Question

In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors A rightarrow = (1,0, -3), rightarrow = (-2,5,1), and C rightarrow = (3,1,1). Calculate the following, and express your answers as ordered triplets of values separated by commas.

Answer 1

Answer:

The calculated vectors are:

$\vec{A}-\vec{B}=(3,-5,-4)$

$\vec{B}-\vec{C}=(-5,4,0)$

$-\vec{A}+\vec{B}-\vec{C}=(-6,4,3)$

$3\vec{A}-2\vec{C}=(-3,-2,-11)$

Explanation:

To operate with vectors, you sum or rest component to component. To multiply scalars with vectors, you distribute the scalar with each component of the vector. These are the following rules you must apply in these cases:

$\vec{V}+\vec{W}=(V_1,V_2,V_3)+(W_1,W_2,W_3)=(V_1+W_1,V_2+W_2,V_3+W_3)$ (1)

$\vec{V}-\vec{W}=(V_1,V_2,V_3)-(W_1,W_2,W_3)=(V_1-W_1,V_2-W_2,V_3-W_3)$ (2)

$\alpha\cdot\vec{V}=\alpha\cdot(V_1,V_2,V_3)=(\alpha\cdot V_1,\alpha\cdot V_2,\alpha\cdot V_3)$ (3)

The operations in these cases are:

$\vec{A}-\vec{B}=(1,0, -3)-(-2,5,1)=(3,-5,-4)$

$\vec{B}-\vec{C}=(-2,5,1)-(3,1,1)=(-5,4,0)$

$-\vec{A}+\vec{B}-\vec{C}=-(1,0, -3)+(-2,5,1)-(3,1,1)=(-6,4,3)$

$3\vec{A}-2\vec{C}=3(1,0, -3)-2(3,1,1)=(3,0, -9)-(6,2,2)=(-3,-2,-11)$

Answer 2

Answer:

Part A: (3, -5, -4)

Part B: (-5, 4, 0)

Part C: (-6, 5, 3)

Part D: (-3, -2, -11)

Part E: (17, -12, -6)

Explanation:

This problem involves addition and subtraction of vectors. This can be done by adding and subtracting the respective components of each vector as the case may be.

The full descriptive solution can be found in the attachment below.