The number of vectors of unit length perpendicular to the vectors \(\rm\vec{a}\) = 2î + ĵ + 2k̂ and \(\rm\vec{b}\) = ĵ + k̂ is

Calculation:

Given two vector are  \(\rm\vec{a}\) = 2î + ĵ + 2k̂ and \(\rm\vec{b}\) = ĵ + k̂

Unit vectors perpendicular to \(\rm\vec{a}\) and \(\rm\vec{b}\) are  \(\pm\frac{\rm\vec{a}\times\rm\vec{b}}{|\vec a\times\vec b|}\)

\(\vec a\times\vec b= \begin{vmatrix}\hat i&\hat j&\hat k\\2&1&2\\0&1&1\end{vmatrix}\)

\(\vec a\times\vec b=\hat i(1-2)-\hat j(2-0)+\hat k(2-0) \)

\(\vec a\times \vec b= -\hat i-2\hat j+2\hat k\)

\(|\vec a\times \vec b| =√{(-1)^2+(-2)^2+(2)^2}=√{1+4+4}\)

= √9  = 3

The unit vectors perpendicular to  \(\rm\vec{a}\) and \(\rm\vec{b}\) are \(\pm(\frac{-\hat i-2\hat j+2\hat k}{3})\)

Hence there are two vectors perpendicular to the vectors \(\vec a\) and \(\vec b\).

The correct answer is option 2.