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Lesson 12 – 7 Geometric Vectors

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1 Lesson 12 – 7 Geometric Vectors
Pg. 643 #1, 5, 10, 13, 18, 20, 27, 29, 30, 32, 35–39 Lesson 12 – 7 Geometric Vectors Pre-calculus To draw vectors To find the norm (magnitude) of vectors To use graphs of vectors

2 Learning Objective To draw vectors
To find the norm (magnitude) of vectors To use graphs of vectors

3 Vectors 𝐴𝐵 Vector: a quantity that has both magnitude and direction. B
head A tail 𝐴𝐵 Vectors can be placed anywhere on a grid, not necessarily starting at the origin Note: the textbook will use bold print v, we will write 𝑣 The magnitude or length or norm of a vector can be found using the Pythagorean Theorem 𝑣 ***Vectors are equivalent if they have the same magnitude and direction no matter where they are located.

4 Vectors If you add two vectors, you get a resultant vector
There are two methods. Note: Subtracting vectors is like adding the opposite of the second vector. 𝑣 1. Find 𝑢 + 𝑣 𝑢 Method 1: Tail–to–head Method 2: Parallelogram 𝑢 𝑢 + 𝑣 𝑣 𝑢 + 𝑣 𝑣 𝑣 𝑢 𝑢

5 Vectors 2. Find 𝑢 − 𝑣 = 𝑢 +(− 𝑣 ) 𝑣 𝑢 Method 1: Tail–to–head Method 2:
= 𝑢 +(− 𝑣 ) Vectors 𝑣 𝑢 Method 1: Tail–to–head Method 2: Parallelogram 𝑢 𝑢 − 𝑣 𝑢 − 𝑣 − 𝑣 𝑢 − 𝑣 − 𝑣 𝑢

6 Zero Vector: If 𝑣 = 𝑢 , then 𝑣 − 𝑢 =0 is the zero vector (magnitude = 0)
Vectors A scalar will change the length of a vector If 𝑎 is a scalar and 𝑣 is a vector, 𝑎 𝑣 is longer if 𝑎 >1 𝑎 𝑣 is shorter if 𝑎 <1 * If 𝑎 is negative, the direction reverses

7 Vectors 3. Use vector 𝑣 to draw & find the norm of 0.5 𝑣 4 4 0.5 𝑣
= 𝑣 =0.5( ) =0.5( 25 ) =0.5(5) =2.5

8 We can find the horizontal & vertical components of a vector using trig.
Vectors 𝑣 𝑦=𝑟 sin 𝜃 𝜃 𝑥=𝑟 cos 𝜃

9 4. Given that vector 𝑤 has a magnitude of 10 and a direction of 135 𝑜 , find the 𝑥 and 𝑦 components.
Vectors 5 2 10 135° −5 2 =10 − 𝑥=10 cos 135 𝑜 =−5 2 = 𝑦=10 sin 135 𝑜 =5 2

10 Vectors can be used together with the Law of Cosines or the Law of Sines to solve problems involving physical quantities such as force, velocity, and acceleration Vectors N heading air velocity course wind velocity ground velocity Norm of air velocity = air speed Direction of air velocity = plane’s heading Measured CLOCKWISE from the north Ground velocity = vector sum of air velocity and wind velocity Course = direction of ground velocity (angle from the north)

11 5. Aviation: An airplane has an air speed of 520 mph and a heading of 115 𝑜 . Wind blows from the east at 37 mph. Find the plane’s ground speed (norm of ground velocity) and course. Vectors N 115° 90° 25° 𝛼 𝑜 520 𝑣 25° 37 Find 𝑣 and 𝛼 𝑜 3 sides … Law of Cosines 𝑣 2 = − cos 25 𝑜 Plane’s Ground Speed = 𝑣 =487 𝑚𝑝ℎ So course is 115 𝑜 + 2 𝑜 = 117 𝑜 sin 𝛼 𝑜 = sin 25 𝑜 487 𝛼= 𝑠𝑖𝑛 −1 ( 37sin 25 𝑜 ) 𝛼= 2 𝑜

12 Assignment Pg. 643 #1, 5, 10, 13, 18, 20, 27, 29, 30, 32, 35–39 all


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