Respuesta :
Answer:
a) The magnitude of the vector is changed my a multiplication factor of 2
b) The direction of the vector is changed by a multiplication factor of 1
Explanation:Let the first component of a vector be x
Let the second component of the vector be y
The vector, A = xi + yj
The magnitude of the vector A will be:
[tex]|A|\text{ = }\sqrt[]{x^2+y^2}[/tex]The direction of the vector A will be:
[tex]\theta_A\text{ = }\tan ^{-1}\frac{y}{x}[/tex]If each component of the vector is doubled
Double of the first component = 2x
Double of the second component = 2y
The new vector, B = 2xi + 2yj
The magnitude of the new vector B will be:
[tex]\begin{gathered} |B|\text{ = }\sqrt[]{(2x)^2+(2y)^2} \\ |B|\text{ = }\sqrt[]{4x^2+4y^2} \\ |B|\text{ = }\sqrt[]{4(x^2+y^2)} \\ |B|\text{ = 2}\sqrt[]{x^2+y^2} \end{gathered}[/tex]The direction of the new vector B will be:
[tex]\begin{gathered} \theta_B=\text{ }\tan ^{-1}\frac{2y}{2x} \\ \theta_B=\text{ }\tan ^{-1}\frac{y}{x} \end{gathered}[/tex]Since:
[tex]|A|\text{ = }\sqrt[]{x^2+y^2}\text{ and }|B|=2\sqrt[]{x^2+y^2}[/tex]then:
|B| = 2|A|
The maginitude of the vector is double. That is, it is changed by a multiplication factor of 2
Since:
[tex]\theta_A=\text{ }\tan ^{-1}\frac{y}{x}\text{ and }\theta_B=\text{ }\tan ^{-1}\frac{y}{x}[/tex]then:
The direction of the vector does not change. That is, the multiplicative factor is 1
