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Question: If an unit vector \(a\hat{i}+b\hat{j}\)is perpendicular to \(\left( \hat{i}-\hat{j} \right)\) then t...

If an unit vector ai^+bj^a\hat{i}+b\hat{j}is perpendicular to (i^j^)\left( \hat{i}-\hat{j} \right) then the value of aa and bb 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

Explanation

Solution

We use the condition of two vectors a=a1i^+a2j^,b=b1i^+b2j^\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j},\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j} perpendicular that is a1b1+a2b2=0{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}=0 to find a relation between to express bb in terms of aa. We use the magnitude of a unit vector as 1 to make the equation , put bb in terms of aa and then solve for aa. We then find bb. $$$$

Complete step-by-step solution:
We know that the unit vector is a vector with magnitude 1. We also know that i^\hat{i} and j^\hat{j} are unit vectors (vectors with magnitude 1) along x,yx,y axes respectively. So the magnitudes of these vectors are i^=j^=1\left| {\hat{i}} \right|=\left| {\hat{j}} \right|=1.
We know that any vector a\overrightarrow{a} on a plane can be expressed in terms of unit vectors with its components a1{{a}_{1}} along xx-axis and a2{{a}_{2}} along yy-axis respectively as
a=a1i^+a2j^\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}
The magnitude of the vector a\overrightarrow{a} is given as
a=a12+a22\left| \overrightarrow{a} \right|=\sqrt{a_{1}^{2}+a_{2}^{2}}
We know that if two vectors a=a1i^+a2j^,b=b1i^+b2j^\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 ;
a1b1+a2b2=0{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}=0
We are given in the question that the unit vector ai^+bj^a\hat{i}+b\hat{j}is perpendicular to (i^j^)\left( \hat{i}-\hat{j} \right). Lee us denote the unit vector as n^=ai^+bj^\hat{n}=a\hat{i}+b\hat{j} and the vector as v=i^j^\overrightarrow{v}=\hat{i}-\hat{j}. The xx-components of n^\hat{n} and v\overrightarrow{v} are aa and 1 respectively. The yy-components of f n^\hat{n} and v\overrightarrow{v} are bb and 1 respectively. We use the perpendicular condition of two vectors and have;

& a\left( 1 \right)+b\left( -1 \right)=0 \\\ & \Rightarrow a-b=0 \\\ & \Rightarrow a=b \\\ \end{aligned}$$ The magnitude of $\hat{n}$ is 1 since $\hat{n}$ is a unit vector. So we have; $$\sqrt{{{a}^{2}}+{{b}^{2}}}=1$$ We put previously obtained $a=b$ in the above step to have; $$\begin{aligned} & \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{aligned}$$ We use the property of modulus function to have; $$\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 $\theta ={{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{2}$.