Answer:
[tex]\dfrac{1}{\sqrt{42}}(i+4j-5k)[/tex] and [tex]-\dfrac{1}{\sqrt{42}}(i+4j-5k)[/tex]
Step-by-step explanation:
The two vectors are
[tex]a=i+j+k[/tex]
[tex]b=5i+k[/tex]
Unit vector is given by
[tex]\dfrac{a\times b}{|a\times b|}[/tex]
[tex]a\times b=\begin{vmatrix} i & j & k\\1 & 1 & 1\\5 & 0 & 1\end{vmatrix}\\ =i(1-0)-j(1-5)+k(0-5)\\ =i+4j-5k[/tex]
[tex]|a\times b|=\sqrt{1^2+4^2+(-5)^2}=\pm\sqrt{42}[/tex]
[tex]\dfrac{a\times b}{|a\times b|}=\dfrac{i+4j-5k}{\sqrt{42}}\\\Rightarrow \dfrac{a\times b}{|a\times b|}=\pm\dfrac{1}{\sqrt{42}}(i+4j-5k)[/tex]
The two unit vectors that are orthogonal to both given vectors is [tex]\dfrac{1}{\sqrt{42}}(i+4j-5k)[/tex] and [tex]-\dfrac{1}{\sqrt{42}}(i+4j-5k)[/tex].
