ME 2304: 3D Geometry & Vector Calculus Dr. Faraz Junejo

b = m a Magnitude m a Direction along a Multiplication of a Vector b a By a scalar b = m a b Magnitude m a Direction along a a We can increase or decrease the magnitude of a vector by multiplying the vector by a scalar

Scalar Multiplication In the examples, vector B (2 units) is half the size of vector A (which is 4 units) . We can write: B = 0.5 A This is an example of a scalar multiple. We have multiplied the vector A by the scalar 0.5.

Exercise Given a = (15,-6,24) , b = (5,-2, 8) and c = (-15/2, 3, -12). Show that a & b have the same direction. Show that a & c have opposite direction By inspecting a & b, it can be seen that: a = 3b or b = 1/3 a Since the scalar 3 or 1/3 is positive, this implies a & b have the same direction. Whereas, by inspecting a & c, it can be seen that: a = -2c or c = -1/2 a Since the scalar -2 or -1/2 is negative, this implies a & c have opposite directions.

Multiplication of Vectors(contd) By a vector two types Cross (Vector) Product Dot (Scalar) Product

Scalar Product multiplication SCALAR a b a . b DOT PRODUCT

a . b = a b cos  Dot Product (contd) b a  360 - In words, a.b is the length of a times the length of b times the cosine of the angle between a and b

Dot Product (contd) X b b a a Magnitude of Component of either one of the vectors Component of the other along the direction of the first X a b  a b  acos bcos 

Dot Product: Summary If we have any two vectors, say a and b, the dot product of a and b is given by: Where, |a| and |b| are the magnitudes of a and b respectively, and θ is the angle between the 2 vectors. The dot product of the vectors a and b is also known as the scalar product since it always returns a scalar value.

Examples in Physics W = F . d Work done on a particle by a constant force F during a displacement d is W = F . d

Exercise The corresponding vector PQ is (0i-100j) A small cart weighing 100 lb is pushed up an incline that makes an angle of 30 Degrees with the horizontal. Find the work done against gravity in pushing the cart a distance of 80ft. y R (a1,a2) The vector PQ represents the force of gravity acting vertically downward with a magnitude of 100 lb. 80 ft Θ=30 P T The corresponding vector PQ is (0i-100j) x From triangle PTR, we conclude that Q (0,-100) Therefore, the work done against gravity is PQ.PR = 4000 ft-lb

Find the dot product of force vectors F1= 4N and F2= 6N acting at 40 Degrees to each other ? θ F1=4N

Scalar Product: Applications To test whether two vectors are perpendicular. If the dot product of two non zero vectors is zero, the vectors are perpendicular.

Scalar Product: Applications Finding the angle between two vectors.

What is the angle between a and b ?

Exercises Find the dot product of vectors P and Q, if they are acting at right angle to each other and Answer: 0 Find P.Q , if P= 6i+5j and Q= 2i-8j Answer: -28 Find the angle between P=3i-5j and Q=4i+6j. Draw a rough sketch to verify your answer. Answer: 115.3 Degrees If, the work done by F in acting from P to q is Answer: 60 Joules

Scalar Product: Applications Finding the projection of one vector on another vector. Projection of a on b is XY a The projection of a onto b will be given by:  Q P In summary, the proja b has length b X Y , and direction It is called the scalar component of a in the direction of b

Vector Projection (contd) Therefore, scalar component of a in the direction of b

Vector Projection (contd) What is the scalar component of a in the direction of b ? Scalar component of a on b is

Since, projection of a onto b is: we know, Therefore, Similarly, projection of b onto a is given by:

If b is a vector not equal to zero, then any vector “a” can be projected onto “b” (as mentioned earlier) as well as onto a vector It can be seen from figure shown below that using vector addition it can be written: a b

Find the vector projection of b = (6i+3j+2k) onto a=i-2j-2k and the scalar component of b in the direction of a Now the scalar component of b in the direction of a will be given by:

Cross Product multiplication VECTOR a b a  b CROSS PRODUCT

Vector Product (contd) a b  Magnitude : Area of the parallelogram generated by a and b.

Vector Product (contd) b  a Magnitude :

Vector Product (contd) b  a Direction : Perpendicular to both a and b.

Right-Hand Rule The order of vector multiplication is important.

Geometrical Interpretation A = a  b b sin  Area of the parallelogram formed by a being the base and height being b sin

Cross Product: Summary Suppose we have 2 vectors a and b. These 2 vectors lie on a plane and the unit vector n is normal (at right angles) to that plane The cross product (also known as the vector product) of a and b is given by: The right hand side represents a vector at right angles to the plane containing vectors A and B.

Cross Product: Properties If and a and b are not null vectors, then a is parallel to b.

Cross Product: Determinant

Vector Products Using Determinants The cross product can be expressed as Expanding the determinants gives

Example A simple cross product -5 9 17

Example (Contd) is perpendicular to C H E K

Example -2i-6j+10K B x A = 2i+6j-10k, which implies B x A = - (A x B) Find A x B and B x A, if -2i-6j+10K B x A = 2i+6j-10k, which implies B x A = - (A x B)

Applications Finding a unit vector perpendicular to a plane. Find a unit vector perpendicular to the plane containing two vectors A vector perpendicular to a and b is Corresponding unit vector

Example Determine a unit vector perpendicular to the plane of and Cross product Magnitude Unit vector is

Find the area of the triangle with vertices A(1,1,2) B(-1,3,2) and C(4,1,5). Find two sides of the triangle, the vectors AB and AC. The area is given as follows: and Now find the cross product of the two vectors:

Find the area of the triangle with vertices A(1,1,1) B(2,3,4) and C(3,0,-1).

