Imagine one vector, say v = (1, 2, 3). It’s a lonely arrow. It wants friends, but only perpendicular friends. No tangents allowed.
Perpendicular means the dot product is zero. Dot product? It’s just a fancy way of saying: multiply matching coordinates, then add them up. If the sum equals zero, you’ve got a right-angle buddy.
So, how do we find not one, but two such buddies? And they must be linearly independent. That’s just math-speak for “not pointing in the same direction.” No clones allowed.
Step 1: Guess a Random Arrow
Here’s the fun part. You can make up a vector. Seriously. Pick numbers out of thin air. Let’s call it u = (a, b, c). We need it to be perpendicular to our poor vector v.
Set up the dot product equation: 1a + 2b + 3c = 0. This is one equation with three unknowns. That’s like having a party with two free guests. Massive wiggle room.
Pick a = 1 and b = 0. Then 11 + 20 + 3c = 0. Solve: 1 + 3c = 0, so c = -1/3. Boom: u = (1, 0, -1/3) is perpendicular. But it’s a weird fraction. Who wants fractions? Let’s multiply everything by 3 to get u = (3, 0, -1). Much cleaner.
Step 2: Get a Second, Different Arrow
Now we need another vector, call it w = (x, y, z). It must also be perpendicular to v. That’s the same equation: 1x + 2y + 3z = 0. But it must not be a multiple of u.
Solved (1 point) Find two linearly independent vectors | Chegg.com
Easy fix. This time, pick x = 0 and y = 2. Then 10 + 22 + 3z = 0 gives 4 + 3z = 0, so z = -4/3. So our vector is (0, 2, -4/3). Multiply by 3 to get (0, 6, -4). That’s valid, but check: is it a multiple of (3, 0, -1)? Nope. No common factor. They point in different directions. Linearly independent!
You’ve got two unique perpendicular arrows. Congratulations. You’re now a vector-whisperer.
