Find two unit vectors orthogonal to both (3 , 2, 1) and (- 1, 1, 0 ).

Solution:

Given, a = (3, 2, 1)

b = (-1, 1, 0)

We have to find the two unit vectors orthogonal to both a and b.

Finding cross product,

\(a\times b=\begin{vmatrix} i &j &k \\ 3 &2 &1 \\ -1 & 1 &0 \end{vmatrix}\)

\(\\=i(0 - 1)-j(0+1)+k(3+2)\\=-i-j+5k\)

= (-1, -1, 5)

\(\left | \vec{a}\times \vec{b} \right |=\sqrt{(-1)^{2}+(-1)^{2}+(5)^{2}}=\sqrt{1+1+25}=\sqrt{27}\)

Now, unit vector perpendicular to \(\vec{a}\, and\, \vec{b}=\pm \frac{(\vec{a}\times \vec{b})}{\left | \vec{a}\times \vec{b} \right |}\)

= \(\pm \frac{1}{\sqrt{27}}(-1, -1, 5)\)

Therefore, the two unit vectors are \(\pm \frac{1}{\sqrt{27}}(-1, -1, 5)\).

Find two unit vectors orthogonal to both (3 , 2, 1) and (- 1, 1, 0 ).

Summary:

The two unit vectors orthogonal to both 3, 2,1 and -1,1,0 are \(\pm \frac{1}{\sqrt{27}}(-1, -1, 5)\).