what is 1537km[N37W] - 853km[W]

Question

Can you show me how you would do this subtraction 2d vector problem

Answer 1

V₁ - V₂ = V₃

V₁ + -V₂ = V₃ where -V₂ is the V₂ in the opposite direction.

Here is a picture of the problem:

In order to add/subtract vectors we need them to be in the same direction. However, these are not. Let's break V₁ into 2 pieces, one in the x-direction and one in the y-direction like this:

That way, we can add V_1x with -V_2x to get V_3x then we can add V_1y + -V_2y to get V_3y (but this is easy because V_2y = 0

Using trig, V_1x = V₁sin(37°) = 1537sin(37°) = 924.9897 but this points in the -x direction

And V_1y = V₁cos(37°) = 1537cos(37°) = 1227.5028

-V_2x is in the positive x-direction so:

So V_3x = V_1x + -V_2x

V_3x = -924.9897 + 853 = -71.9897

And V_3y = V_1y + -V_2y

V_3y = V_1y + -V_2y

V_3y = 1227.5028 + 0 = 1227.5028

We can now combine V_3x and V_3y into a single vector V₃ using the Pythagorean Theorem:

V_3x² + V_3y² = V₃²

V₃ = √(V_3x² + V_3y²)

V₃ = √(-71.9897)² + (1227.5028)²)

V₃ = √(5182.5156 + 1506763.072)

V₃ = √1511945.588

V₃ = 1229.6 km

The direction is found using the inverse tangent function:

tan(θ) = -71.9897/1227.5028 = -0.058647 (I'm going to drop the negative for the moment and remember that this is an angle in Quadrant 2 (NW sector)

θ = tan^-1(0.058647)

θ = 3.36°

So the resulting vector is 1229.6 km[N3.36W]

Answer 2

We consider traditional coordinates system xoy (axis x has E direction, axis y has N direction). Then A_x ⁼ 1537cos(90° + 37°) - (- 853) = - 72 km; A_y = 1537cos37°=1227.50 km;

Then absA = ((- 72)² + 1227.50²)^1/2 = 1229.61 km;

So bearin tanβ = A_y/A_x = 925/(- 72) β = tan^-1(1229.61/(- 72)) = 93.35°

Vector A = 928 km[N3.35°W]

Answer 3

A good first step for any vector problem is to draw a picture, drawing the vectors to scale and at the proper angles. This helps with visualizing the problem/solution. Draw the first vector starting at the origin; the second vector will be drawn starting at the head of the first vector. Since the second vector is subtracted, also draw (use a dashed or different colored line) in a direction 180 degrees from the original direction.

At this point you will also define your positive and negative directions in the horizontal and vertical. It is traditional to define East as positive horizontal and North as positive vertical, so this is what I will do.

Second step is to break each vector into its horizontal and vertical components. I am not 100% sure what your notation (N37W) means - does it mean the angle is 37 degrees north of west or 37 degrees west of north? Which is correct affects the way the trig functions are applied.

Vector 1: 1537 km (N37W)

Angle definition horizontal component vertical component

37 degrees north of west 1537 * cos 37 = 1227.5km (w) 1537 * sin 37 = 958.5 (N)

37 degrees west of north 1537 * sin 37 = 958.5 (w) 1537 * cos 37 = 1227.5km (N)

Vector 2: 853 km (W) - since this vector is only to the west, the vertical component is 0

Third step: Perform the specified operations on the horizontal and vertical components SEPARATELY. At this point remember to apply the correct sign based on the coordinate system defined in step 1. In this example, I will be using the assumption that the specified angle is 37 degrees north of west. If this is not correct, you can repeat the step using the other components.

Horizontal component: X = -1228 - (-853) = - 375 km. Since the result came out negative and west is negative, the horizontal component is: 375 km (W)

Vertical component: Y = 959 - 0 = 959 km (N)

step 4: Find the magnitude and angle of the resultant vector using the Pythagorean theorem and inverse tangent function:

Magnitude:

R = √(X² + Y²) = √[(375)² + (959)²] = 1030 km

Direction: NOTE: I am finding the angle wrt the West direction, so use the absolute value of the horizontal component.

Θ = tan^-1(Y/X) = tan^-1(959/375) = 69 degrees north of west.

If the original angle was wrt the north direction, then there are two options from this point - solve the equation as written and then subtract from 90 degrees to get the angle wrt the north direction OR use Θ = tan^-1(X/Y)

Answer 4

Hello Bella

Here we add vector P and -Q (853E)

R² = P² +Q² +2PQ Cos theta

Theta in this case is 90-37 = 53 degree

P= 1537 km,

Q= 853 km

Plug in P and Q

get R = resultant ( In N to E direction )