Question

Consider the following three vectors:

A = (2,−1,1), B = (3,0,5), and C = (1,4,−2).
Calculate the following combinations. Express your answers as ordered triplets [e.g., (9,4,−2)].
a. A+Bb. B+Cc. A+B+Cd. 3A+2Ce. 2A+ 3B+Cd. 2A+3(B+C)

Answer 1

Vector combinations involve adding or scalar-multiplying the respective components of each vector given. The results for the different combinations are expressed as ordered triplets using standard vector arithmetic.

Step-by-step explanation:

To calculate the vector combinations for vectors A, B, and C, we perform vector addition and scalar multiplication as follows:

• a. A + B: We add the corresponding components of vectors A and B, which gives us (2+3, -1+0, 1+5) = (5, -1, 6).
• b. B + C: Similarly, adding B and C gives us (3+1, 0+4, 5-2) = (4, 4, 3).
• c. A + B + C: Adding all three vectors gives us (2+3+1, -1+0+4, 1+5-2) = (6, 3, 4).
• d. 3A + 2C: This combination is obtained by multiplying vector A by 3 and vector C by 2 before adding them: (3*2+2*1, 3*-1+2*4, 3*1+2*-2) = (8, 5, 1).
• e. 2A + 3B + C: Multiply each vector by its respective scalar and add the results: (2*2+3*3+1, 2*-1+3*0+4, 2*1+3*5-2) = (13, 2, 17).
• f. 2A + 3(B + C): We first add vectors B and C, then multiply the result by 3, and finally add it to 2 times vector A: 2A + 3*(4, 4, 3) = (2*2, 2*-1, 2*1) + (3*4, 3*4, 3*3) = (4, -2, 2) + (12, 12, 9) = (16, 10, 11).

The algebra of vectors allows us to manipulate vector equations using the commutative, associative, and distributive laws, which makes calculating the above combinations straightforward.

Answer 2

To make the sum of the vectors, we must make the sum of each of its components. Note that each comma indicates a different component and that component must be the same as that added to the other vector. If there is a number multiplying the vector, we will first proceed to perform the internal multiplication and then the sum.

A)

`A+B = (2,-1,1)+(3,0,5)`

`A+B = (5,-1,6)`

B)

`B+C = (3,0,5)+(1,4,-2)`

`B+C = (4,4,3)`

C)

`A+B+C =(2,-1,1)+(3,0,5)+(1,4,-2)`

`A+B+C = (6,3,4)`

D)

`3A+2C = 3(2,-1,1)+2(1,4,-2)`

`3A+2C = (8,5,-1)`

E)

`2A+3B+C = 2(2,-1,1)+3(3,0,5)+)+(1,4,-2)`

`2A+3B+C = (14,2,15)`

F)

`2A+3(B+C) = 2(2,-1,1)+3((3,0,5)+(1,4-2))`

`2A+3(B+C) = (16,10,11)`