Question

Given two displacement vectors →A=(3.00ˆi−4.00ˆj+4.00ˆk)m and →B=(2.00ˆi+3.00ˆj−7.00ˆk)m, find the displacements and their magnitudes for (a) →C=→A+→B and (b) →D=2→A−→B.

a) →C: (5.00ˆi−1.00ˆj−3.00ˆk) m, Magnitude: 5.916 m; →D: (4.00ˆi−11.00ˆj+15.00ˆk) m, Magnitude: 18.028 m
b) →C: (6.00ˆi−1.00ˆj−3.00ˆk) m, Magnitude: 6.928 m; →D: (5.00ˆi−10.00ˆj+10.00ˆk) m, Magnitude: 14.730 m
c) →C: (6.00ˆi−1.00ˆj−3.00ˆk) m, Magnitude: 7.616 m; →D: (4.00ˆi−9.00ˆj+13.00ˆk) m, Magnitude: 15.620 m
d) →C: (5.00ˆi−1.00ˆj−3.00ˆk) m, Magnitude: 6.708 m; →D: (3.00ˆi−10.00ˆj+12.00ˆk) m, Magnitude: 14.422 m

Answer 1

The high school physics problem involves determining the resultant displacement vectors →C and →D by vector addition and subtraction, and subsequently finding their magnitudes using the Pythagorean theorem in three-dimensional space.

Step-by-step explanation:

The question is related to the addition and subtraction of displacement vectors, and the calculation of their magnitudes in the subject of Physics for a high school level student. Vector addition is done by adding the corresponding components of the vectors, whereas for vector subtraction, components of one vector are subtracted from the respective components of the other vector. The magnitude of a vector is found using the Pythagorean theorem for the three-dimensional space, which is the square root of the sum of the squares of its components.

For vector →C=→A+→B, addition is performed by adding each i, j, and k component of vectors →A and →B. For vector →D=2→A-→B, each component of →B is subtracted from twice the respective component of →A. After determining the vector components, their magnitudes are calculated as sqrt(Cx^2+Cy^2+Cz^2) and sqrt(Dx^2+Dy^2+Dz^2) respectively.