To find the magnitude and direction angle of the sum of the vectors $y + z$, we need to use vector addition in terms of their components.

Given:

• Vector $y$ has a magnitude of 4 and a direction angle of 90°.
• Vector $z$ has a magnitude of 2 and a direction angle of 0°.

We can break each vector into its horizontal ($x$-axis) and vertical ($y$-axis) components using trigonometric functions:

1. For vector $y$: $y_x = 4 \cos(90^\circ) = 0$ $y_y = 4 \sin(90^\circ) = 4$ So, the components of $y$ are $(0, 4)$.

2. For vector $z$: $z_x = 2 \cos(0^\circ) = 2$ $z_y = 2 \sin(0^\circ) = 0$ So, the components of $z$ are $(2, 0)$.

Add the components:

$x_{\text{total}} = y_x + z_x = 0 + 2 = 2$ $y_{\text{total}} = y_y + z_y = 4 + 0 = 4$

Now, the vector sum $y + z$ has components $(2, 4)$.

Magnitude of $y + z$:

The magnitude of the resultant vector is given by: $|y + z| = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.472$

Direction angle $\theta$:

The direction angle $\theta$ can be found using the inverse tangent: $\theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right) = \tan^{-1}\left(\frac{4}{2}\right) = \tan^{-1}(2) \approx 63.435^\circ$

Final answer:

• Magnitude of $y + z$: $\approx 4.472$
• Direction angle $\theta$: $\approx 63.435^\circ$

Would you like more details or have any questions?

Related Questions:

1. How do you convert between polar and Cartesian coordinates?
2. How would the result change if $y$ had a different direction angle?
3. What happens if both vectors had the same direction angle?
4. How can we apply this concept in 3D vectors?
5. How does vector subtraction work in component form?

Tip:

Always break vectors into components for easier addition, especially when angles are involved.