Lec6-12.pdf - Euclidean and Complex Vector Spaces Introduction to Linear Algebra Math115 Ghazal Geshnizjani \u2014\u2014 Reference for some of the slides

Lec6-12.pdf - Euclidean and Complex Vector Spaces...

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Unformatted text preview: Euclidean and Complex Vector Spaces Introduction to Linear Algebra, Math115 Ghazal Geshnizjani —— Reference for some of the slides: Introduction to Linear Algebra for Science and Engineering by Norman and Wolczuk Chapter 1, 9 Euclidean Vectors by one number Scalar Quantity:defined Example: distance, speed, mass, etc. Vector quantity: defined by a number and direction Example: displacement, velocity, force, etc. A P ~ AB ~ PQ B Q x y ~ and AB ~ have the same PQ magnitude and direction ~ = AB ~ =⇒ PQ vectors can be translated we only need one label to ~ and AB; ~ for identify PQ example just ~v . Vectors in R2 x2 Q P Definition (R ) 2 O x1   x is the set of all vectors of the form 1 , where x1 , x2 ∈ R are called x2 the coordinates(or components) of the vector.    x1 2 R = x ,x ∈ R x2 1 2 R2 Notation We use the notation   x ~x = 1 x2 to denote vectors in R2 . The first component is referred to as x1 Vectors in R3 In three dimensions, we always pick our coordinate axes so that they form a right-handed system. x3 x3 x2 O O x2 x1 Definition (R3 ) x1 R3 = x2 | x1 , x2 , x3 ∈ R x3 x1 Extending Vectors to n dimensions Definition (Rn ) x1 .. n R = . | x1 , . . . , xn ∈ R xn Definition (Zero vector in Rn ) n Zero vector in Rn is the vector in R that all of its components are zero 0 .. ~ and it is denoted by 0 = . 0 0 0 ~0 = ∈ R4 , · · · 0 0 we have many different zero vectors!   ~0 = 0 ∈ R2 , 0 0 ~0 = 0 ∈ R3 , 0 Vectors Definition (Equal Vectors) x1 y1 .. .. ~ = . in Rn are equal if and only if Two vectors, ~v = . and w xn yn ( ⇐⇒ ) x1 = y1 , x2 = y2 , · · · , xn = yn Examples:   3 3 6= 2 2 0     1 a =⇒ a = 1, b = 3 = b−a 2 Standard position of Vectors p1 We can identify the vector ~p = ... in Rn , with the point P(p1 , p2 , · · · , pn ). pn ~ where O is the origin, O(0, · · · , 0) in Rn . So we have ~p = OP ~ is the standard vector or position vector representation for ~p . When ~p OP has its tail at the origin, we say ~p is in standard position. We represent the standard vector graphically by drawing an arrow from (0, · · · , 0) to (p1 , · · · , pn ). Example for R2 : x2 P = (p1 , p2 )   p p = 1 p2 Addition in Rn Definition (Vector Addition) Suppose ~a , ~b ∈ Rn : x1 y1 .. .. ~ ~a = . , b = . . xn yn ~ = AC ~ ~ + BC AB A then x1 y1 x1 + y1 ~a + ~b = ... + ... = ... . xn yn xn + yn Examples: 1 0 1 −4 + 5 = 1 , 2 −3 −1 C ~ AC ~ AB ~ BC B 1 3 + 4 = not defined! −1 −3   Scalar Multiplication in Rn Definition (Scalar Multiplication) Suppose x1 .. ~x = . , xn t ∈ R. The scalar multiplication of ~x vector by the factor of t, called a scalar, is defined by x1 tx1 .. .. t~x = t . = . . xn txn Examples: 1 2 2 −4 = −8 , 2 4 0     3 0 = = ~0 −1 0 −~v o ~v 2~v Definition (Parallel Vectors) Two non-zero vectors are parallel if and only if ( ⇐⇒ ) one is a scalar multiple of the other. Examples:     3 −9 ~v = ~ = , w 2 −6 1 ~v = − w ~ =⇒ ~v and w ~ are parallel 3     3 −8 ~v = ~ = , w 2 −6 @k ∈ R such that 3 = −8k and 2 = −6k ~ are not parallel =⇒ ~v and w Subtraction in Rn Definition (Vector difference) Suppose ~a , ~b ∈ Rn : x1 .. ~a = . , xn ~ = AC ~ − AB ~ BC y1 . ~b = .. . yn A then x1 y1 x1 − y1 ~a − ~b = ... + (−1) ... = ... . xn yn xn − yn Examples: 1 0 1 −4 − 5 = −9 , 2 −3 5 C ~ AC ~ AB ~ BC B 1 3 − 4 = not defined! −1 −3   Properties of vectors under addition and scalar multiplication Theorem ~ , ~x , ~y ∈ Rn and s, t ∈ R, we have the following. For all w 1 ~ x + ~y ∈ Rn (closed under addition) 2 ~ x + ~y = ~y + ~x (addition is commutative) 3 (~ ~ = ~x + (~y + w ~) x + ~y ) + w (addition is associative) n 4 There exists a vector ~ 0 ∈ R such that ~z + ~0 = ~z for all ~z ∈ Rn . (zero vector) 5 There exists a vector −~ x ∈ Rn such that ~x + (−~x ) = ~0. (additive inverses) 6 t~ x ∈ Rn (closed under scalar multiplication) 7 s(t~ x ) = (st)~x (scalar multiplication is associative) 8 (s + t)~ x = s~x + t~x (a distributive law) 9 t(~ x + ~y ) = t~x + t~y (another distributive law) 10 1~ x = ~x (scalar multiplicative identity) Linear combination of vectors Definition (Linear Combination) Given vectors ~v1 , . . . , ~vk ∈ Rn and numbers t1 , . . . , tk ∈ R, we call t1~v1 + · · · + tk ~vk , a linear combination of ~v1 , . . . , ~vk . ~v1 , . . . , ~vk ∈ Rn , property 1 and 6 =⇒ t1~v1 + · · · + tk ~vk ∈ Rn (closed under linear combination) Examples: 1 1 0 1 2 −4 − 5 = −7 2 2 −3 4 1 1 0 2 =⇒ −7 is linear combination of −4 and 5 . 2 −3 4 Standard Basis 1 1 0 0 1 Consider −4 ∈ R3 , −4 = 1 0 + (−4) 1 + 2 0 2 0 0 1 2 if we call 1 0 0 1 ~e1 = 0 ,~e2 = 1 ,~e3 = 0 =⇒ −4 = 1~e1 + (−4)~e2 + 2~e3 0 0 1 2 Definition (Standard Basis for Rn ) In Rn , let ~ei be the vector whose i-th component is 1 and all other components are 0. The set {~e1 , . . . ,~en } is called the standard basis for Rn . x1 x1 .. .. n For . ∈ R we get . = x1~e1 + x2~e2 + · · · + xn~en xn xn Definition (Norm) x1 .. The norm or magnitude of ~x = . is xn v u n q uX 2 2 k~x k = x1 + · · · + xn = t xi2 . i=1 1   2 √ 3 =⇒ k~ ~ = Examples: ~v = =⇒ k~v k = 5, w w k = 6 1 4 0 Theorem Let ~x , ~y ∈ Rn and t ∈ R. Then 1 k~ x k ≥ 0, and k~x k = 0 if and only if ~x = ~0, 2 kt~ x k = |t|k~x k, 3 k~ x + ~y k ≤ k~x k + k~y k.Triangle Inequality Example Find the distance from A(1, −1, 2) to B(3, 2, 1) Show k~x k − k~y k ≤ k~x − ~y k. Definition (Unit Vector) A vector ~x ∈ Rn such that k~x k = 1 is called a unit vector. Examples: ~e2 ∈ R2 is a unit vector. ( ∀ ~ei ∈ standard basis of Rn =⇒ ~ei is a unit vector.) 