Question

Find all the vectors in R4 that are perpendicular to the three vectors <1,1,1,1>, <1,2,3,4>, and <1,9,9,7>

Answer 1

Let <x,y,z,w> be an arbitrary vector in R which is perpendicular to the three vectors <1,1,1,1>, <1,2,3,4>, and <1,9,9,7>.

Then < x, y, z, w >. <1,1,1,1> = 0 or, x+y+z+w = 0…(1).

Also, < x, y, z, w >. <1,2,3,4> = 0 or, x+2y+3z+4w = 0…(2)

and < x, y, z, w >. <1,9,9,7> = 0 or, x+9y+9z+7w = 0…(3).

The coefficient matrix of the above linear system of homogeneous equations is A (say) =

1	1	1	1
1	2	3	4
1	9	9	7

In order to solve the above linear system of homogeneous equations, we will reduce A to its RREF as under:

Add -1 times the 1st row to the 2nd row

Add -1 times the 1st row to the 3rd row

Add -8 times the 2nd row to the 3rd row

Multiply the 3rd row by -1/8

Add -2 times the 3rd row to the 2nd row

Add -1 times the 3rd row to the 1st row

Add -1 times the 2nd row to the 1st row

Then the RREF of A is

1	0	0	¼
0	1	0	-3/2
0	0	1	9/4

It implies that the above linear system of homogeneous equations is equivalent to x+w/4 = 0 or, x = -w/4, y-3w/2=0 or, y=3w/2 and z+9w/4 = 0 or, z = -9w/4. Then, v = <-w/4,3w/2,-9w/4,w> = w/4< -1,6,--9,4> = t<-1,6,-9,4>, where t = w/4 is an arbitrary real number.

Hence all the vectors of the form t<-1,6,-9,4>, where t is an arbitrary real number, are perpendicular to the three vectors <1,1,1,1>, <1,2,3,4>, and <1,9,9,7>