# a vector = [k+2,k^2-4], a≠0 b vector = [3,-3] Decide k so that a vector ⊥ b vector

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The 2 vectors are orthogonal if their dot product is 0

`(k+2)*3-3(k^2-4)=0`

`3k+6-3k^2+12=0`

divide the equation by 3

`-k^2+k+6=0`

`(k+2)(-k+3)=0`

**2 answers** : k=3 or K=-2

two vector are said to be orthogonal if their dot product=0

if a,b are two vector then

a.b=(a*1**b*1*)+(a*2**b*2*)+................+(a*n*b**n*)

so by given problem

a.b=[(k+2)3]+[(k^2-4)(-3)]

=[3k+6]+[-3k^2+12]

=-3k^2+3k+18

since a.b should be =0 as they are orthogonal

-3k^2+3k+18=0

so when we take -3 ascommon in the expression the expression is

k^2-k-6=0

this expression can be witten as

k^2-3k+2k-6=0 [since -3 and 2 are factors of -6]

this expression can be rewitten as

k(k-3)+2(k-3)=0

(k-3)(k+2)=0

so the values of k can be k=3 or -2

but k=-2 makes a=0 but as per the condition a!=0

so k=3