1 √ 3 ~ = √13 , =⇒ k~ wk = w √1 3 q 1 3 + 1 3 + 1 3 =1 Theorem ~ = Let ~v ∈ Rn with ~v 6= ~0. Then w 1 ~ k~v k v Proof: k~ w k = k~v1k ~v = k~v1k k~v k = is a unite vector in direction of ~v . 1 vk k~v k k~ =1 ~ is a scalar multiplication of ~v by a positive factor w in the same direction as ~v . 1 k~v k so it’s parallel and Example: 4 Find the unit vector in direction of ~v = 5. 6 Solution:k~v k = so √ 42 + 52 + 62 = √ 77 √4 4 77 1 ~ = √ 5 = √577 w 77 6 √6 77 ~ w ~v The Dot Product Dot Product (or scalar product ) x1 y1 .. .. The dot product of vectors ~x = . , ~y = . in Rn is xn yn ~x · ~y = x1 y1 + x2 y2 + · · · + xn yn . Examples: 1 −3 1 · −4 = 1(−3) + 1(−4) + 2(5) = −3 − 4 + 10 = 3. 2 5     a −3 · = −2 =⇒ a(−3) + 5a = −2 =⇒ 2a = −2 =⇒ a = −1 5 a Properties of Dot Product Theorem Let ~x , ~y ,~z ∈ Rn and t ∈ R. Then 1 ~ x · ~y ∈ R √ 2 ~ x · ~x = k~x k2 (k~x k = ~x · ~x ) 3 ~ x · ~x ≥ 0, and ~x · ~x = 0 if and only if ~x = ~0, 4 ~ x · ~0 = 0, 5 ~ x · ~y = ~y · ~x , 6 ~ x · (~y ± ~z ) = ~x · ~y ± ~x · ~z , 7 (t~ x ) · ~y = t(~x · ~y ) = ~x · (t~y ). Exercise: prove above properties. For (7), 1st part: (t~x ) · ~y = (tx1 )y1 + (tx2 )y2 + · · · + (txn )yn = t(x1 y1 + x2 y2 + · · · + xn yn ). = t(~x · ~y ) Dot Product and Angle between Vectors Let ~v , ~u be non-zero vectors in Rn with a common tail. Then they determine a unique angle, θ with 0 ≤ θ ≤ π. ~u ~u θ ~v ~v Acute: 0 ≤ θ < ~u θ π 2 Obtuse: π 2 <θ≤π θ ~v Orthogonal: θ = π 2 Theorem For two non-zero vectors ~v , ~u ∈ Rn and the determining angle, θ, where 0 ≤ θ ≤ π the following relation holds: ~v · ~u = k~v kk~u k cos θ ~u − ~v ~u θ ~v k~u − ~v k k~u k θ k~v k Proof:From Cosine Law we know how to relates the lengths of the sides of a triangle to the cosine of one of its angles: k~u − ~v k2 = k~u k2 + k~v k2 − 2k~u kk~v k cos θ, (1) From properties of dot product we know k~u −~v k2 = (~u −~v ) · (~u −~v ) = ~u ·~u +~v ·~v − 2~v ·~u = k~u k2 + k~v k2 − 2~v ·~u , (2) (1) and (2) =⇒ −2k~u kk~v k cos θ = −2~u .~v =⇒ ~v · ~u = k~u kk~v k cos θ Corollaries based on ~v · ~u = k~v kk~u k cos θ: Cauchy-Schwarz Inequality: For any two vectors ~v , ~u ∈ Rn , the size of dot product can not exceed the product of their norms: |~v · ~u | ≤ k~v kk~u k Proof: ~v · ~u = k~v kk~u k cos θ =⇒ |~v · ~u | = k~v kk~u k| cos θ| we know| cos θ| ≤ 1, =⇒ |~v · ~u | ≤ k~v kk~u k Calculating angles between vectors For any two non-zero vectors ~v , ~u ∈ Rn , we can calculate the angle θ between them (0 ≤ θ ≤ π ): cos θ = ~v · ~u k~v kk~u k Example: 2 1 ~ = −1. Compute the angle between ~v = 1 and w −1 −2 Solution: cos θ = ~v · w ~ 2(1) − 1 + 2 3 1 √ =√ =√ √ = k~v kk~ wk 2 4+1+1 1+1+4 6 6 π 1 θ = cos−1 ( ) = 3 2 Shape of the angle 0 ≤ θ ≤ π between vectors: k~v kk~u k > 0 =⇒ cos θ and ~v · ~u have the same sign. π ~v · ~u > 0 ⇐⇒ 0 ≤ θ < , (ACUTE) 2 ~v · ~u = 0 ⇐⇒ θ = ~v · ~u < 0 ⇐⇒ ~u ~u θ π < θ ≤ π, (OBTUSE) 2 ~u θ ~v ~v ACUTE: 0 ≤ θ < π , (ORTHOGONAL) 2 π 2 OBTUSE: π 2 <θ≤π θ ~v ORTHOGONAL: θ = π2 Examples: Determine type of the angle in between the two vectors:     1 −1 1 For ~ ~ = v= and w . 