Find a nonzero vector x perpendicular to the vectors v 9 4

Find a nonzero vector x perpendicular to the vectors v = [-9 -4 -5 -7] and u = [-1 -3 7 3]

Solution

Let x= (a,b,c,d)^T. Since x is perpendicular to v, we have x.v = 0 or, (a,b,c,d)^T , ( -9,-4,-5,-7)^T = 0 or, -9a-4b-5c-7d = 0 or, 9a+4b+5c+7d=0....(1).

Similarly, x.u = 0 or, ( a,b,c,d)^T.(-1,-3,7,3)^T = 0 or, -a-3b+7c+3d = 0 or, a+3b-7c-3d = 0...(2)

Let A be a matrix with rows (9,4,5,7) and (1,3,-7,-3). The rows of the RREF of this matrix are (1,0, 43/23, 33/23) and (0,1, -68/23, -34/23). a + 43c/23 + 33d/23 = 0...(3) and b - 68c/23-34d/23= 0..(4). Then x = (a,b,c,d)^T = ( -43c/23 -33d/23, 68c/23+34d/23,c,d)^T. We can assign arbitrary values (not both 0) to c and d to get corresponding values of x. For example, if c = 0, d = 23, then x = ( -33, 34, 0,23)^T. There will be infinite solutions for x.