Question

Let the angle between two non-zero vectors $\overrightarrow A$ and $\overrightarrow B$ be $120^\circ$ and its resultant be$\overrightarrow C$. Then which of the following is correct?
(a) $C$ Must be equal to $\left| {\overrightarrow A - \overrightarrow B } \right|$
(b) $C$ Must be less than $\left| {\overrightarrow A - \overrightarrow B } \right|$
(c) $C$ Must be greater than $\left| {\overrightarrow A - \overrightarrow B } \right|$
(d) $C$ May be equal to $\left| {\overrightarrow A - \overrightarrow B } \right|$

Answer 1

In this type of problem first we have to find the angle between two vectors. There is a formula for finding we have to use it. After that we have to identify the vector and also write all the information you concerning the two vectors. Then you have to determine the length of each of the vectors. This problem also asks directly in multiple choice question types.
Formula/Concept Used: The angle between two non-zero vectors $\overrightarrow A$ and $\overrightarrow B$ and the angle between them is $\theta$ then,
$A.B = \left| A \right|\left| B \right|\cos \theta$.

Complete step by step solution:
Let $A$ and $B$ are the two angles, then according to the question
The angle between two non-zero vectors $\overrightarrow A$ and $\overrightarrow B$ with angle between them $\theta$ is
$A.B = \left| A \right|\left| B \right|\cos \theta$…………………. (i)
Now as we know that $\theta = 120^\circ$as per given in the question, we will place the value of $\theta$in the equation (i)
$A.B = \left| A \right|\left| B \right|\cos (120^\circ )$
Now we know that value of $\cos (120^\circ ) = - \dfrac{1}{2}$
$A.B = \dfrac{1}{2}\left| A \right|\left| B \right|$……………………… (ii)
Now according to the question
$\overrightarrow C = \overrightarrow A + \overrightarrow B$
Now we will take square on both sides then the resulted equation will be
${C^2} = \left( {A + B} \right).\left( {A + B} \right)$
Now as know that $(A + B).(A + {\kern 1pt} B) = {A^2} + {B^2} + 2A.{\kern 1pt} B$
$\Rightarrow {A^2} + {B^2} + 2A.B$
Now we know that $A.B = - \left| A \right|\left| B \right|$
$\Rightarrow {A^2} + {B^2} - 2\left| A \right|\left| B \right|$
$\Rightarrow {\left| {A - B} \right|^2}$
Now we will open the square of the above equation
$\Rightarrow \left( {A - B} \right).\left( {A - B} \right)$
Now as know that $(A - B).(A - {\kern 1pt} B) = {A^2} + {B^2} - 2A.{\kern 1pt} B$
$\Rightarrow {A^2} + {B^2} - 2A.B$
$\Rightarrow {A^2} + {B^2} + \left| A \right|\left| B \right|$
$\Rightarrow {\left| {A - B} \right|^2} - {C^2} = 2\left| A \right|\left| B \right|$

According to the above equation we can say that
${C^2} < {\left| {A - B} \right|^2}$ Given the condition that $A,B \ne 0$
Now we will cancel the squares on both sides and the resultant equation will be
$\Rightarrow \left| C \right| < \left| {A - B} \right|$

So, the correct answer is “Option B”.

Note: In this type of problem it also asks to convert a from direction and magnitude. While solving the problem check all the possibilities that are given in the problem and what we have to find. Then you have to analyze the problem of which formula is used there. There is always a word that is present in the question that tells us about which formula you have to use. These are some important points which you have to remember before solving the problem.