The scalar product of vector $\mathbf{A} = 3.00 \hat{i} + 2.00 \hat{j}$ and vector $\mathbf{B}$ is 12.0. Which of the following vectors could be vector $\mathbf{B}$?

A. $5.00 \hat{i} + 5.00 \hat{j}$
B. $17.0 \hat{i}$
C. $2.00 \hat{i} + 3.00 \hat{j}$
D. $3.00 \hat{i} + 4.00 \hat{j}$
E. $4.00 \hat{i} + 5.00 \hat{j}$

To find the vector B that results in a scalar product of 12.0 with the given vector A (3.00 i^ + 2.00 j^), you would need to calculate the dot products for each option provided. The only vector that yields the desired scalar product is vector c) 2.00 i^ + 3.00 j^.

Explanation:

The question relates to finding which vector B, when dotted with vector A (3.00 i^ + 2.00 j^), gives a scalar product of 12.0. The scalar product (also known as the dot product) of two vectors A and B in two-dimensional space is given by A.B = AxBx + AyBy, where Ax and Ay are the components of vector A, and Bx and By are the components of vector B. Thus, to find vector B that results in a scalar product of 12.0, we substitute the given values and solve for Bx and By.

Given vector A is 3.00 i^ + 2.00 j^. To solve for B, we need to consider the possible vectors provided:

Option a) 5.00 i^ + 5.00 j^: A.B = (3.00)(5.00) + (2.00)(5.00) = 15 + 10 = 25
Option b) 17.0 i^: A.B = (3.00)(17.00) + (2.00)(0) = 51 + 0 = 51
Option c) 2.00 i^ + 3.00 j^: A.B = (3.00)(2.00) + (2.00)(3.00) = 6 + 6 = 12
Option d) 3.00 i^ + 4.00 j^: A.B = (3.00)(3.00) + (2.00)(4.00) = 9 + 8 = 17
Option e) 4.00 i^ + 5.00 j^: A.B = (3.00)(4.00) + (2.00)(5.00) = 12 + 10 = 22

The only vector that results in a scalar product of 12.0 with vector A is option c) 2.00 i^ + 3.00 j^.