VECTORS Eugene, Yelim, Victoria and Matt
Contents • Addition and subtraction of vectors • Position vectors • The scalar Product • Vectors equations of a line
Introduction to Vectors (and position vectors) (the little pointy guys)
What is a Vector? (Revision) • A vector is a quantity specified by a magnitude and a direction. They can exist in any dimensions. You can compare a vector quantity to velocity or acceleration, and the scalar quantity to speed. • The magnitude is the of the path going in that direction. (the notation for it is |x|. The | | modulus denotes x as an absolute value which is always positive.) y A (4,5) 5 Vector: will be the position vector of (4,5) = OP 4 x
What does a vector need? • Without a direction, the quantity will be a scalar. The speed of an object is doesn’t have a direction, so is a scalar. • Without a magnitude…there would be no direction in the first place. • So, all vectors would require magnitude and direction.
Finding Magnitude • The magnitude can visualized as the length of the vector. The vector can be found by applying Pythagoras's theorem on it (given that the vector is a line). Magnitude of vector “v”= Where
The Unit vector • A unit vector, as its name suggests, is a basic unit. • It is a vector, like any other, only that it has a length, or magnitude of 1. • The unit vectors in two-dimensional space are Magnitude: And/or 3 Dimensional Space… Magnitude: or or
Why the Unit Vector? • Because a unit vector has a CONSTANT magnitude (of 1), the “meaningful” part of the vector is only the direction. • A unit vector can point in any direction. It can be used to define a direction for a vector. • Therefore, the use of the unit vector is to be the “unit of measurement” of other vectors. It forms the basis for some vector equations. • To find out how many unit vectors are in a vector, we use a method called VECTOR NORMALIZATION.
Vector Normalization a = [a1, a2, a3] The vector to be normalized. The ^ sign on top of the “a” means that it is the unit vector of “a”. Find the magnitude To find the unit vector, the vector “a” must be divided by its magnitude. Because vector “a” is a 3D vector, there are three directives. The e1 and others are the Standard unit vectors for 3-dimensions. Ignore the fact that ||a|| and substitute by |a|.
Worked Example of Vector Normalization Reminder: The original vector The vector can be written like this, in this “pseudo-scalar” form Finding magnitude • Applying to find unit vector in the original vector’s direction The vector now has a length of 1 and is in the direction of original vector “v”.
Another Example The original vector The vector can be written like this, in this “pseudo-scalar” form Finding magnitude • Applying to find unit vector in the original vector’s direction The vector now has a length of 1 and is in the direction of original vector “v”.
What is a Position vector? A position vector is a vector that has one fixed “reference” point. The reference point is usually denoted as “O”. A P B a b Q OA = a (Position vector of point A is a ) O c C OP = OA + AP = a + 1/2 (b-a) = ½ a +1/2 b = ½ (a+b) O is the origin, which is a fix point.
