How to Find Unit Vector Perpendicular to 2 Vectors :
Here we are going to see 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
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
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
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
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
= ± 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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