HOW TO FIND UNIT VECTOR PERPENDICULAR TO 2 VECTORS

The unit vectors perpendicular to both and a vector and b vector are

Vectors of magnitude λ, perpendicular to both and a vector b vector are

Finding unit vector perpendicular to two vectors - Examples

Question 1 :

Find the vectors of magnitude 10√3 that are  perpendicular to the plane which contains i vector + 2j vector + k vector and i vector + 3j vector + 4k vector

Solution :

Let a vector  =  i vector + 2j vector + k vector

b vector  =  i vector + 3j vector + 4k vector

required vector perpendicular to given vectors 

=  ± μ [(a x b)/ |a x b|]

  =  i[8-3] - j[4-1] + k[3-2]

a x b  =  5i - 3j + k

|a x b|  =  √5² + (-3)² + 1²

  =  √(25+9+1)

  =  √35

Required vector  = ± (10√3/√35) (5i - 3j + k)

= ± (10√3/√35) (5i - 3j + k)

Question 2 :

Find the unit vectors perpendicular to each of the vectors a vector + b vector and a vector - b vector  where a vector  =  i vector + j vector + k vector and b vector =  i vector + 2j vector + 3k vector

Solution :

Let a vector + b vector  =  c vector and

a vector - b vector  =  d vector

c vector  =  (i + j + k) + (i + 2j + 3k)

c vector  =  2i + 3j + 4k  ---(1)

d vector  =  (i + j + k) - (i + 2j + 3k)

d vector  =  -j - 2k ---(2)

Unit perpendicular vector to both c vector and d vector.

  =  ±(c vector x d vector)/|c vector x d vector|

c x d  =  i[-6+4] -j[-4-0]+k[-2+0]

c x d  =  -2i+4j-2k

|c x d|  =  √(-2)² + 4² + (-2)²  =  √(4+16+4)  =  √24

  =  -2i+4j-2k/√24

  =  ± 2(-i-2j+k)/2√6

  = ± (-i-2j+k)/√6

Hence the required vector perpendicular to the given vectors is ± (-i-2j+k)/√6.

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