# The vectors vector u = <2,-2,3> , vector v = <-2,0,-2> and vector w = <-14,6,-17> are linearly dependent. Find a non-trivial linear relation 0 = a(vector u)+b(vector v)+c(vector...

The vectors vector u = <2,-2,3> , vector v = <-2,0,-2> and vector w = <-14,6,-17> are linearly dependent. Find a non-trivial linear relation

and enter your answer in vector form as <a,b,c>.

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You need to write the following equation that links the linearly dependent vectors, such that:

`a*<2,-2,3> + b*<-2,0,-2> + c*<-14,6,-17> = 0`

Extracting three equations from above relation, yields:

`{(2a - 2b - 14c = 0),(-2a + 6c = 0),(3a - 2b - 17c = 0):}`

`{(2a - 2b - 14c = 0),(2a = 6c),(3a - 2b - 17c = 0):}`

`{(2a - 2b - 14c = 0),(a = 3c),(3a - 2b - 17c = 0):}`

`{(6c - 2b - 14c = 0),(a = 3c),(9c - 2b - 17c = 0):}`

`{(-8c - 2b = 0),(a = 3c),(-8c - 2b = 0):}`

`{(b = 4c),(a = 3c):}`

Considering c = 1 yields:

`b = 4c => b = 4`

`a = 3c => a = 3`

**Hence, evaluating the linear combination of the given vectors, yields**

`a*<2,-2,3> + b*<-2,0,-2> + c*<-14,6,-17> = 3*<2,-2,3> + 4*<-2,0,-2> + 1*<-14,6,-17> = 0.`

Let a<2,-2,3)+b<-2,0,-2>+c<-14,6,-17>=0

`a[2,-2,3],+b[-2,0,-2]+c[-1,6,-17]=[[0,0,0]]`

`2a-2b-14c=0`

`-2a+6c=0`

`3a-2b-17c=0`

`` Solving above system of equations for a,b,c . System is homogeneous

equation so one of its solution is trivial. But rank of coefficient matrix A= `[[2,-2,3],[-2,0,-2],[-14,6,-17]]` is 2. So it has infinite no. of solutions.

Solving above system we get

`a=3k,b=-4k , c=k ,` `k!=0` , k is real number.

So we have

<3k,-4k,k>

Ans.