Linear independent vectors and bases in a room

Question

I have a two-part question.

Part 1:

Choose 2 linear independent vectors out of v,u & w.

u= (6, -6, 6) v= (-12, 12, -14) w= (3, -3, 4)

Part 2:

Find a vector that together with the 2 can become a base for the room.

I guess I should use that linear independence is when aV_1 + bV_2 = 0. where a & b are scalars and V_1 & V_2 are vectors.

What I have tried is to take the vectors 2 at a time and change it to a matrix and look for the solution for when they equal 0, but when a & b are not zero. I find that to be the case for vectors v and w, which I guess is the answer for part 1. However, I don't understand how to solve part 2.

Please ask if something was unclear & how I can improve the question. Thank you for your time.

How I thought about part 2 is that I'll use aV_1 + bV_2 + cV_3 = 0, since we are looking for a base for the room.

So ill search for another vector to v and w.

(-12, 12, -14) + (3, -3, 4) + (a, b, c) = (0, 0, 0)

a = 9, b = -9, c = -10

This however, is incorrect and I don't understand why.

/Andreas

Answer 1

HINT

To determine a pair of linear independent vectors among u, v and w, consider the matrix with the vectors as rows and obtain the RREF.

Once we have two linearly independent vectors the third can be found by orthogonality condition by dot product.

Answer 2

For part one $$u= (6, -6, 6), v= (-12, 12, -14)$$ are linearly independent.

Note that if you multiply the first one by $-2$ you get $(-12,12,12)$ so it is not a scalar multiple of $v$

For part $2$, I will find the cross product of the first two, because it is a common perpendicular to $u$ and $v$ .

You know perpendicular vectors are linearly independent.