Answer
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Hint: The given question requires us to find all unit vectors perpendicular to both the vectors given in the problem. This can be done easily by applying the concepts of vectors as the cross product of two vectors is always perpendicular to both the vectors. Unit vectors can easily be found by dividing a vector by its magnitude.
Complete step by step solution:
Let the required vector be given by $\vec c$.
In the given problem, we are required to first find the position vectors corresponding to the given coordinates.
Position vector of given point $(1,2, - 1)$ is \[\vec a = \hat i + 2\hat j - \hat k\]
Position vector of given point $(3,3, - 4)$is $\vec b = 3\hat i + 3\hat j - 4\hat k$
Now, $\vec c$is orthogonal to both $\vec a$and$\vec b$.
Thus, $\vec c$$ = \vec a \times \vec b$. Now, we have to find the cross product of two vectors. Cross product is found by solving the determinant of coefficients of the rectangular components of the two vectors.
Now, computing the determinant along the first row, we get,
$\vec c$$ = \left| {\text{ }}\hat i{\text{ }}\hat j{\text{ }}\hat k \\
{\text{ }}1{\text{ }}2{\text{ }} - 1 \\
{\text{ }}3{\text{ }}3{\text{ }} - 4 \\ \right|$
$\vec c$$ = \hat i( - 8 - ( - 3)) - \hat j( - 4 - ( - 3)) + \hat k(3 - 6)$
$ \Rightarrow \vec c$$ = \hat i( - 5) - \hat j( - 1) + \hat k( - 3)$
$ \Rightarrow \vec c$$ = - 5\hat i + \hat j - 3\hat k$
Now the required unit vector is $\hat c$. We will find $\hat c$ by dividing $\vec c$ by its magnitude.
Thus, $\hat c = \dfrac{{ - 5\hat i + \hat j - 3\hat k}}{{\sqrt {{{( - 5)}^2} + {{(1)}^2} + {{( - 3)}^2}} }}$
$ \Rightarrow \hat c = \dfrac{{ - 5\hat i + \hat j - 3\hat k}}{{\sqrt {35} }}$
$ \Rightarrow \hat c = \dfrac{{ - 5\hat i}}{{\sqrt {35} }} + \dfrac{{\hat j}}{{\sqrt {35} }} - \dfrac{{3\hat k}}{{\sqrt {35} }}$
So, unit vector perpendicular to both the vectors given by $(1,2, - 1)$ and $(3,3, - 4)$ is $\left( {\dfrac{{ - 5\hat i}}{{\sqrt {35} }} + \dfrac{{\hat j}}{{\sqrt {35} }} - \dfrac{{3\hat k}}{{\sqrt {35} }}} \right)$.
Note: Such type of questions involves concepts of cross product of two vectors. We need to have a strong grip on topics like Vector algebra and Dot and cross product of two vectors so as to solve typical questions from these topics.
Complete step by step solution:
Let the required vector be given by $\vec c$.
In the given problem, we are required to first find the position vectors corresponding to the given coordinates.
Position vector of given point $(1,2, - 1)$ is \[\vec a = \hat i + 2\hat j - \hat k\]
Position vector of given point $(3,3, - 4)$is $\vec b = 3\hat i + 3\hat j - 4\hat k$
Now, $\vec c$is orthogonal to both $\vec a$and$\vec b$.
Thus, $\vec c$$ = \vec a \times \vec b$. Now, we have to find the cross product of two vectors. Cross product is found by solving the determinant of coefficients of the rectangular components of the two vectors.
Now, computing the determinant along the first row, we get,
$\vec c$$ = \left| {\text{ }}\hat i{\text{ }}\hat j{\text{ }}\hat k \\
{\text{ }}1{\text{ }}2{\text{ }} - 1 \\
{\text{ }}3{\text{ }}3{\text{ }} - 4 \\ \right|$
$\vec c$$ = \hat i( - 8 - ( - 3)) - \hat j( - 4 - ( - 3)) + \hat k(3 - 6)$
$ \Rightarrow \vec c$$ = \hat i( - 5) - \hat j( - 1) + \hat k( - 3)$
$ \Rightarrow \vec c$$ = - 5\hat i + \hat j - 3\hat k$
Now the required unit vector is $\hat c$. We will find $\hat c$ by dividing $\vec c$ by its magnitude.
Thus, $\hat c = \dfrac{{ - 5\hat i + \hat j - 3\hat k}}{{\sqrt {{{( - 5)}^2} + {{(1)}^2} + {{( - 3)}^2}} }}$
$ \Rightarrow \hat c = \dfrac{{ - 5\hat i + \hat j - 3\hat k}}{{\sqrt {35} }}$
$ \Rightarrow \hat c = \dfrac{{ - 5\hat i}}{{\sqrt {35} }} + \dfrac{{\hat j}}{{\sqrt {35} }} - \dfrac{{3\hat k}}{{\sqrt {35} }}$
So, unit vector perpendicular to both the vectors given by $(1,2, - 1)$ and $(3,3, - 4)$ is $\left( {\dfrac{{ - 5\hat i}}{{\sqrt {35} }} + \dfrac{{\hat j}}{{\sqrt {35} }} - \dfrac{{3\hat k}}{{\sqrt {35} }}} \right)$.
Note: Such type of questions involves concepts of cross product of two vectors. We need to have a strong grip on topics like Vector algebra and Dot and cross product of two vectors so as to solve typical questions from these topics.
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