Find all the vectors that are perpendicular to (1,4,4,1) and (2,9,8,2).

Answer 1

Two vectors are orthogonal to each other when their inner product equals zero.
For example,
(1,-3) and (3,1) are orthogonal because
(1,-3).(3,1)=1*3+(-3)*1=0

If we are looking for a vector, (p,q,r,s) which is orthogonal to the given vectors (1,4,4,1) and (2,9,8,2), we just have to form the dot products and equate them to zero, and subsequently solve for the values of p,q,r and s.

So
(1,4,4,1).(p,q,r,s)=0
=> p+4q+4r+s = 0 ....(1)
(2,9,8,2).(p,q,r,s)=0
=> 2p+9q+4r+s = 0 ....(2)

Now solving the system of 2 equations in 4 unknowns, we arrive at the reduced echelon form:

p 0 4r s=0 ... (1A)
0 q 0 0=0 ...(2A)

which gives us q=0, and p=-4r-s
q and r are free variables (can take on any value) while p is a dependent variable.

Thus the solution vector is
(-4r-s, 0, r, s)

Check:
(1,4,4,1).(-4r-s,0,r,s)
=-4r-s +4*0 +4r +s
=0

(2,9,8,2).(-4r-s,0,r,s)
=-8r-2s +9*0 + 8r + 2s
=0

So the solution vector is orthogonal to the two given vectors.
Note: r and s can take on any finite value.