
Find the unit vectors orthogonal to both \[\left( {2,3,5} \right)\;\]and \[\left( {2, - 1,4} \right).\]
Answer
600.6k+ views
Hint: A vector which is both perpendiculars to two vectors is denoted by their cross product. And its unit vector can be found by dividing the resultant vector with its magnitude.
Complete step by step answer:
Since, we have two vectors as, \[\left( {2,3,5} \right)\;\]and \[\left( {2, - 1,4} \right),\]by taking \[\bar x = (2,3,5)\]
and \[\bar y = (2, - 1,4)\] we will start the given problem.
Now, any vector which is perpendicular to both of them always has to be in a perpendicular plane of both vectors. So, we now find, \[\bar x \times \bar y\] to get the vector which is perpendicular to them.
\[\bar x \times \bar y = \] \[\left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k} \\
2&3&5 \\
2&{ - 1}&4
\end{array}} \right|\]
\[ = \hat i(12 + 5) - \hat j(8 - 10) + \hat k( - 2 - 6)\]
\[ = 17\hat i + 2\hat j - 8\hat k\]
\[ = (17,2, - 8)\]
So, now, we do have a point as \[(17,2, - 8)\] which is perpendicular to both \[\left( {2,3,5} \right)\;\]and \[\left( {2, - 1,4} \right).\]
But, now, we are here trying to find a unit vector along that direction, so we need to divide the vector with its magnitude value. That is the value of square root of the sum of the squares of all the components in the result of \[\bar x \times \bar y = \]. Here, 3 components are given as, \[(17,2, - 8)\]. Now, we get,
\[\left| {\bar x \times \bar y} \right| = \sqrt {{{17}^2} + {2^2} + {{( - 8)}^2}} \]
\[ \Rightarrow \left| {\bar x \times \bar y} \right| = \]\[\sqrt {289 + 4 + 64} \]
\[ = \sqrt {357} \]
\[\therefore \;Unit{\text{ }}perpendicular{\text{ }}vector{\text{ }}to\;\bar x\;and\;\]\[\bar y\],\[\dfrac{{\bar x \times \bar y}}{{\left| {\bar x \times \bar y} \right|}}\]
\[ = \dfrac{1}{{\sqrt {357} }}(17,2, - 8).\]
Note: We say that 2 vectors are orthogonal if they are perpendicular to each other. i.e. the dot product of the two vectors is zero. We can verify the resultant with the given vectors. A unit vector is a vector that has a magnitude equal to one.
Complete step by step answer:
Since, we have two vectors as, \[\left( {2,3,5} \right)\;\]and \[\left( {2, - 1,4} \right),\]by taking \[\bar x = (2,3,5)\]
and \[\bar y = (2, - 1,4)\] we will start the given problem.
Now, any vector which is perpendicular to both of them always has to be in a perpendicular plane of both vectors. So, we now find, \[\bar x \times \bar y\] to get the vector which is perpendicular to them.
\[\bar x \times \bar y = \] \[\left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k} \\
2&3&5 \\
2&{ - 1}&4
\end{array}} \right|\]
\[ = \hat i(12 + 5) - \hat j(8 - 10) + \hat k( - 2 - 6)\]
\[ = 17\hat i + 2\hat j - 8\hat k\]
\[ = (17,2, - 8)\]
So, now, we do have a point as \[(17,2, - 8)\] which is perpendicular to both \[\left( {2,3,5} \right)\;\]and \[\left( {2, - 1,4} \right).\]
But, now, we are here trying to find a unit vector along that direction, so we need to divide the vector with its magnitude value. That is the value of square root of the sum of the squares of all the components in the result of \[\bar x \times \bar y = \]. Here, 3 components are given as, \[(17,2, - 8)\]. Now, we get,
\[\left| {\bar x \times \bar y} \right| = \sqrt {{{17}^2} + {2^2} + {{( - 8)}^2}} \]
\[ \Rightarrow \left| {\bar x \times \bar y} \right| = \]\[\sqrt {289 + 4 + 64} \]
\[ = \sqrt {357} \]
\[\therefore \;Unit{\text{ }}perpendicular{\text{ }}vector{\text{ }}to\;\bar x\;and\;\]\[\bar y\],\[\dfrac{{\bar x \times \bar y}}{{\left| {\bar x \times \bar y} \right|}}\]
\[ = \dfrac{1}{{\sqrt {357} }}(17,2, - 8).\]
Note: We say that 2 vectors are orthogonal if they are perpendicular to each other. i.e. the dot product of the two vectors is zero. We can verify the resultant with the given vectors. A unit vector is a vector that has a magnitude equal to one.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

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

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Chemical formula of Bleaching powder is A Ca2OCl2 B class 11 chemistry CBSE

Name the part of the brain responsible for the precision class 11 biology CBSE

The growth of tendril in pea plants is due to AEffect class 11 biology CBSE

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

