Bella C.

asked • 07/03/20# what is 1537km[N37W] - 853km[W]

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

## 4 Answers By Expert Tutors

William W. answered • 07/03/20

Experienced Tutor and Retired Engineer

V_{1} - V_{2} = V_{3}

V_{1} + -V_{2} = V_{3} where -V_{2} is the V_{2} 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 V_{1} 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} = V_{1}sin(37°) = 1537sin(37°) = 924.9897 but this points in the -x direction

And V_{1y} = V_{1}cos(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 V_{3} using the Pythagorean Theorem:

V_{3x}^{2} + V_{3y}^{2} = V_{3}^{2}

V_{3} = √(V_{3x}^{2} + V_{3y}^{2})

V_{3} = √(-71.9897)^{2} + (1227.5028)^{2})

V_{3} = √(5182.5156 + 1506763.072)

V_{3} = √1511945.588

V_{3} = 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]

William W.

Kind of funny that you got 4 different answers from 4 tutors. Some are missing the fact that navigation is noted with the primary direction first, then the angle, then the direction the angle is toward. So N37W means primarily north with an offset towards west of 37 degrees. Please make sure you account for that in looking at the answers.07/04/20

Yefim S. answered • 07/03/20

Math Tutor with Experience

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} + 1227.50^{2})^{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]

William W.

Looks like you made an arithmetic error Yefim, You specify Ay = 1537cos(37), which is correct) but then equate it to 925. It should be 1227.507/04/20

Heidi T. answered • 07/03/20

MS in Mathematics, PhD in Physics, 7+ years teaching experience

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 = √(X^{2} + Y^{2}) = √[(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)

Dr Gulshan S. answered • 07/03/20

PhD In Physics and experience of teaching IB Physics and Math High sc

Hello Bella

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

R^{2 } = P^{2 }+Q^{2} +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 )

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Bella C.

why is there 4 different answers??07/04/20