Worked Example(s) C M D • Given that… • OA is same as CD • MD is 1/3 of vector a • Point N is the midpoint of OC. • Find… • OC • DO • OM • MN • MA N O A
Answers: OC = b-a DO= -b OM= b -1/3 a MN= -b + 1/3 a + ½ (b-a) = -1/2 b – 1/6 a
Addition of Vectors B • There are two possible routes that could be taken to get from A to C: • From A to B to C • Directly from A to C • Therefore AC = AB + BC • This means that AC is the resultant vector of AB and BC. • Because x = AB y = BC z = AC x y C z A It can be written as z = x + y
Subtraction of Vectors B • Because BA has the same magnitude as ABbut just in the opposite direction,It can be said that whenAB = xBA = -x • So CA, which is – z, would be - y - x x y C z A
questions c: questions c: Two vectors, a and b are EQUAL if they have the same magnitude and direction, even if they do not have the same starting points. 1. Given that AB = 5i + 2j – 7k and BC = -3i +6j –k, find AC 2. Given that BC = 9i – j + 3k and AC = 2j – 4k, find BA
Basics • The scalar product (v· w) of the two vectors v and w is defined by the expression: where theta is the angle between the two vectors
Angles and Values • When the two vectors of v and w are perpendicular, the angle between them is a right angle and therefore cos90°= 0 so v· w = 0 • When the angle between the vectors v and w is acute, the cosine value is greater than zero so v· w > 0 • When the angle between the vectors v and w is obtuse, then the cosine value is lower than 0 so v· w < 0
Proof of the Scalar Product • We want to find the scalar product of v· w: • Let v = v1i + v2j and let w = w1i + w2j. Then: v· w = (v1i + v2j)· (w1i + w2j) • Next we expand the brackets: v· w = v1w1 i·i + v1w2 i·j + v2w1 i·j + v2w2 j·j • i·i = j·j = 1 because they are perpendicular and because of this i·j = j·i = 0. Therefore: v· w = v1w1 + v2w2 • In three dimensions we have to add the z plain: v· w = v1w1 + v2w2 + v3w3
Using Scalar Products • We can use scalar to find the angle between two vectors. • First you calculate v· w: v· w = v1w1 + v2w2 + v3w3 • Next you find |v| and |w| and then input those values into the equation: • Rearrange the formula and use inverse cosine to find the value of the angle.
Using Scalar Products (cont’d) • You can also use this to find possible values of a constant e.g.: • Given that the two vectors v = (3t + 1)i + j + k and w = (t + 3)i + 3j – 2k are perpendicular, find the possible values of the constant t. • Because the two vectors are perpendicular, v· w = 0 so… (3t + 1)(t + 3) + (1 x 3) + (-1 x -2) = 0 3t² + 10t + 8 = 0 • Factorizing gives us: (3t + 4)(t + 2) • Solving it gives us the values t = -4/3 or -2 which are the solutions we need
B Y b y A a O Vector Equations Vector Equations are NOT the same as direction vectors. Direction vectors give the position of the vector. e.g. The direction vector of line AB is: b – a Equation vectors are used to identify whether a point lies on line connecting 2 points. e.g. To find whether point Y lies on line AB, the equation vector: y = a + t(b – a) [=
B Y b t y A a O Finding the Vector Equation To find the vector equation of line AB using point Y : OY = OA + AYAY = t(AB) So… OY = OA + t(AB) Using these, we can find the vector equation of line AB using vectors a, b and y: OY = yOA = aAB = b – a y = a + t(b – a) [=
(4,3) O i – 2j Finding Vector Equations Find a vector equation for the line passing through the point (4,3) and parallel to the vector i – 2j. Do textbook question: 13D # 2 [=
Finding Vector Equations Points P and Q have coordinates (3,5) and (-3,-7). Find a vector equation for the line which passes through the point P, and which is perpendicular to the line PQ. Find PQ: Thus, find line passing through P which is perpendicular to PQ: Do textbook question: 13D # 6,8
Concurrent Lines Concurrent lines are two or more line which share a common point. p = (6 - 2s)i + (s – 5)j q = ti +3(1 – t)j r = (5 – u)i + (2u – 9)j Show that the lines are concurrent and find the position vector of their point of intersection.
Application of Vectors Because vectors have both magnitude and direction, they are extremely applicable to everyday situations, such as the velocity of an aircraft. Everyday situation 1: A tractor is moving across a field at 10kmh-1 on a bearing of 150 degrees. Express its velocity as a column vector.
Everyday situation 2: • A cruise ship leaves a harbour and its position (x,y) at time t hours is given by the vector • ( ) = t( ) • Find the speed of the boat. • The position (x,y) of a cargo ship at time t hours is given by the vector • ( ) = ( ) + t( ) • b) Find the distance of the cargo ship from the harbour when the cruise ship starts its journey. • c) If both the cruise ship and the cargo ship continued on the same courses, after what period of time would they collide? x y 10 7 x y 12 12 4 1