Question

# If $a=\hat{i}-2\hat{j}+3\hat{k}$ and $b=-3\hat{i}+\hat{j}-\hat{k}$ and $r\times a=b\times a,r\times b=a\times b$, then a unit vector in the direction of $r$ is(a) $\pm \dfrac{1}{3}\left( -2\hat{i}+\hat{j}-\hat{k} \right)$(b) $\pm \dfrac{1}{3}\left( -2\hat{i}-\hat{j}+2\hat{k} \right)$(c) $\pm \dfrac{1}{3}\left( -2\hat{i}-\hat{j}-2\hat{k} \right)$(d) None of these

Hint: Use a cross product of two vectors to find the value of $b\times a$ and $a\times b$. Assume the vector $r$ as $r=x\hat{i}+y\hat{j}+z\hat{k}$ and write the equations satisfying the given conditions. Solve those equations to get the value of the vector $r$.

We have two vectors $a=\hat{i}-2\hat{j}+3\hat{k}$ and $b=-3\hat{i}+\hat{j}-\hat{k}$. There exists a third vector $r$ such that $r\times a=b\times a,r\times b=a\times b$. We will form equations satisfying the given properties to find a unit vector in the direction of the vector $r$.
Let’s assume that we can write the vector $r$ as $r=x\hat{i}+y\hat{j}+z\hat{k}$.
We will now find the value of $b\times a$ where $a=\hat{i}-2\hat{j}+3\hat{k}$ and $b=-3\hat{i}+\hat{j}-\hat{k}$.
We know that if there are two vectors $a={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and $b={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$, then we have $a\times b=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ \end{matrix} \right|$ as the value of cross multiplication of vectors $a={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and $b={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$.
Thus, when $a=\hat{i}-2\hat{j}+3\hat{k}$ and $b=-3\hat{i}+\hat{j}-\hat{k}$, we have $a\times b=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -2 & 3 \\ -3 & 1 & -1 \\ \end{matrix} \right|=\hat{i}\left( 2-3 \right)-\hat{j}\left( -1+9 \right)+\hat{k}\left( 1-6 \right)=-\hat{i}-8\hat{j}-5\hat{k}$.
Now, we will evaluate the value of $r\times b$ where $r=x\hat{i}+y\hat{j}+z\hat{k}$ and $b=-3\hat{i}+\hat{j}-\hat{k}$. Thus, we have $r\times b=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ x & y & z \\ -3 & 1 & -1 \\ \end{matrix} \right|=\hat{i}\left( -y-z \right)-\hat{j}\left( -x+3z \right)+\hat{k}\left( x+3y \right)$.
As we have $r\times b=a\times b$, we have $-\hat{i}-8\hat{j}-5\hat{k}=\hat{i}\left( -y-z \right)-\hat{j}\left( -x+3z \right)+\hat{k}\left( x+3y \right)$.
Comparing the coefficients on both sides, we get $y+z=1,3z-x=8,x+3y=-5$.
We know that $b\times a=-\left( a\times b \right)$. Thus, we have $b\times a=-\left( -\hat{i}-8\hat{j}-5\hat{k} \right)=\hat{i}+8\hat{j}+5\hat{k}$.
Now, we will evaluate the value of $r\times a$. Thus, we have $r\times a=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ x & y & z \\ 1 & -2 & 3 \\ \end{matrix} \right|=\hat{i}\left( 3y+2z \right)-\hat{j}\left( 3x-z \right)+\hat{k}\left( -2x-y \right)$.
As we have $r\times a=b\times a$, we have $\hat{i}+8\hat{j}+5\hat{k}=\hat{i}\left( 3y+2z \right)-\hat{j}\left( -z+3x \right)+\hat{k}\left( -2x-y \right)$.
Comparing the coefficients on both sides, we get $1=3y+2z,8=z-3x,5=-2x-y$.
Thus, we have the equations $y+z=1,3z-x=8,x+3y=-5$ and $1=3y+2z,8=z-3x,5=-2x-y$. We will now solve these equations by elimination method.
Multiplying equation $y+z=1$ with $2$ and subtracting it from equation $1=3y+2z$, we get $3y+2z-2\left( y+z \right)=1-2$.
$\Rightarrow y=-1$
Substituting the value $y=-1$ in equation $5=-y-2x$ , we have $5=1-2x$. Further solving, we get $2x=-4\Rightarrow x=-2$.
Substituting the value $x=-2$ in equation $3z-x=8$ , we have $3z-\left( -2 \right)=8$. Further solving, we get $3z=6\Rightarrow z=2$.
Thus, we have $r=x\hat{i}+y\hat{j}+z\hat{k}=-2\hat{i}-\hat{j}+2\hat{k}$.
We want to find the unit vector in the direction of the vector $r$. We will divide it by the value of $\left| r \right|$.
We have $\left| r \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}=\sqrt{9}=\pm 3$.
Hence, the unit vector in direction of $r$ is $\dfrac{r}{\left| r \right|}=\pm \dfrac{1}{3}\left( -2\hat{i}-\hat{j}+2\hat{k} \right)$, which is option (b).

Note: We should observe that the vector $r$ lies in the plane perpendicular to both $a$ and $b$. Also, one must keep in mind that we have to find the unit vector, not just any vector satisfying the conditions. We can also find the value of $b\times a$ by substituting the values in the matrix and finding its value. However, it will be time consuming. Thus, it’s better to use $b\times a=-\left( a\times b \right)$.