Question

For any two vectors $\vec{a}$ and $\vec{b}$ we have $|\vec{a}.\vec{b}|\_\_\_|\vec{a}|.|\vec{b}|$. Fill in the blank with appropriate inequality.
 \[\begin{align}
  & a)\le \\
 & b)\ge \\
 & c)< \\
 & d)> \\
\end{align}\]

Answer 1

Hint: We know that $\vec{a}.\vec{b}=|\vec{a}||\vec{b}|\cos \theta $ . Hence we can take modulus on both side and find $|\vec{a}.\vec{b}|=||\vec{a}||\vec{b}|\cos \theta |$. Now we will use the property that $|a.b|=|a|.|b|$. Now we also know that the range of $\cos \theta $ is from -1 to 1. Hence taking mod we can find that $|\cos \theta |\le 1$ . Hence we get relation between $|\vec{a}.\vec{b}|$ and $|\vec{a}|.|\vec{b}|$

Complete step by step answer:
Now let us say $\vec{a}$ and $\vec{b}$ are two vectors.
Now we will take dot product or scalar product of the vectors $\vec{a}$ and $\vec{b}$
We know that the dot product $\vec{a}.\vec{b}$ is given by $|\vec{a}|.|\vec{b}|.\cos \theta $
Hence now we have $\vec{a}.\vec{b}=|\vec{a}|.|\vec{b}|\cos \theta $
Now taking modulus on both side we get $|\vec{a}.\vec{b}|=||\vec{a}|.|\vec{b}|.\cos \theta |$
Now the values $|\vec{a}|,|\vec{b}|,\cos \theta $ are all scalar quantities and for scalars we have $|a.b|=|a|.|b|$
$|\vec{a}.\vec{b}|=||\vec{a}||.||\vec{b}||.|\cos \theta |$
Now we know that $|\vec{a}|$ is a scalar which is non negative since modulus is nothing but distance and distance is never negative and for any positive scalar we know that $|a|=a$ hence we have $||\vec{a}||=|\vec{a}|$.
Similarly $|\vec{b}|$ is a scalar which is non negative since modulus is nothing but distance and distance is never negative and for any positive scalar we know that $|a|=a$ hence we have $||\vec{b}||=|\vec{b}|$
Hence, we get
$|\vec{a}.\vec{b}|=|\vec{a}|.|\vec{b}|.|\cos \theta |................(1)$
Now we know that for $\cos \theta $ the range is from $\left[ -1,1 \right]$
This means $-1\le \cos \theta \le 1$
Now taking modulus we get \[0\le |\cos \theta |\le 1\]
Multiplying the equation with $|\vec{a}|.|\vec{b}|$ we get
$0\le |\vec{a}|.|\vec{b}||\cos \theta |\le |\vec{a}|.|\vec{b}|...........(2)$
Now from equation (1) and equation (2) we get
$|\vec{a}.\vec{b}|\le |\vec{a}||\vec{b}|$

So, the correct answer is “Option A”.

Note: Now dot product of two vectors is defined as \[\vec{a}.\vec{b}=|\vec{a}|.|\vec{b}|\cos \theta \] and magnitude of cross product of vector is $|\vec{a}\times \vec{b}|=|\vec{a}|.|\vec{b}|\sin \theta $ Note that the two similar looking quantities have quite different meanings.