To find a unit vector that is perpendicular to the given vectors
+2− and
3−+2, we need to find their cross product. Then, we normalize the resulting vector to make it a unit vector.
Let:
a=+2−=(1,2,−1)b=3−+2=(3,−1,2) The cross product of
a and
b, denoted as
a×b, can be computed using the determinant of the following matrix:
a×b=|| To compute this determinant, we expand as follows:
Calculating each of the minors, we have:
||=(4−1)=3||=(2+3)=5||=(−1−6)=−7 Putting these together:
a×b=3−5−7=(3,−5,−7) To convert this into a unit vector, we need to divide by its magnitude. The magnitude of vector
a×b is:
√32+(−5)2+(−7)2=√9+25+49=√83Thus, the unit vector in the direction of
a×b is:
u=(,−,−)Comparing this with the given options, the correct answer is Option C:
(,−,−)