Question

If an unit vector $a\hat{i}+b\hat{j}$is perpendicular to $\left( \hat{i}-\hat{j} \right)$ then the value of $a$ and $b$ are : \[\]
A).$\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}$\[\]
B). $\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}$\[\]
C). $-\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}$\[\]
D). Cannot be determined \[\]

Answer 1

Hint: We use the condition of two vectors $\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j},\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}$ perpendicular that is ${{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}=0$ to find a relation between to express $b$ in terms of $a$. We use the magnitude of a unit vector as 1 to make the equation , put $b$ in terms of $a$ and then solve for $a$. We then find $b$. \[\]

Complete step-by-step solution:
We know that the unit vector is a vector with magnitude 1. We also know that $\hat{i}$ and $\hat{j}$ are unit vectors (vectors with magnitude 1) along $x,y$ axes respectively. So the magnitudes of these vectors are $\left| {\hat{i}} \right|=\left| {\hat{j}} \right|=1$.
We know that any vector $\overrightarrow{a}$ on a plane can be expressed in terms of unit vectors with its components ${{a}_{1}}$ along $x-$axis and ${{a}_{2}}$ along $y-$axis respectively as
\[\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}\]
The magnitude of the vector $\overrightarrow{a}$ is given as
\[\left| \overrightarrow{a} \right|=\sqrt{a_{1}^{2}+a_{2}^{2}}\]
We know that if two vectors $\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j},\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}$ are perpendicular to each other then their sum of products of corresponding components is zero which means ;
\[{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}=0\]
We are given in the question that the unit vector $a\hat{i}+b\hat{j}$is perpendicular to $\left( \hat{i}-\hat{j} \right)$. Lee us denote the unit vector as $\hat{n}=a\hat{i}+b\hat{j}$ and the vector as $\overrightarrow{v}=\hat{i}-\hat{j}$. The $x-$components of $\hat{n}$ and $\overrightarrow{v}$ are $a$ and 1 respectively. The $y-$components of f $\hat{n}$ and $\overrightarrow{v}$ are $b$ and 1 respectively. We use the perpendicular condition of two vectors and have;
\[\begin{align}
  & a\left( 1 \right)+b\left( -1 \right)=0 \\
 & \Rightarrow a-b=0 \\
 & \Rightarrow a=b \\
\end{align}\]
The magnitude of $\hat{n}$ is 1 since $\hat{n}$ is a unit vector. So we have;
\[\sqrt{{{a}^{2}}+{{b}^{2}}}=1\]
We put previously obtained $a=b$ in the above step to have;
\[\begin{align}
  & \Rightarrow \sqrt{{{a}^{2}}+{{a}^{2}}}=1 \\
 & \Rightarrow \sqrt{2{{a}^{2}}}=1 \\
 & \Rightarrow \sqrt{{{a}^{2}}}=\dfrac{1}{\sqrt{2}} \\
 & \Rightarrow \left| a \right|=\dfrac{1}{\sqrt{2}} \\
\end{align}\]
We use the property of modulus function to have;
\[\Rightarrow a=\pm \dfrac{1}{\sqrt{2}}\]
So the values of $a$ and $b$ are;
\[\Rightarrow a=\dfrac{1}{\sqrt{2}}\Rightarrow b=a=\dfrac{1}{\sqrt{2}}\]
Since we have obtained 2 values of $a$,
\[\Rightarrow a=-\dfrac{1}{\sqrt{2}}\Rightarrow b=a=-\dfrac{1}{\sqrt{2}}\]
So the correct options are A and C.

Note: We should remember the relation between square root and modulus as $\sqrt{{{x}^{2}}}=\left| x \right|$ and also the property of modulus $\left| x \right|=a\Rightarrow x=\pm a$. We have also used here the property of square root $\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}$.If the angle between two vectors is given by $\theta ={{\cos }^{-1}}\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)$. So when ${{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}=0$ we get the perpendicular condition as $\theta ={{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{2}$.