# Let V be the set of vectors vector v is an element of  R^4that are perpendicular to < 1,3,-2,1 >.

sciencesolve | Certified Educator

You need to remember that the scalar product of two perpendicular vectors is zero, hence, you need to perform the vectors multiplication and then you need to impose the restriction of scalar product of perependicular vectors.

Since `bar v` is an element of `R^4` yields:

`bar v = <x_1,x_2,x_3,x_4>`

Performing the vector multiplication yields:

`{(<1,3,-2,1>*<x_1,x_2,x_3,x_4> = 1*x_1 + 3*x_2 - 2*x_3 + 1*x_4),(<1,3,-2,1>*<x_1,x_2,x_3,x_4> = 0):} => x_1 + 3x_2 - 2x_3 + x_4 = 0`

Hence, the vector `bar v` is perpendicular to the given vector `<1,3,-2,1>` if the following linear combination `x_1 + 3x_2 - 2x_3 + x_4 = 0` holds.

rakesh05 | Certified Educator

The vector v is perpendicular to the vector <1,3,-2,1>=u (say).

Let v=<v_1, v_2, v_3, v_4>.

The condition for two vectors u and v to be perpendicular is that u.v=0 .

so   <1,3,-2,1>.<v_1, v_2, v_3, v_4>=0

or,       1v_1+3v_2-2v_3+1v_4=0