Answer

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**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.

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