Question

Let $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$ be two non-null vectors such that$\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{a}}\,-2\overset{\to }{\mathop{b}}\, \right|$. Then value of $\dfrac{\left| \overset{\to }{\mathop{a}}\, \right|}{\left| \overset{\to }{\mathop{b}}\, \right|}$ may be:
$A.\,\dfrac{1}{4}$
$B.\,\dfrac{1}{8}$
$C.\,1$
$D.\,2$

Answer 1

This is a question based on the concept of the vectors. So, we will make use of the properties of the vectors to solve this problem. The condition of the equality is given, so, we will proceed with the same to start the calculation and will arrive at the solution by using the formula for computing the magnitude of the vectors.

Complete step by step solution:
From the data, we have the data as follows.
$\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$are the two non-null vectors. Here, the term ‘non-null’ vectors refers to the non empty vectors.
The condition is,
$\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{a}}\,-2\overset{\to }{\mathop{b}}\, \right|$.
As the magnitudes of the above vectors are equal, we will find the magnitudes of the above vectors first.
Taking square on the both sides of the equation, we get,
${{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\,-2\overset{\to }{\mathop{b}}\, \right|}^{2}}$
Continue the further calculations.
Now will make use of the algebraic identity to continue with the calculation.
${{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta ={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+4{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}-2\left| \overset{\to }{\mathop{a}}\, \right|\left| 2\overset{\to }{\mathop{b}}\, \right|\cos \theta$
Cancel out the common terms and continue the calculations.
$2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta +2\left| \overset{\to }{\mathop{a}}\, \right|\left| 2\overset{\to }{\mathop{b}}\, \right|\cos \theta =3{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}$
Upon further solving, we get,
$2\left| \overset{\to }{\mathop{a}}\, \right|\cos \theta =\left| \overset{\to }{\mathop{b}}\, \right|$
Now rearrange the terms to represent the equations in terms of $\dfrac{\left| \overset{\to }{\mathop{a}}\, \right|}{\left| \overset{\to }{\mathop{b}}\, \right|}$.

& 2\left| \overset{\to }{\mathop{a}}\, \right|\cos \theta =\left| \overset{\to }{\mathop{b}}\, \right| \\\ & \dfrac{1}{2}\le \dfrac{\left| \overset{\to }{\mathop{a}}\, \right|}{\left| \overset{\to }{\mathop{b}}\, \right|}=\dfrac{1}{2\cos \theta }<\infty \\\ \end{aligned}$$ Therefore, the range can be calculated as follows. $$1\le \dfrac{\left| \overset{\to }{\mathop{a}}\, \right|}{\left| \overset{\to }{\mathop{b}}\, \right|}\le 2$$ Thus, the value of $$\dfrac{\left| \overset{\to }{\mathop{a}}\, \right|}{\left| \overset{\to }{\mathop{b}}\, \right|}$$ranges between 1 and 2. $$\therefore $$ The value of $$\dfrac{\left| \overset{\to }{\mathop{a}}\, \right|}{\left| \overset{\to }{\mathop{b}}\, \right|}$$ may be 1 or 2. As, the value of value of $$\dfrac{\left| \overset{\to }{\mathop{a}}\, \right|}{\left| \overset{\to }{\mathop{b}}\, \right|}$$ may be 1 or 2. **Thus, the options (C) and (D) are correct.** **Note:** The method of finding the magnitude of the vectors should be known to solve this type of problems. From the given question statement, the condition using which the problem can be solved should be known. The properties of the vectors should also be known.