The vector that is NOT perpendicular to the vectors (i + j + k) and (i + 2j + 3k) is . . . . . . . . A. (i – 2j + k) B. (-i + 2j – k) C. (0i + 0j + 0k) D. (4i + 3j + 5k)

The correct answer is $\boxed{\text{(D)}}$.

A vector is perpendicular to another vector if the dot product of the two vectors is zero. The dot product of two vectors is defined as follows:

$$\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$

where $\mathbf{a} = (a_1, a_2, a_3)$ and $\mathbf{b} = (b_1, b_2, b_3)$.

For the vectors $\mathbf{a} = (i + j + k)$ and $\mathbf{b} = (i + 2j + 3k)$, the dot product is:

$$\mathbf{a} \cdot \mathbf{b} = 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 = 7$$

Therefore, the vectors $\mathbf{a}$ and $\mathbf{b}$ are not perpendicular.

For each of the other options, the dot product with $\mathbf{a}$ is zero. Therefore, those vectors are perpendicular to $\mathbf{a}$.

• Option $\text{(A)}$: $\mathbf{a} \cdot (i – 2j + k) = 1 \cdot 1 – 2 \cdot 2 + 1 \cdot 1 = -3$
• Option $\text{(B)}$: $\mathbf{a} \cdot (-i + 2j – k) = -1 \cdot 1 + 2 \cdot 2 – 1 \cdot 1 = 1$
• Option $\text{(C)}$: $\mathbf{a} \cdot (0i + 0j + 0k) = 0 \cdot 1 + 0 \cdot 2 + 0 \cdot 3 = 0$