Question

Consider a vector 4.08 + 3.0 , wherex, are the unit vectors in x-, y-directions, respectively. (a) What is the magnitude of the vector A? (b) What are the angles vector A makes with the x and y axes, respectively?

Answer 1

Answer:

Part a)

Magnitude = 5.06 unit

Part b)

$\theta = 36.2 ^0$

Explanation:

Part a)

Vector is given as

$\vec A = 4.08 \hat x + 3.0 \hat y$

now from above we can say that

x component of the vector is 4.08

y component of the vector is given as 3.0

so the magnitude of the vector is given as

$|A| = \sqrt{4.08^2 + 3^2}$

$|A| = 5.06 unit$

Part b)

Now the angle made by the vector is given as

$\theta = tan^{-1}(\frac{y}{x})$

$\theta = tan^{-1}(\frac{3}{4.08})$

$\theta = 36.3 degree$