Answer
398.4k+ 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
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)