Answer

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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 \[r\].

Complete step-by-step answer:

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 \[b\times a=-\left( a\times b \right)\].

Complete step-by-step answer:

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 \[b\times a=-\left( a\times b \right)\].

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