
What is unit vector in the perpendicular to the following vectors $2\hat i + 2\hat j - \hat k$ and $6\hat i - 3\hat j + 2\hat k$
A. $\dfrac{{\hat i + 10\hat j - 18\hat k}}{{5\sqrt {17} }}$
B. $\dfrac{{\hat i - 10\hat j + 18\hat k}}{{5\sqrt {17} }}$
C. $\dfrac{{\hat i - 10\hat j - 18\hat k}}{{5\sqrt {17} }}$
D. $\dfrac{{\hat i + 10\hat j + 18\hat k}}{{5\sqrt {17} }}$
Answer
499.8k+ views
Hint: In order to solve this question we need to understand the vector or cross product of two vectors. So a vector product of two vectors is defined as a vector when cross multiplied with another vector then the resultant vector is in perpendicular direction from both the vectors. So in this question we are going to find the cross product of given two vectors and then later find the unit vector. A unit vector is defined as the vector direction and it is mathematically expressed as the ratio of vectors by its magnitude.
Complete step by step answer:
Consider vector A as, $\vec A = 2\hat i + 2\hat j - \hat k$. And vector B as, $\vec B = 6\hat i - 3\hat j + 2\hat k$. So the cross product of two vectors is defined as, $\vec N = \vec A \times \vec B$. Here, $\vec N$ is the result of cross product of both vectors and is perpendicular to both vectors.
We use the determinant method to find the cross product solution,So putting values to get $\vec N$ as,
\[\;\vec N = (2\hat i + 2\hat j - \hat k) \times (6\hat i - 3\hat j + 2\hat k)\]
\[\Rightarrow \vec N = \left( {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k} \\
2&2&{ - 1} \\
6&{ - 3}&2
\end{array}} \right)\]
\[\Rightarrow \vec N = \hat i[\{ 2 \times 2\} - \{ - 3 \times - 1\} ] - \hat j[\{ 2 \times 2\} - \{ 6 \times - 1\} ] + \hat k[\{ 2 \times - 3\} - \{ 6 \times 2\} ]\]
\[\Rightarrow \vec N = \hat i[4 - 3] - \hat j[4 + 6] + \hat k[ - 6 - 12]\]
\[\Rightarrow \vec N = \hat i - 10\hat j - 18\hat k\]
So magnitude of this vector is given as,
\[\left| {\vec N} \right| = \sqrt {{1^2} + {{10}^2} + {{18}^2}} \]
\[\Rightarrow \left| {\vec N} \right| = \sqrt {425} \]
\[\Rightarrow \left| {\vec N} \right| = \sqrt {25 \times 17} \]
\[\Rightarrow \left| {\vec N} \right| = 5\sqrt {17} \]
So the unit vector in this direction is defined as,
\[\hat N = \dfrac{{\vec N}}{{\left| {\vec N} \right|}}\]
Putting values we get,
\[\therefore \hat N = \dfrac{{\hat i - 10\hat j - 18\hat k}}{{5\sqrt {17} }}\]
Therefore, the correct option is C.
Note: It should be remembered that vectors are those quantities which have both magnitude and direction, also it must follow the triangle rule of vector addition. This law is a compulsory condition for quantity to be considered as a vector, for example current is not a vector as it has both magnitude and direction but it does not follow the triangle rule of vector addition, because of Kirchhoff’s junction rule.
Complete step by step answer:
Consider vector A as, $\vec A = 2\hat i + 2\hat j - \hat k$. And vector B as, $\vec B = 6\hat i - 3\hat j + 2\hat k$. So the cross product of two vectors is defined as, $\vec N = \vec A \times \vec B$. Here, $\vec N$ is the result of cross product of both vectors and is perpendicular to both vectors.
We use the determinant method to find the cross product solution,So putting values to get $\vec N$ as,
\[\;\vec N = (2\hat i + 2\hat j - \hat k) \times (6\hat i - 3\hat j + 2\hat k)\]
\[\Rightarrow \vec N = \left( {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k} \\
2&2&{ - 1} \\
6&{ - 3}&2
\end{array}} \right)\]
\[\Rightarrow \vec N = \hat i[\{ 2 \times 2\} - \{ - 3 \times - 1\} ] - \hat j[\{ 2 \times 2\} - \{ 6 \times - 1\} ] + \hat k[\{ 2 \times - 3\} - \{ 6 \times 2\} ]\]
\[\Rightarrow \vec N = \hat i[4 - 3] - \hat j[4 + 6] + \hat k[ - 6 - 12]\]
\[\Rightarrow \vec N = \hat i - 10\hat j - 18\hat k\]
So magnitude of this vector is given as,
\[\left| {\vec N} \right| = \sqrt {{1^2} + {{10}^2} + {{18}^2}} \]
\[\Rightarrow \left| {\vec N} \right| = \sqrt {425} \]
\[\Rightarrow \left| {\vec N} \right| = \sqrt {25 \times 17} \]
\[\Rightarrow \left| {\vec N} \right| = 5\sqrt {17} \]
So the unit vector in this direction is defined as,
\[\hat N = \dfrac{{\vec N}}{{\left| {\vec N} \right|}}\]
Putting values we get,
\[\therefore \hat N = \dfrac{{\hat i - 10\hat j - 18\hat k}}{{5\sqrt {17} }}\]
Therefore, the correct option is C.
Note: It should be remembered that vectors are those quantities which have both magnitude and direction, also it must follow the triangle rule of vector addition. This law is a compulsory condition for quantity to be considered as a vector, for example current is not a vector as it has both magnitude and direction but it does not follow the triangle rule of vector addition, because of Kirchhoff’s junction rule.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

