Question

A unit vector perpendicular to two vectors 3i+j+2k and 2i-2j+4k

Answer 1

A unit vector perpendicular upon A and B is given by
$\bold{n=\frac{\pm( A\times B)}{|A \times B|}}$

Here , A = 3i + j + 2K
B = 2i - 2j + 4 k
So, A × B =i(4 + 4) -j(12 - 4) + k(-6 - 2) = 8i - 8j - 8k = 8(i - j - k)
⇒ A × B = 8(i - j - k)

Now, |A × B | = $\bold{\sqrt{8^2+(-8)^2+(-8)^2}}$
|A ×B| = 8√3 unit

Now, unit vector , n =± 8(i - j - k)/8√3
=±(i - j - k)/√3

Answer 2

The vector perpendicular to the two vectors would be given by cross product of two
A=3i+j+2k
B=2i-2j+4k

Unit vector=AxB / IAXBI

AxB=  i    j   k 
           3     1  2   
           2     -2   4

=i(8)-j(12-4)+k(-6-2)

=8i-8j-8k

AXB=8[i-j-k]

now IAxBI=√[8²+(-8)²+(-8)²=√3 [8]²
=8 √3

Now unit vector perpendicular to given two vectors = 8[i-j-k] / 8 √3
=[i-j-k]/√3