Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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^.