Answer
A unit vector perpendicular to both the given vectors in 1.41 is $-\frac{13}{18}\widehat{i}$ + $\frac{1}{3} \widehat{j}$ - $\frac{11}{18}\widehat{k}$
Work Step by Step
The vectors mentioned are
$\vec{A}$ = -2.00$\widehat{i}$ + 3.00$\widehat{j}$ + 4.00$\widehat{k}$
$\vec{B}$ = 3.00$\widehat{i}$ + 1.00$\widehat{j}$ - 3.00$\widehat{k}$
We know that the cross product of two vectors is always perpendicular to both of them
Obtaining that cross product, we get
$\vec{R}$ = (-9-4)$\widehat{i}$ - (6-12)$\widehat{j}$ + (-2-9)$\widehat{k}$
= -13$\widehat{i}$ + 6$\widehat{j}$ - 11$\widehat{k}$
To obtain a unit vector in the same direction, we need to divide the vecor by its magnitude
To get the magnitude,
$R$ = $\sqrt {(-13)^{2}+6^2 + (-11)^2}$ = 18
Dividing our resultant by its magnitude, we get
$\bar{R}$ = $-\frac{13}{18}\widehat{i}$ + $\frac{1}{3} \widehat{j}$ - $\frac{11}{18}\widehat{k}$