2 2 2 3 ~v · w ~ = −1 + 4 = 3 > 0 =⇒ acute angle. 3 1 ~ = −5. For ~v = 0 and w 5 −1 ~v · w ~ = 3 − 5 = −2 < 0 =⇒ obtuse angle. 3 1 0 −5 ~ = For ~v = 5 and w −1. 2 1 ~v · w ~ = 3 − 5 + 2 = 0 =⇒ orthogonal. Exmple: Evaluate the angle for number (1) above: ~v · w ~ 3 3 −1 3 Orthogonal Vectors     1 0 ~ Example: What is the angle between ~v = and 0 = ? 2 0 ~v · ~0 = 0 + 0 = 0 =⇒ orthogonal? cos θ? Note: ∀ ~v ∈ Rn =⇒ ~v · ~0 = 0. So we can define ~0 ∈ Rn to be orthogonal to every ~v ∈ Rn . Orthogonal vectors Two vectors, ~v , ~u ∈ Rn are orthogonal ⇐⇒ ~v · ~u = 0. Complex Vectors Definition (Cn ) The vector space Cn is defined to be the set z 1 .. n C = . | z1 , . . . , zn ∈ C zn with addition of vectors and scalar multiplication as for Rn . Definition (Complex Conjugate) The complex conjugate of z1 .. ~z = . ∈ Cn zn z1 .. is ~z = . . zn Definition (Addition and Scalar multiplication in Cn ) ~ ∈ Cn and k ∈ C: Suppose ~z , w z1 .. ~z = . , zn w1 ~ = ... w wn then z1 w1 z1 ± w1 ~z ± w ~ = ... ± ... = ... , zn wn z n ± wn   −1 Example:What is (1 − 3j) ? 1+j kz1 k~z = ... kzn Inner Product in Complex Vector Spaces How to define norm or magnitude for complex vectors, such that we preserve k~z k ∈ R, k~z k ≥ 0, etc.? √ Can we use k~z k = ~z · ~z ? Example:     z 1+j ~z = 1 = z2 2−j ~z · ~z = (1 + j)(1 + j) + (2 − j)(2 − j) = 3 − 2j ∈ /R but ~z · ~z = (1 − j)(1 + j) + (2 + j)(2 − j) = 1 + 1 + 4 + 1 = 7 ∈ R ≥ 0 p We can use k~z k = ~z · ~z instead! Definition (Standard Inner Product on Cn ) In Cn the standard inner product h , i is defined by ~ i = ~z · w ~ = z1 w1 + · · · + zn wn , h~z , w ~ ,~z ∈ Cn . for w Definition (Norm for vectios in Cn ) For any ~z ∈ Cn the norm k~z k is defined by p k~z k = h~z ,~z i. Remarks: k~z k = √ z1 z1 + · · · + zn zn = p |z1 |2 + · · · + |zn |2 dot product is not the same as standard inner product. ~ ∈ Rn , h~z , w ~ i = ~z · w ~ For ~z , w Example    2 − 2j 2+j ~ = ~ i and h~ Let ~z = and w . Evaluate h~z , w w ,~z i. 1+j 3  Properties of Complex Inner Product ~ .~z ∈ Cn , α ∈ C Let ~v , w 1 h~ z ,~z i ≥ 0 and h~z ,~z i = 0 if and only if ~z = ~0, 2 h~ ~ i = h~ z, w w ,~z i, 3 4 5 6 (i) (ii) (i) (ii) ~i h~v + ~z , w ~ + ~v i h~z , w ~i = hα~z , w h~z , α~ wi = ~ i + h~z , w ~ i, = h~v , w ~ i + h~z , ~v i, = h~z , w ~ i, αh~z , w ~ i. αh~z , w ~ i| ≤ k~z k k~ |h~z , w w k, (Cauchy-Schwarz Inequality) ~ k ≤ k~z k + k~ k~z + w w k. (Triangle Inequality) Proof for (4i): ~ = αz1 w1 + · · · + αzn wn ~ i = α~z · w hα~z , w ~i = α[z1 w1 + · · · + zn wn ] = αh~z , w Back to real vectors: Cross-Products in R3 Definition (Cross-Product) Let ~u , ~v ∈ R3 . The cross product of ~u and ~v is u1 v1 u2 v3 − u3 v2 ~u × ~v = u2 × v2 = u3 v1 − u1 v3 . u3 v3 u1 v2 − u2 v1 1 −1 Example: ~u = 6 , ~v = 3 . Find ~u × ~u , ~u × ~v , ~v × ~u , ~u · (~u × ~v ) 3 2 and ~v · (~u × ~v ). Properties of Cross Product Theorem For ~x , ~y ,~z ∈ R3 and t ∈ R, we have 1 ~ x × ~y ∈ R3 2 ~ x × ~y = −~y × ~x , 3 ~ x × ~y is orthogonal to both ~x and ~y . 