Scalar Triple Product: Example In here, scalar triple product can be utilized to determine whether the given points lie in same plane or not

The 3-dimensional Co-ordinate System The x-y plane is horizontal in diagram and shaded green. It can also be described using the equation z = 0, since all points on that plane will have 0 for their z-value.

Normally the 'right-hand orientation’ is used for the 3 axes, with the positive x-axis pointing in the direction of the first finger of our right hand, the positive y-axis pointing in the direction of our second finger and the positive z-axis pointing up in the direction of our thumb.

Example The vector OP has initial point at the origin O (0, 0, 0) and terminal point at P (2, 3, 5). We can draw the vector OP as follows: For the vector OP above, the magnitude of the vector is given by: | OP | = √(22 + 32 + 52) = 6.16 units

Find the distance between the points P (2, 3, 5) and Q (4, -2, 3). Distance Between 2 Points in 3 Dimensions If we have point A (x1, y1, z1) and another point B (x2, y2, z2) then the distance AB between them is given by the formula: Distance AB =sqrt[(x2-x1)2+(y2-y1)2+(z2-z1)2] Find the distance between the points P (2, 3, 5) and Q (4, -2, 3). Answer: 5.74 Units

Example Implying it is not a right triangle. Thus it is neither. Find the lengths of the sides of triangle with vertices (0, 0, 0), (5, 4, 1) and (4, -2, 3). Then determine if the triangle is a right triangle, an isosceles triangle or neither. Solution: First find the length of each side of the triangle by finding the distance between each pair of vertices. (0, 0, 0) and (5, 4, 1) (0, 0, 0) and (4, -2, 3) (5, 4, 1) and (4, -2, 3) These are the lengths of the sides of the triangle. Since none of them are equal we know that it is not an isosceles triangle and since Implying it is not a right triangle. Thus it is neither.

Equation of a Sphere If we square both sides of this equation we get: A sphere is the collection of all points equal distance from a center point. To come up with the equation of a sphere, keep in mind that the distance from any point (x, y, z) on the sphere to the center of the sphere, is the constant r which is the radius of the sphere. Using the two points (x, y, z), and r, the radius in the distance formula, we get: If we square both sides of this equation we get: The standard equation of a sphere is where r is the radius and is the center.

Exercises Find the equation of the sphere with radius, r = 5 and center, (2, -3, 1). By Just plugging above values into the standard equation of a sphere we will get: Find the equation of the sphere with endpoints of a diameter (4, 3, 1) and (-2, 5, 7). Using the midpoint formula we can find the center and using the distance formula we can find the radius. Thus the equation is:

Exercises (Contd) Thus the center is (2, -3, -4) and the radius is 6. Find the center and radius of the sphere, . To find the center and the radius we simply need to write the equation of the sphere in standard form, . Then we can easily identify the center, and the radius, r. To do this we will need to complete the square on each variable. Thus the center is (2, -3, -4) and the radius is 6.

Adding 3-dimensional Vectors Earlier we saw how to add 2-dimensional vectors. We now extend the idea for 3-dimensional vectors. We simply add the i components together, then the j components and finally, the k components. If A = 2i + 5j − 4k and B = −2i − 3j − 5k, then determine A+B. Answer: 2 j − 9 k

Angle Between 3-Dimensional Vectors Earlier, we saw how to find the angle between 2-dimensional vectors. We use the same formula for 3-dimensional vectors, i.e. Find the angle between the vectors a = 4i + 0j + 7k and b = -2i + j + 3k. a• b = (4 i + 0 j + 7 j) • (-2 i + j + 3 k ) a• b = (4 × -2) + (0 × 1) + (7 × 3) a• b = 13

Angle Between 3-Dimensional Vectors (contd) And now for denominator i.e. magnitudes So, θ = arccos(13 ÷ 30.166) Therefore the angle between the vectors a and b is θ = 64.47° Find the angle between the vectors a = 3i + 4j − 7k and b = -2i + j + 3k. Answer: θ = 135.6°

Example We have a cube ABCO PQRS which has a string along the cube's diagonal B to S and another along the other diagonal C to P. What is the angle between the 2 strings? θ

Example (contd) We will assume that we have a unit cube i.e. each side has length 1 unit, with 0 being the origin. The unit vectors i, j, and k act in the x-, y-, and z-directions respectively. So in our diagram, since we have a unit cube OA = I, OC = j and OS = k From the diagram, we see that to move from B to S, we need to go -1 unit in the x direction, -1 unit in the y-direction and up 1 unit in the z-direction. Since we have a unit cube, we can write: BS = −i − j + k Similarly, to move from C to P CP = i − j + k

Example (contd) The dot (scalar) product for the vectors BS and CP is: BS • CP = |BS| |CP| cos θ BS • CP = (−i − j + k) • (i − j + k) = 1 |BS| |CP|  = 3 So , θ = arccos (1/3) θ = 70.5° Therefore, the angle between the strings is 70.5°