VECTORS IN COMPONENT FORM

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Erik Cameron
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1 VECTORS IN COMPONENT FORM In Cartesian coordinates any D vector a can be written as a = a x i + a y j + a z k a x a y a x a y a z a z where i, j and k are unit vectors in x, y and z directions. i = j = k = a = a x + a y + a z a is called magnitude, length, modulus or norm example: a = 7 = = + 7 = i + 7 j k.

2 Unit vector Definition A unit vector is a vector whose length is. It gives direction only! For a vector a, a unit vector is in the same direction as a and is given by: a = a a a = a x i + a y j + a z k a x + a y + a z = a x + a y + a z a x a y a z

3 VECTOR BETWEEN TWO POINTS AB = BA = x B x A y B y A = (x B x A ) i + (y B y A ) j + (z B z A ) k z B z A x A x B y A y B = (x A x B ) i + (y A y B ) j + (z A z B ) k z A z B modulus length: AB = BA = (x B x A ) +(y B y A ) +(z B z A )

4 PARALLEL VECTORS Definition Two vectors a and b are parallel if and only if a = kb where k is a scalar. (where kεr) a = 6 9 b = 6 9 = a b

5 How can you check it: ARE POINTS COLLINEAR?. Form two vectors with these three points. They will definitely have one common point.. Check if these two vectors are parallel. If two vectors have a common point and are parallel (or antiparallel) they are collinear. Show that P(,, 4), Q(,, ) and R(5,, ) are collinear. PR = 5, QR = 5 QR = PR QR and PR have a common direction and a common point. Therefore P, Q and R are collinear.

6 THE DIVISION OF A LINE SEGMENT X divides [AB] AB in the ratio a: b means AX: XB = a b A = (, 7, 8) B = (,, ) INTERNAL DIVISION P divides [AB] internally in ratio :. Find P AP: PB = : AP = 4 AB AP = 4 AB x y 7 z point P is (, 6, 9) EXTERNAL DIVISION X divide [AB] externally in ratio :, or X divide [AB] in ratio : AQ: QB = : BQ = AB x y z = 4 4 point Q is (,,6)

DOT/SCALAR PRODUCT Scalar: ± number Definition The dot/scalar product of two vectors a and b is: a b = b a = a b cos θ b θ a or: Product of the length of one of them and projection of the other one

7 DOT/SCALAR PRODUCT Scalar: ± number Definition The dot/scalar product of two vectors a and b is: a b = b a = a b cos θ b θ a or: Product of the length of one of them and projection of the other one on the first one b a b a a b = a x a y a z b x b y b z = a x b x + a y b y + a z b z Ex: Find the angle between Dot product of perpendicular vectors is zero. a = 5 i j + k and b = θ = arccos a b = a b θ = arc cos = π/ i + j k

8 a b = a x a y a z b x b y b z = a x b x + a y b y + a z b z a x i + a y j + a z k b x i + b y j + b z k =

9 Properties of dot product a b = b a if a and b are parallel, then a b = a b if a and b are antiparallel, then a b = a b a a = a a b + c = a b + a c a + b c + d = a c + a d + b c + b d a b = a, b a and b are perpendicular

a b = a b sin θ Direction of the vector a b is given by right hand rule:

10 CROSS / VECTOR PRODUCT Definition The magnitude of the vector a b is equal to the area determined by both vectors. a b = a b sin θ Direction of the vector a b is given by right hand rule: Point the fingers in direction of a; curl them toward b. Your thumb points in the direction of cross product. b a = a b

11 Properties of vector/cross product a b = b a if a and b are perpendicular, then a b = a b a b + c = a b + a c a + b c + d = a c + a d + b c + b d a b = a, b a and b are parallel For parallel vectors the vector product is. => i i = j j = k k = i j = k j k = i k i = j a b = a y b z a z b y a z b x a x b z a x b y a y b x = i j k a x a y a z b x b y b z

12 a = 5 i j + k b = i + j k (a) Find the angle between them (b) Find the unit vector perpendicular to both (a) θ = arc sin a b = a b a b i j k 5 = 5 i + 6 j + 7 k a b = 5 i + 6 j + 7 k = Find all vectors perpendicular to both a = a b = and b = i j k vector of the form k = i j + k where k is any non zero real number. a = b = θ = arc sin = π/ (b) n = a b = a b 5 6 7

13 Find the area of the triangle with vertices A(,,), B(4,-,), and C(,,8) It is one-half the area of the parallelogram determined by the vectors AB = and AC = 5 AB BC = i j k 5 = ( ) +( ) +( ) = 8.6 units

Conclusions: To find angle between vectors the easiest way is to use dot product, not vector product.

acute angle between them, so θ = arccos a b a b acute angle!