4 ~ x × ~0 = ~0 5 ~ x × ~x = ~0, 6 ~ x × (~y ± ~z ) = ~x × ~y ± ~x × ~z , 7 (~ x ± ~y ) × ~z = (~x × ~z ) ± (~y × ~z ), 8 (t~ x ) × ~y = t(~x × ~y ). Exercise for you: prove these properties :) x2 x3 − x2 x3 0 pause Proof for (5): ~x × ~x = x3 x1 − x1 x3 = 0 x1 x2 − x2 x1 0 Remark:Order of multiplication matters for cross product! 1 0 0 ~ = 0, show Example: ~u = 1 , ~v = 1 , w 0 0 1 ~ 6= ~u × (~v × w ~) (~u × ~v ) × w Remark:We can use cross product to find a vector orthogonal to two non-zero vectors. 1 3 Example: Find a vector orthogonal to ~u = −2 and ~v = 0 1 1 Lagrange Identity ∀ ~u , ~v ∈ R3 =⇒ k~u × ~v k2 = k~u k2 k~v k2 − (~u .~v )2 . Theorem If θ is the angle between two vectors ~u , ~v ∈ R3 where 0 ≤ θ ≤ π, then k~u × ~v k = k~u k k~v k sin θ. Proof: k~u × ~v k2 = k~u k2 k~v k2 − (~u .~v )2 = k~u k2 k~v k2 − k~u k2 k~v k2 cos2 θ = k~u k2 k~v k2 (1 − cos2 θ) = k~u k2 k~v k2 sin2 θ We also know 0 ≤ θ ≤ π =⇒ sin θ ≥ 0 if we take the the square root we get: k~u × ~v k = k~u k k~v k sin θ Applictions of Cross Product: Area of a Parallelogram h k~u k ~u θ θ k~v k = b ~v h = height, b = base, h = k~u k sin θ Area of parallelogram: A = b · h = k~v kk~u k sin θ = k~v × ~u k = k~u × ~v k The area of the parallelogram formed by ~v , ~u ∈ R3 is A = k~v × ~u k = k~u × ~v k. Example 3 1 Let ~u = 0 and ~v = −1. 5 1 find the area of the parallelogram determined by ~u and ~v . 5 3 1 ~u × ~v = 0 × −1 = 2 −3 5 1 √ √ k~u × ~v k = 25 + 4 + 9 = 38 find the area of the triangle determined by ~u and ~v . 1 1√ A = k~u × ~v k = 38 2 2 ~u θ ~v A Line in R2 Back in high school: representing a line in R2 : y =m x +b slope= m y-intercept= b In Math 115: We use the notation, x1 , x2 ! , We can also determine all the point on a Straight Line, if we have   ~ = ~p = p1 one of its points: P(p1 , p2 ) =⇒ OP p2 the orientation or the direction, ~d Q(q1 , q2 ) We can use vectors to represent a line! ~d P(p1 , p2 )     q1 p = 1 + 2~d q2 p2 The Vector Equation of a Line in R2 The line through ~p with direction vector ~d is {~p + t ~d | t ∈ R}, which has vector equation ~x = ~p + t ~d, t ∈ R. This line is parallel to the line through the origin with equation ~x = t ~d, x2 t ∈ R. d + p line x = td + p p d x1 We say the line  has been translated by~p.   2 0 2 Example: ~x = t is parallel to ~x = +t 1 1 1 Parametric Equations and Scalar Form Parametric Equations Sometimes the components of a vector equation are written separately: ( x1 = p1 + td1 , ~x = ~p + t ~d becomes t ∈ R. x2 = p2 + td2 , This is the parametric equation of the line. (     x1 = 2t, 0 2 Example: ~x = +t =⇒ 1 1 x2 = 1 + t, t ∈ R. If we eliminate t we get(the scalar form:     x1 = 2t, 0 2 ~x = =⇒ t ∈ R. =⇒ x2 = 1 + 12 x1 +t 1 1 x2 = 1 + t, Exercise for you: Scalar Form The familiar scalar form of the equation of the line is obtained by eliminating the parameter t. If d1 6= 0, d2 6= 0, we have x1 − p1 x2 − p2 =t= d1 d2 or x2 = p2 + d2 (x1 − p1 ). d1 Line in Higher Dimensions Definition (Line in Rn ) Let ~p , ~d ∈ Rn , with ~d 6= ~0. Then the set with vector equation ~x = ~p + t ~d, t ∈ R, is called a line in Rn that passes through ~p . To find the parametric equations we carry out the scalar multiplication and addition. x1 = p1 + td1 , .. ~x = ~p + t ~d becomes t ∈ R. . x = p + td , n n n Remark: There are infinite ways to write a vector equation for a line and similarly parametric equations. Use different P on the line or a non-zero scalar multiple of ~d. Example Example: Find the line through the points A(1, 1, −1) and B(4, 0, −3). Present both vector equation and parametric equations. x3 Solution: x2 ~d x1 B A 4 1 3 ~ OA ~ = 0 − 1 = −1 ~d = AB ~ = OB− −3 −1 −2 ~ + t ~d ~x = OA 1 3 =⇒ Vector equation: ~x = 1 + t −1 , t ∈ R −1 −2 x1 = 1 + 3t, parametric equations: x2 = 1 − t, t ∈ R. Plane in R3 and Higher Dimensions Definition (Plane in Rn ) Let ~v1 , ~v2 , ~p ∈ Rn , with ~v1 , ~v2 non-zero and not parallel to each other. Then the set with vector equation ~x = ~p + t1~v1 + t2~v2 , t1 , t2 ∈ R, is called a plane in Rn that passes through ~p . x3 v~2 P ~p x1 v~1 x2 Again note that the representation is not unique. Depends on choice of P, ~v1 and ~v2 . Exercise: Show if ~v1 and ~v2 are parallel, then we get a line instead of plane. x3 C Example Find an equation of the plane containing points A(1, 1, 1), B(1, 2, 3) and C(−1, 1, 2). B x2 A x1 x1 1 0 −2 Solution: x2 = 1 + r 1 + s 0 , r , s ∈ R x3 1 2 1 If we sets = 0 then we get a line: x1 1 0 x2 = 1 + r 1 , r ∈ R x3 1 2 If we set r = 0 then we getanother x1 1 −2 line: x2 = 1 + s 0 , s ∈ R x3 1 1 Both lines are on the plane. x3 C B A x1 x2 To find the parametric equations we carry out the scalar multiplication and addition. x1 = p1 + rd1 + sv1 , .. ~x = ~p + r ~d + s~v becomes r , s ∈ R. . x = p + rd + sv , n n n n Example: x1 1 0 −2 x2 = 1 + r 1 + s 0 , r , s ∈ R x3 1 2 1 x1 = 1 − 2s =⇒ x2 = 1 + r x3 = 1 + 2r + s Again not unique! r , s ∈ R. Example Find the vector equation theplane, containing the point (1, −1, −2) of 1 1 and the line ~x = 3 + t 1 , t ∈ R. −1 4 B ~v ~d A 1 1 0 ~ = −1 − 3 = −4 Solution: ~v = AB −2 −1 −1 x1 1 1 0 =⇒ x2 = 3 + t 1 + s −4 x3 −1 4 −1 Exercise: Write parametric equations and then eliminate t and s to find a scalar equation of the plane. Answer: 5x1 + x2 − 4x3 = 22 Scalar equation of a plane in R3 R3 In we can also determine a plane if we have one point on the plane and the direction of a line orthogonal to it. ~n P0 Normal Vector to a plane A vector ~n 6= 0 is a normal vector for a plane, if ~n is orthogonal to every vector in the plane. It is not unique; every