Find two unit vectors orthogonal to both 7, 5, 1 and -1, 1, 0 .

Solution:

Given, a = (7, 5, 1)

b = (-1, 1, 0)

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

Finding cross product,

\(a\times b=\begin{vmatrix} i &j &k \\ 7 &5 &1 \\ -1 & 1 &0 \end{vmatrix}\)

\(\\=i(0 - 1)-j(0+1)+k(7-(-5))\\=-i-j+12k\)

= (-1, -1, 12)

\(\left | \vec{a}\times \vec{b} \right |=\sqrt{(-1)^{2}+(-1)^{2}+(12)^{2}}=\sqrt{1+1+144}=\sqrt{146}\)

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

= \(\pm \frac{1}{\sqrt{146}}(-1, -1, 12)\)

Therefore, the two unit vectors are \(\pm \frac{1}{\sqrt{146}}(-1, -1, 12)\).

Find two unit vectors orthogonal to both 7, 5, 1 and -1, 1, 0 .

Summary:

The two unit vectors orthogonal to both 7,5,1 and -1,1,0 are \(\pm \frac{1}{\sqrt{146}}(-1, -1, 12)\).