v={x, y,z} such that the points satisfies x-2y+3z=0. is v vector space? if not, find all for which v is not a victor space

Yes, the set of vectors

V = {(x, y, z) : x - 2y + 3z = 0}

is indeed a vector space.

Let u = (x, y, z) and v = (r, s, t) be any two vectors in V. Then

x - 2y + 3z = 0

and

r - 2s + 3t = 0

Their vector sum is

u + v = (x + r, y + s, z + t)

We need to show that u + v also belongs to V - in other words, V is closed under summation. This is a matter of showing that the coordinates of u + v satisfy the condition on all vectors of V:

(x + r) - 2 (y + s) + 3 (s + t) = (x - 2y + 3z) + (r - 2s + 3t) = 0 + 0 = 0

Then V is indeed closed under summation.

Scaling any vector v by a constant c gives

cv = (cx, cy, cz)

We also need to show that cv belongs to V - that V is closed under scalar multiplication. We have

cx - 2cy + 3cz = c (x - 2y + 3z) = 0c = 0

so V is need closed under scalar multiplication.

step-by-step explanation:

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