let Hspanv1 v2 k spanv3 v4 where v15 3 8 v2 1 3 4 v3 2 1 5

let H=span(v1, v2) k = span(v3, v4) where v1=(5 3 8) v2= (1 3 4) v3 = (2 -1 5) v4 = (0 -12 -28) Then H and K are subspaces of R3. In fact, H and K are planes in R3 through the origin, and they intersect in a line through 0 vertor. Find a nonzero vercor w that generates that line.

Solution

Let p = ( a, b, c) be the vector normal to the plane containing the vectors v₁ and v₂ . Then p.v₁ = 0 and p.v₂ = 0 so that 5a +3b + 8c = 0 and a + 3b + 4c = 0. If c = t, then 5a + 3b = - 8t and a + 3b = - 4t so that 5a + 3b – (a + 3b) = - 8t – ( -4t) or, 4a = - 4t so that a = -t. Then – t + 3b = - 4t or, 3b = - 4t + t = -3t so that b = -t. Then, p, the vector normal to the plane containing the vectors v₁ and v₂ is t ( -1 , -1, 1) or, -t ( 1,1, -1) or, ( 1,1, -1).

Similarly, let q = (d , e, f) be a vector normal to the plane containing the vectors v₃ and v₄. Then q.v₃ = 0 and q.v₄ = 0 so that 2d - e + 5f = 0 and – 12e – 28 f = 0. Now, if f = s, then 12 e = - 28s or, e = (-28/12) s = (- 7/3)s. Then 2d = e – 5f = ( -7/3)s – 5s = (- 22/3)s. Then q, the vector normal to the plane containing the vectors v₃ and v₄ is s( -22/3 , -7/3 , 1) = -s/3( 22, 7, -3) or ( 22, 7, -1).

Now, if (x,y,z) is a vector on the plane through ( 0,0,0) containing v₁ and v₂, then (x,y,z) . ( 1,1,-1) = 0 or, x + y – z = 0…(1)

Similarly, if (x,y,z) is a vector on the plane through ( 0,0,0) containing v₃ and v₄ , then (x,y,z) . ( 22, 7,-3) = 0 or, 22x + 7y – 3z = 0…(2)

The line of intersection of the planes (1) and (2) can be obtained by solving these two equations. If z = n, then from (1), we have x + y = n … (3) and from (2), we have 22x + 3y = 3n… (4). On solving equations (3) and (4), we get x = (-4/15) n and y = (19/15)n. Thus the line of intersection of the two planes is w = ( x, y, z) = n( - 4/15, 19/15, 1).