Question

100 points! Given p with magnitude of 40 and a direction of 70°, q with magnitude of 50 and a direction of 270°, and r with magnitude of 60 and a direction of 235°, what is the magnitude of p + q + r? Round to the thousandths place.

Answer 1

To find the magnitude of p + q + r, we need to add the vectors p, q, and r component-wise.

First, let's find the horizontal and vertical components of each vector:

Vector p has a magnitude of 40 and a direction of 70°. The horizontal component of p, px, is given by px = 40 * cos(70°), and the vertical component, py, is given by py = 40 * sin(70°).

Vector q has a magnitude of 50 and a direction of 270°. The horizontal component of q, qx, is given by qx = 50 * cos(270°), and the vertical component, qy, is given by qy = 50 * sin(270°).

Vector r has a magnitude of 60 and a direction of 235°. The horizontal component of r, rx, is given by rx = 60 * cos(235°) and the vertical component, ry, is given by ry = 60 * sin(235°).

Now, let's calculate the horizontal and vertical components for each vector:

px = 40 * cos(70°) = 40 * 0.34202 ≈ 13.6808
py = 40 * sin(70°) = 40 * 0.93969 ≈ 37.5876

qx = 50 * cos(270°) = 50 * 0 ≈ 0
qy = 50 * sin(270°) = 50 * -1 = -50

rx = 60 * cos(235°) ≈ 60 * 0.81915 ≈ 49.089
ry = 60 * sin(235°) ≈ 60 * -0.57358 ≈ -34.415

Now, let's add up the horizontal and vertical components:

Sum of horizontal components = px + qx + rx ≈ 13.6808 + 0 + 49.089 ≈ 62.7698
Sum of vertical components = py + qy + ry ≈ 37.5876 + (-50) + (-34.415) ≈ -47.827

Finally, we can find the magnitude of the vector p + q + r using the Pythagorean theorem:

Magnitude of p + q + r = sqrt((Sum of horizontal components)^2 + (Sum of vertical components)^2)
= sqrt((62.7698)^2 + (-47.827)^2)
≈ sqrt(3939.732 + 2286.776)
≈ sqrt(6226.508)
≈ 78.92

Therefore, the magnitude of p + q + r is approximately 78.92 (rounded to the thousandths place).

Answer 2

64.959

Explanation:

To calculate the magnitude of the vector sum p + q + r, combine the vector components in both the horizontal (x) and vertical (y) directions and then use the Pythagorean theorem to find the magnitude.

Break down the vectors into their horizontal and vertical components.

For vector p:

• Horizontal component: 40 cos(70°)
• Vertical component: 40 sin(70°)

For vector q:

• Horizontal component: 50 cos(270°)
• Vertical component: 50 sin(270°)

For vector r:

• Horizontal component: 60 cos(235°)
• Vertical component: 60 sin⁡(235°)

Now, add up the horizontal components and vertical components separately:

Horizontal component sum:

`40 \cos70^(\circ) + 50 \cos270^(\circ)+ 60 \cos235^(\circ)`

`= 40 \cos70^(\circ) + 0 + 60 \cos235^(\circ)`

`= 40 \cos70^(\circ) + 60 \cos235^(\circ)`

Vertical component sum:

`40 \sin 70^(\circ) + 50 \sin270^(\circ) + 60 \sin235^(\circ)`

`= 40 \sin70^(\circ) - 50 + 60 \sin235^(\circ)`

Finally, use the Pythagorean theorem to find the magnitude of the vector sum:

`\textsf{Magnitude}=\sf √((Horizontal\; component \;sum)^2+(Vertical \;component\; sum)^2)`

`\textsf{Magnitude}=\sqrt{(40 \cos70^(\circ) + 60 \cos235^(\circ))^2+(40 \sin70^(\circ) - 50 + 60 \sin235^(\circ))^2}`

`\textsf{Magnitude}=64.959201168...`

`\textsf{Magnitude}=64.959\; \sf (3\;d.p.)`

Therefore, the magnitude of the given vector sum is 64.959 (rounded to the nearest thousandth).