Chapter 3–2: Vector Operations. Physics Coach Kelsoe Pages 86–94. Objectives. Identify appropriate coordinate systems for solving problems with vectors. Apply the Pythagorean theorem and tangent function to calculate the magnitude and direction of a resultant vector.
The magnitude of the vertical displacement is the height of the pyramid.
The magnitude of the horizontal displacement is the distance from one edge of the pyramid to the middle, or half the width.
These two vectors are perpendicular.
The Pythagorean theorem states that for any right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides, or legs.
So in this illustration, we see thatd2 = Δx2 + Δy2
Once you find the displacement, it is important to know the direction.
We can use the inverse tangent function to find the angle, which denotes the direction.
We can use sine or cosine functions, but our tangent values are given.
Tan θ =
An archaeologist climbs the Great Pyramid in Giza, Egypt. The pyramid’s height is 136 m and its width is 2.30 x 102 m. What is the magnitude and the direction of the displacement of the archaeologist after she has climbed from the bottom of the pyramid to the top?
d2 = Δx2 + Δy2
tan θ =
d = √Δx2 + Δy2
θ = tan-1
d = √Δx + Δy
θ = tan-1
d = √1152 + 1362
θ = tan-1
d = 178 m
θ = 49.8°
In the previous example, the horizontal and vertical parts that add up to give the actual displacement are called components.
The x component is parallel to the x-axis and the y component is parallel to the y-axis.
You can often describe an object’s motion more conveniently by breaking a single vector into two components, a process we call resolving the vector.
Sine of angle θ =
Cosine of angle θ =
The goal of resolving vectors is to allow us to have components that are aligned along the x- and y-axis.
Once the components are set up on these axes, our math becomes simple– we add our x components together and y components together to find our new x and y components.
With our new-found x and y components, we can find our new resultant vector.
A hiker walks 27.0 km from her base camp at 35° south of east. The next day, she walks 41.0 km in a direction of 65° north of east and discovers a forest ranger’s tower. Find the magnitude and direction of her resultant displacement.
d1 = 27.0 km θ1 = -35°
d2 = 41.0 km θ2 = 65°
d = ? θ = ?
Δx1 = d1cos θ1 = (27.0 km)(cos -35°) = 22 km
Δy1 = d1sin θ1 = (27.0 km)(sin -35°) = -15 km
Δx2 = d2cos θ2 = (41.0 km)(cos 65°) = 17 km
Δy2 = d2sin θ2 = (41.0 km)(sin 65°) = 37 km
Δxtotal = Δx1+ Δx2 = 22 km + 17 km = 39 km
Δytotal = Δy1+ Δy2 = -15 km + 37 km = 22 km
d2 = (Δxtotal)2 + (Δytotal)2
d = √(39 km)2 + (22 km)2
d = 45 km
Tan θ = opposite/adjacent
Tan θ = Δy/Δx = 22 km/39 km
Tan θ = 0.564
θ = 29° north of east