Vector A is in the direction 34.0 clockwise from the y-axis. The x-component of Ais A_x= –16.0 m.

(a) What is the y-component of A?

(b) What is the magnitude of A?

The vector's various coordinates are presented here. We know that in coordinates, the x component of a vector comes first, followed by the y component. To calculate the angle of the vector with respect to the x-axis, just multiply the y component by the x component and then take the inverse of that result.

This will tell you the vector's angle with the x-axis.

tan^-1 (y/x) is one way to express it.

Here

• $\theta$= 90º – 34º = 56º (Clockwise)
• A_x = 16 m

a)

Here from the direction of the vector the component can be calculated as:

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ \tan \theta =\frac{A_y}{A_x} \\ {A}_x\tan \theta =A_y \end{gathered}$$

Here we will use a negative sign because the vector is in the second quadrant:

A_y = 16 tan 56º

= 23.7 m

Hence, the y coordinate is 23.7 m.

b)

The magnitude of the vector can be calculated by the sum of the square of the magnitude of the individual coordinates as:

$$\begin{gathered} A=\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2} \\ =\sqrt{(23.7{)}^2+(16{)}^2} \\ =28.6\mathrm{m} \end{gathered}$$

Hence, the magnitude of the vector is 28.6 m.