ab-3bc+4cd a(3;-1;1) b(1;3;2) c(1;-1;-1) d(4;0;3)

Let's evaluate the expression \( ab - 3bc + 4cd \) using the given vectors \( a(3, -1, 1) \), \( b(1, 3, 2) \), \( c(1, -1, -1) \), and \( d(4, 0, 3) \).

First, we will compute the required vector products:

1. Finding \( ab \): \[ ab = a \cdot b = (3, -1, 1) \cdot (1, 3, 2) = 3 \times 1 + (-1) \times 3 + 1 \times 2 = 3 - 3 + 2 = 2 \]

2. Finding \( bc \): \[ bc = b \cdot c = (1, 3, 2) \cdot (1, -1, -1) = 1 \times 1 + 3 \times (-1) + 2 \times (-1) = 1 - 3 - 2 = -4 \]

3. Finding \( cd \): \[ cd = c \cdot d = (1, -1, -1) \cdot (4, 0, 3) = 1 \times 4 + (-1) \times 0 + (-1) \times 3 = 4 + 0 - 3 = 1 \]

Now, substitute these results into the expression \( ab - 3bc + 4cd \): \[ ab - 3bc + 4cd = 2 - 3(-4) + 4(1) \] \[ = 2 + 12 + 4 = 18 \]

Thus, the result of \( ab - 3bc + 4cd \) is \( 18 \).