To show that two lines are perpendicular use the dot product with line direction vectors.

14 Conclusions: To find angle between vectors the easiest way is to use dot product, not vector product. Angle between vector can be positive or negative θ = arccos a b a b Angle between lines is by definition acute angle between them, so θ = arccos a b a b acute angle! Dot product of perpendicular vectors is zero. For perpendicular vectors the dot/scalar product is. To show that two lines are perpendicular use the dot product with line direction vectors. To show that two planes are perpendicular use the dot product on their normal vectors. To find the angle between two lines a b cos θ = a b a and b are direction vectors

15 Volume of a parallelepiped = scalar triple product V = c a b = c x c y c z a x a y a z units b x b y b z Volume of a tetrahedron = 6 scalar triple product V = 6 c a b = 6 c x c y c z a x a y a z units b x b y b z TEST FOR FOUR COPLANAR POINTS If the volume of the tetrahedron is zero points are coplanar.

16 LINE EQUATION IN D and D COORDINATE SYSTEM Vector equation of a line The position vector r of any general point P on the line passing through point A and having direction vector b is given by the equation r = a + t b t R r = a i + a j + a k + λ b i + b j + b k or IB Convention: x y z = a a a use t for D line λ for D line b + λ b b Parametric equation of a line λ is called a parameter λ R x y z = a a a + λ b b b x = a + λb y = a + λb z = a + λb Cartesian equation of a line x = a + λb λ = (x a )/b y = a + λb λ = (y a )/b x a = y a = z a b z = a + λb λ = (z a )/b b b (= λ)

17 Find a) vector b) parametric c) Cartesian and d) general line equation of a line passing through A = & B = 6. choosing point A = and direction AB = 4 a) x y = + t 4 t R b) x = + t y = + 4t t R c) x = y 4 d) 4x 4 = y 4 y 4x =

18 Shortest distance from a point to a line Point P is at the shortest distance from the line when PQ is perpendicular to b PQ b = Find the shortest distance between r = + λ and point P (,,). (The goal is to find Q first, and then PQ ) Point Q is on the line, hence its coordinates must satisfy line equation: x Q y Q zq = + λ + λ + λ PQ = λ + λ + λ λ + λ + λ = 4λ + + 9λ 4 + 4λ = 7 λ = λ = 7 PQ = /7 /7 /7 find PQ

19 Relationship between lines D: D: the lines are coplanar (they lie in the same plane). They could be: intersecting parallel coincident the lines are not coplanar and are therefore skew (neither parallel nor intersecting)

20 r = + λ and r = + μ. Are the lines the same?.check by inspection parallel? check by inspection skew or do they have one point in common? solving r = r will give equations in and µ. Solve two of the equations for and µ. if the values of and µ do not satisfy the third equation then the lines are skew, and they do not intersect. If these values do satisfy the three equations then substitute the value of or µ into the appropriate line and find the point of intersection.

21 Line : x = + s, y = s, z = + 4s Line : x = t, y = t, z = t Line : x = + u, y = u, z = 4 + u a) Show that lines and intersect and find angle between them b) Show that line and are skew. a) t = + u t = u, t = u u = u u = & t = checking with z: t = 4 + u = 4 + confirmed intersection (,, 7) direction vectors for line and line are: b = and d = cos θ = b d b d = = 9 84 θ.9o b) + s = + u s u =, s = u s + u =, u = & s = checking with z: + 4s = 4 + u + 4 = no simultaneous solution to all equations the line do not meet, and as they are not parallel b k d, kεr they must be skew.

