To determine whether or not the plane 2x+4y+3z=1 is a subspace of R3, we need to check if it satisfies the three properties of a subspace: closure under addition, closure under scalar multiplication, and containing the zero vector.

1. Closure under addition: Let's assume we have two vectors (x1, y1, z1) and (x2, y2, z2) that satisfy the equation 2x+4y+3z=1. We need to check if their sum, (x1+x2, y1+y2, z1+z2), also satisfies the equation.

Substituting the values into the equation, we get:
2(x1+x2) + 4(y1+y2) + 3(z1+z2) = 2x1 + 2x2 + 4y1 + 4y2 + 3z1 + 3z2 = (2x1 + 4y1 + 3z1) + (2x2 + 4y2 + 3z2) = 1 + 1 = 2

Since the sum satisfies the equation 2x+4y+3z=1, the plane is closed under addition.

2. Closure under scalar multiplication: Let's assume we have a vector (x, y, z) that satisfies the equation 2x+4y+3z=1. We need to check if its scalar multiple, k(x, y, z), also satisfies the equation for any scalar k.

Substituting the values into the equation, we get:
2(kx) + 4(ky) + 3(kz) = k(2x + 4y + 3z) = k(1) = k

Since the scalar multiple satisfies the equation 2x+4y+3z=1, the plane is closed under scalar multiplication.

3. Containing the zero vector: To check if the plane contains the zero vector, we substitute x=0, y=0, and z=0 into the equation 2x+4y+3z=1.

2(0) + 4(0) + 3(0) = 0 + 0 + 0 = 0

Since the zero vector satisfies the equation 2x+4y+3z=1, the plane contains the zero vector.

Therefore, since the plane satisfies all three properties of a subspace, we can conclude that the plane 2x+4y+3z=1 is a subspace of R3.

Question

To determine whether or not the plane 2x+4y+3z=1 is a subspace of R3, we need to check if it satisfies the three properties of a subspace: closure under addition, closure under scalar multiplication, and containing the zero vector.

1. Closure under addition: Let's assume we have two vectors (x1, y1, z1) and (x2, y2, z2) that satisfy the equation 2x+4y+3z=1. We need to check if their sum, (x1+x2, y1+y2, z1+z2), also satisfies the equation.

Substituting the values into the equation, we get:
2(x1+x2) + 4(y1+y2) + 3(z1+z2) = 2x1 + 2x2 + 4y1 + 4y2 + 3z1 + 3z2 = (2x1 + 4y1 + 3z1) + (2x2 + 4y2 + 3z2) = 1 + 1 = 2

Since the sum satisfies the equation 2x+4y+3z=1, the plane is closed under addition.

2. Closure under scalar multiplication: Let's assume we have a vector (x, y, z) that satisfies the equation 2x+4y+3z=1. We need to check if its scalar multiple, k(x, y, z), also satisfies the equation for any scalar k.

Substituting the values into the equation, we get:
2(kx) + 4(ky) + 3(kz) = k(2x + 4y + 3z) = k(1) = k

Since the scalar multiple satisfies the equation 2x+4y+3z=1, the plane is closed under scalar multiplication.

3. Containing the zero vector: To check if the plane contains the zero vector, we substitute x=0, y=0, and z=0 into the equation 2x+4y+3z=1.

2(0) + 4(0) + 3(0) = 0 + 0 + 0 = 0

Since the zero vector satisfies the equation 2x+4y+3z=1, the plane contains the zero vector.

Therefore, since the plane satisfies all three properties of a subspace, we can conclude that the plane 2x+4y+3z=1 is a subspace of R3.

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