Question

find a unit vector perpendicular to each of the vector 3i+j+2k and 2i-2j+4k ​

Answer 1

Step-by-step explanation:

A perndicular vector in the direction of given two vector can be obtained by their cross product:

$(3i + j + 2k) \times (2i - 2j + 4k)$

Solving it gives:

$8i - 8j - 8k$

Now unit vector in that direction is given by

$\dfrac{8i - 8j - 8k}{ \sqrt{8 {}^{2} + {8}^{2} + {8}^{2} } } \\ = \dfrac{8i - 8j - 8k}{ \sqrt{3 \times 64} } \\ = \dfrac{8i - 8j - 8k}{8 \sqrt{3 } } \\ = \dfrac{i}{ \sqrt{3} } - \dfrac{j}{ \sqrt{3} } - \dfrac{k}{ \sqrt{3} }$

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