Question

If $\vec{a}$ and $\vec{b}$ are two vectors then which of the following options is true

& a)|\vec{a}.\vec{b}|>|\vec{a}||\vec{b}| \\\ & b)|\vec{a}.\vec{b}|<|\vec{a}||\vec{b}| \\\ & c)|\vec{a}.\vec{b}|\ge |\vec{a}||\vec{b}| \\\ & d)|\vec{a}.\vec{b}|\le |\vec{a}||\vec{b}| \\\ \end{aligned}$$

Answer 1

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 consider $\vec{a}$ and $\vec{b}$ are two vectors.
Now we will take $\vec{a}.\vec{b}$ that is nothing but dot product or scalar product of the vectors.
We know that $\vec{a}.\vec{b}$ is given by $|\vec{a}|.|\vec{b}|.\cos \theta$
Note that the value on RHS is a scalar and not a vector
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 we now that for any scalars a and b we get $|a.b|=|a||b|$.
Now the values $|\vec{a}|,|\vec{b}|,\cos \theta$ are all scalar quantities
Hence using the above result we get
$|\vec{a}.\vec{b}|=||\vec{a}||.||\vec{b}||.|\cos \theta |$
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 the range of $\cos \theta$ is from -1 to 1.
Hence we get $-1\le \cos \theta \le 1$
Taking modulus we get $0\le |\cos \theta |\le 1$
Now 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 D”.

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 $|\vec{a}\times \vec{b}|=|\vec{a}|.|\vec{b}|\sin \theta$ , not to be confused among the two.