Question

If we are given two vectors $\overrightarrow{a}\,and\,\overrightarrow{b}$ such that $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|$, then prove that the vector $2\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to the vector $\overrightarrow{b}$.

Answer 1

Hint: We are given one condition i.e. $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|$ so to solve this question we need to manipulate this condition to find some useful result.so, we will square both sides of the equation $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|$ to get some useful result. And, we know that if dot product of any two vectors is 0 then they are perpendicular to each other, given that none of them is zero. So we will find \[\left( 2\overrightarrow{a}+\overrightarrow{b} \right).\overrightarrow{b}\] and then prove it to be zero.

Complete step by step answer:
We are given two vectors $\overrightarrow{a}\,and\,\overrightarrow{b}$, such that
$|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|$
Now if we square both sides of the above equation, we get
${{\left( |\overrightarrow{a}+\overrightarrow{b}| \right)}^{2}}={{\left( |\overrightarrow{a}| \right)}^{2}}$
$|\overrightarrow{a}{{|}^{2}}+2\overrightarrow{a}.\overrightarrow{b}+|\overrightarrow{b}{{|}^{2}}=|\overrightarrow{a}{{|}^{2}}$
So we get,
$2\overrightarrow{a}.\overrightarrow{b}+|\overrightarrow{b}{{|}^{2}}=0\,\,...\left( 1 \right)$
Now if the vector $2\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to the vector $\overrightarrow{b}$ then their dot product must be zero,
Hence taking the dot product of the above two vectors, we get,
$\begin{align}
  & \left( 2\overrightarrow{a}+\overrightarrow{b} \right).\overrightarrow{b} \\
 & =2\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{b} \\
 & =2\overrightarrow{a}.\overrightarrow{b}+|\overrightarrow{b}{{|}^{2}} \\
\end{align}$
We can observe that the above expression is equal to the equation 1, hence
$2\overrightarrow{a}.\overrightarrow{b}+|\overrightarrow{b}{{|}^{2}}=0\,$

So, clearly we get dot product of the vector $2\overrightarrow{a}+\overrightarrow{b}$ and vector $\overrightarrow{b}$ as zero hence, they are perpendicular to each other.

Note: Whenever we are given to prove two vectors to be perpendicular or to prove two vectors parallel then remember most of times that kind of questions are solved using either by taking dot product of those two vectors or by taking cross product of those two vectors. And also while squaring the term $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|$ students may miss out the term $2\overrightarrow{a}.\overrightarrow{b}$ on the left side of the equation so, be cautious while squaring.