22 Distance between two skew lines r = a + λ b and r = c + μ d The cross product of b and d is perpendicular to both lines, as is the unit vector: n = b d b d The distance between the lines is then d = n ( c a) (sometimes I see it, sometimes I don t)

23 PLANE EQUATION Vector equation of a plane r = a + λ b + µ c A plane is completely determined by two intersecting lines, what can be translated into a fixed point A and two nonparallel direction vectors The position vector r of any general point P on the plane passing through point A and having direction vectors b and c is given by the equation r = a + λ b + µ c λ, µ R AP = λ b + µ c Parametric equation of a plane: λ, μ are called a parameters λ,μ R x y z = a a a + λ b b b + μ c c c x = a + λb + μc y = a + λb + μc z = a + λb + μc Normal/Scalar product form of vector equation of a plane n r = n a + λ b + µ c r n = a n or n r a = Cartesian equation of a plane r n = a n n x + n y + n z = n a + n a + n a = d n x + n y + n z = d

24 Distance from origin: D = r n = a n = a n n +n +n = n a +n a +n a n +n +n

25 Find the equation of the plane passing through the three points P (,-,4), P (,7,-), and P (5,,-). b = P P = 8 5 one point on the plane is P = vector form: r = n = i j k = c = P P = Any non-zero multiple of n is also a normal vector of the plane. Multiply by -. n = λ Cartesian form: 5x + 5y + z = 44 + µ = 44 Find the equation of the plane with normal vector 5 containing point (-,, 4). 5,,4 = = 7 x + y + 5z = 7 Find the distance of the plane r from the origin, and the unit vector perpendicular to the plane. 4 9 r 4 D = 8 9 = 9 = 8 9 n = = 8

26 It's fairly straightforward to convert a vector equation into a Cartesian equation, as you simply find the cross product of the two vectors appearing in the vector equation to find a normal to the plane and use that to find the Cartesian equation. But this process can't exactly be reversed to go the other way. To convert Cartesian -> vector form, you need either two vectors or three points that lie on the plane! Convert the Cartesian equation of the plane x -y + z = 5 into (a) a vector equation of the form r n = D, where n is a unit vector. (b) vector form (c) State the perpendicular distance of the plane from the origin. (a) n = +4+4 = x y z / / / = 5/ 5/ is the perpendicular distance of the plane from the origin (b) choose three arbitrary random non-collinear points: A(, /, ) B(, /, 4) C(,, ) AB = AC = / r = / + λ + μ / (c) 5/

two planes The angle between two planes is the same as the

27 ANGLES The angle between a line and a plane take acute angle sin θ = cos φ = n d n d θ = arc sin n d n d The angle between two planes The angle between two planes is the same as the angle between their normal vectors cos θ = n m n m θ = arc cos n m n m

28 INTERSECTION of a LINE and a PLANE Line L: r = + μ and plane Π: x + y + z = 5 First check that the line is not contained in the plane, nor parallel to it. i.e = therefore the line and the plane are not parallel and the line will intersect the plane in one point. Substitute the line equation into the plane equation to obtain the value of the line parameter, µ. Substitute for µ into the equation of the line to obtain the co-ordinates of the point of intersection. i.e. Solve + 4μ + 5μ + 6μ = 5 + 4µ - 4+µ - + 8µ = 5. Solve for µ and substitute into the equation of the line to get the point of intersection. If this equation gives you something like = 5, then the line will be parallel and not in the plane, and if the equation gives you something like 5 = 5 then the line is contained in the plane.

29 INTERSECTION of TWO PLANES and the EQUATION of the LINE of INTERSECTION Solve between the two plane equations in terms of a parameter say, λ, plane Π : x + y + z = 5 and plane Π : x y z = First, check by inspection, that the planes are not parallel (normal vectors are not parallel). Find intersection: 5 ~ R = R + R ~ 4 8 x = t y = 4x + 8 z = x + 7 t = x = y = z 7 which is the equation of the common line, which in vector form is r = t 4 (Equally you can write t = y as a function of x and z, or t = z as a function of x and y). OR Find the vector product of both normals to give the direction of the line. Then you need a point on the line do it in the future

30 INTERSECTION OF TWO or MORE PLANES

What does the equation x + 4y = give in and dimensions? https://www.osc-ib.com/ib-videos/default.asp ID: bojanaradja@hotmail.

31 What does the equation x + 4y = give in and dimensions? ID: bojanaradja@hotmail.com PW: kokakoka subject%report/maths%hl%subject%report%%tz.pdf

FINDING THE INTERSECTION OF TWO LINES

FINDING THE INTERSECTION OF TWO LINES REALTIONSHIP BETWEEN LINES 2 D: D: the lines are coplanar (they lie in the same plane). They could be: intersecting parallel coincident the lines are not coplanar