Question

2. Consider the set V = span {v} = (1,0,2), v2 = (2, 1, 2)}. (a) For each choice of numbers for a and b, the set of points of the form (3,2, a) + (6-1,4), te R, is a line L in R'. In set notation: 20| L = {(3,2, a) + t(0, -1,4) € R'|teR} Find all values of a and b, if any, for which the line L is contained in V. b) After replacing a and with the values you found in part (a), express the point (2,y,) - (32.a) + k(6, -1,4) on L as a linear combination of V. and V2. 110
Im not understanding why you let t=0 and t=1 can you explain? thank you!

Answer 1

a)

(3,2, a) + t(0, -1, 4) = C1(1,0,2) + cz(2,1,2)

(bt +3,2-t, a + 4t) = (c1 + 2c2,c2, 2c1 +202)

So that

bt + 3 = 1+2c2, 2-t = c2, a + 4t = 261 +2cm

a/2 + 2t = c +cz

So that

a/2+2t = (1+(2-t) > c=3+ a/2 - t

Meaning

C1 = 3+a/2-t,c2= 2 - +

So that

bt +3=C1 +2c2 = 3+a/2-t+2/2-t) = 7 – 3t+a/2

We
bt + 3 = 7 – 3t+a/2
and this must hold for all $t\in\mathbb{R}$

Which is only possible if

b= -3,3 = 7+ a/2= a= -8

b)

(x, y, z) = (3,2, a) + k (6,-1, 4) =(3, 2, -8) +k(-3,1,4)

Then

C2 = 2-k,C1 = 3+a/2 - k= -k-1

Which means

(x, y, z) = (3, 2, -8) + (-3, 1, 4) =(-k - 1)(1,0,2) + (2 - k) 2,1,2)

$\blacksquare$