\vec { a } =b{ a }_{ x } $$ If \(\theta ={ 0 }^{ o } \), it means that the \(\vec b \) vector will come exactly over vector \(\vec b \) so both will be in the same direction. $$ \vec { a } .\vec { b } =ab $$ Here the direction is not playing the role. If \(\theta =9{ 0 }^{ o }\) $$ \vec { a } .\vec { b } =0 $$ The vector \(\vec b \) is completely perpendicular to vector \(\vec a \) and perpendicular vector is having no projection. The other thing is that if \(- \theta \) then as \(cosine \) is an even function and \(cosine \) of \(- \theta \) will be equal to \(\cos { \theta } \). The limits of this \(\theta \) is in the first and second quadrant. $$ 0\le \theta \le { 180 }^{ o } $$ Or in terms of radians I can write $$ 0\le \theta \le \pi radians $$ Let's say I take one example, my vectors are now not just ordinary vectors but they are some physical quantities. Let's say one physical quantity that I'm considering is force and the other is displacement.
How I would multiply these two vectors? I cannot multiply these two the force and the displacement like ordinary numbers. I will have to multiply them like suppose I'm pushing a block of mass \(m \), then $$ \vec { F } .\vec { x } =Fx\cos { \theta } $$ If \(\theta ={ 0 }^{ o }\), and \(\vec { x } \) are along same direction $$ Fx=W $$ We call this work.
If we take \(\theta ={ 30 }^{ o }\) $$ \frac { Fx }{ 2 } =\frac { W }{ 2 } $$ I will do half of the work in this case.
If \(\theta ={ 90 }^{ o }\), then I know that I'm doing no work. Here the product is work which is a scalar.
Another question is that if we have a mass \(m\) and a crane is lifting this mass up. Is the crane doing any work? The answer is, Yes. The crane is doing maximum work, why? because the \(\vec { F } \) and the \(\vec { x } \) both are in upward direction. Now for example $$ \vec { a } =3\hat { x } +2\hat { y } +4\hat { z } $$ $$ \vec { b } =2\hat { x } +4\hat { y } +\hat { z } $$ What will be \(\vec { a } .\vec { b } \)? $$ \vec { a } .\vec { b } =6+8+4=18 $$ As the inner components are being multiplied this why we call this as inner product as well.
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