Consider the three displacement vectors
A=(3i+3j)meters,
B-(i-4j) m
C=(-2i+5j) m
Use the Component method to determine
a) the magnitude and direction of the vector D= A+B+C
b) the magnitude And direction of E=-A-B+C

Respuesta :

Answer:

Explanation:

[tex]\overrightarrow{A} = 3\widehat{i}+3\widehat{j}[/tex]

[tex]\overrightarrow{B} = \widehat{i}-4\widehat{j}[/tex]

[tex]\overrightarrow{C} = -2\widehat{i}+5\widehat{j}[/tex]

(a)

[tex]\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}[/tex]

[tex]\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}[/tex]

[tex]\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}[/tex]

Magnitude of [tex]\overrightarrow{D}[/tex] = [tex]\sqrt{2^{2}+4^{2}}[/tex]

                                                                     = 4.47 m

Let θ be the direction of vector D

[tex]tan\theta =\frac{4}{2}[/tex]

θ = 63.44°

(b)

[tex]\overrightarrow{E} =

- \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}[/tex]

[tex]\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}[/tex]

[tex]\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}[/tex]

Magnitude of [tex]\overrightarrow{E}[/tex] = [tex]\sqrt{6^{2}+6^{2}}[/tex]

                                                                     = 8.485 m

Let θ be the direction of vector D

[tex]tan\theta =\frac{6}{-6}[/tex]

θ = 135°

ACCESS MORE