non-zero scalar multiple of ~n is also normal vector to the plane. Finding the Normal to a Plane The normal to the plane with vector equation ~x = ~p + s~u + t~v , where ~u and ~v are non-zero and not parallel is ~n = ~u × ~v . Scalar equation of a plane in R3 Suppose P0 (p1 , p2 , p3 ) is a known point on the plane with normal vector n1 ~n = n2 . If Q(x1 , x2 , x3 ) is any other point on the plane: n3 ~ lies in the plane P and Q lie on the plane ⇐⇒ PQ ~ =0 ⇐⇒ ~n · PQ ~ − OP) ~ =0 ⇐⇒ ~n · (OQ ~ = ~n · OP ~ ⇐⇒ ~n · OQ n1 x1 n1 p1 n2 · x2 = n2 · p2 =⇒ n1 x1 + n2 x2 + n3 x3 = n1 p1 + n2 p2 + n3 p3 n3 x3 n3 p3 If we can calculate RHS: d ≡ n1 p1 + n2 p2 + n3 p3 =⇒ n1 x1 + n2 x2 + n3 x3 = d Example Find a scalar equation of the plane containing the points A(3, 1, 2), B(1, 2, 3) and C(−2, 1, 3). Also check if ~n is the normal vector for the plane then ~ = ~n · OC ~ = ~n · OA. ~ ~n · OB Remark: the scalar equation for a plane in R3 is not unique. Why? Parallel planes in R3 Two planes in R3 are parallel, if their normal vectors are parallel. Finding the Line of Intersection of Two Planes The direction vector of the line of intersection of two planes is the cross-product of the normals to those planes. With this direction vector and one point on the line, one can find the line of intersection. Projections ~ as a sum of a We want to write w vector parallel to a non-zero ~v and a vector perpendicular to ~v : ~ = proj~v w ~ + perp~v w ~ w that means: ~ = c~v , proj~v w ~ ) · ~v = 0 (perp~v w ~ =w ~ − c~v =⇒ (~ perp~v w w − c~v ) · ~v = 0 ~v 6=0 ~ · ~v − c ~v · ~v = 0 =⇒ c = =⇒ w ~ = =⇒ proj~v w ~ · ~v w ~v , k~v k2 ~ · ~v ~ · ~v w w = ~v · ~v k~v k2 ~ =w ~ − proj~v w ~ perp~v w Projections Definition (Projection) ~ ∈ Rn , with ~v 6= ~0, the projection of w ~ onto ~v is For vectors ~v , w ~ = proj~v w ~ · ~v w ~v . k~v k2 Definition (Perpendicular of a Projection) ~ ∈ Rn , with ~v 6= 0, the projection of w ~ perpendicular to ~v For vectors ~v , w is ~ =w ~ − proj~v w ~. perp~v w Projections proj~v ~u = ~u ~u θ proj~v ~ u ~v π 0≤θ< 2 ~u · ~v > 0 proj~v ~u and ~v in same direction. ~u · ~v ~v k~v k2 θ proj~v ~ u ~v π <θ≤π 2 ~u · ~v < 0 proj~v ~u and ~v in opposite direction. ~u θ ~v θ= π 2 ~u · ~v = 0 proj~v ~u = ~0 Example 1 −1 Let ~u = 2 and ~d = 1 . 3 2 ~ ~ Find proj~d u , perp~d u , proj~u ~d, (perp~d ~u ) · ~d, proj~d (proj~d ~u ). Solution: ~u · ~d = −1 + 2 + 6 = 7, k~dk2 = 1 + 1 + 4 = 6, k~u k2 = 1 + 4 + 9 = 14 7 − −1 ~ ~u · d ~ 7 76 ~ 1 proj~d u = d= = 6 6 7 k~dk2 2 3 7 13 −6 1 6 13 5 4 perp~d ~u = ~u −proj~d ~u = 2− 76 = 56 , (perp~d ~u )·~d = − + + = 0 6 6 3 7 2 3 3 3 1 7 − 1 −1 ~d · ~u 2 7 7 76 ~ ~u = 2 = 1 , proj~d (proj~d ~u ) = 1 = 6 proj~u d = 14 6 k~u k2 7 3 3 2 3 2 Some Properties of Projections Projection property proj~x (proj~x ~y ) = proj~x ~y for all ~x , ~y ∈ Rn . Proof: proj~x (proj~...
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