
Find all unit vectors orthogonal to both the vectors given by $(1,2, - 1)$ and $(3,3, - 4)$?
Answer
538.5k+ views
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

