**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

**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| = √5^{2} + (-3)^{2} + 1^{2}

= √(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} + 4^{2} + (-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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