Final answer:
The set S is shown to be closed under scalar multiplication by taking an arbitrary scalar k and point (x, y, z) from S, forming the scalar product (kx, ky, kz), and demonstrating that this new point still satisfies the defining equation of the set S.
Step-by-step explanation:
The student has asked to show that the set S is closed under scalar multiplication. The set S is defined in R³ with the condition that for any (x, y, z) belonging to S, the equation xy + xz + yz = 0 is satisfied.
To demonstrate that S is closed under scalar multiplication, let's consider a scalar k and a point (x, y, z) in S. The scalar product will be (kx, ky, kz). We need to show that the condition kx * ky + kx * kz + ky * kz = 0 holds. After distributing the scalar k and canceling it out since it's common in all terms, what remains is the original equation xy + xz + yz, which was given to be zero. Consequently, (kx, ky, kz) satisfies the condition to be in S, thereby proving that S is closed under scalar multiplication