MA1623 Lecture 13 PDF

Vector addition satisfies the following properties: A1 (Existence of identity) For all vectors v we have In the language of abstract algebra, these four properties say that v...

Vector addition satisfies the following properties: A1 (Existence of identity) For all vectors v we have In the language of abstract algebra, these four properties say that v + 0 = v. V (Rn ) forms an abelian group. All of these properties are very easily checked directly from the definition of vector addition. A2 (Existence of inverses) For all vectors v we have Next we define the multiplication of a vector by a scalar. v + (−v ) = 0. Definition 3.1.10: Let u ∈ V (Rn ) be a vector, corresponding to a shift by (u1 ,... , un ), and let λ ∈ R. (We call elements of R scalars to A3 (Commutativity) For all vectors u and v we have differentiate them from vectors.) We define λu ∈ V (Rn ), the product of u + v = v + u. λ and u, to be the vector corresponding to the shift by A4 (Associativity) For all vectors u, v , and w we have (λu1 ,... , λun ). (u + v ) + w = u + (v + w). Joe Chuang (City St George’s) MA1623 Lecture 13 1/9 Joe Chuang (City St George’s) MA1623 Lecture 13 2/9 Scalar multiplication satisfies the following properties (which are all easily verified): S1 (Associativity of multiplication) For all vectors v and scalars λ, µ, The algebraic properties of vectors (A1-A4) and (S1-S4) can be we have abstracted to give the general notion of a vector space. This is a (λµ)v = λ(µ(v )). mathematical object that behaves like it is a collection of vectors, but which can be more general than just the ordinary vectors which we S2 (Distributivity) For all vectors u, v and scalars λ we have have seen so far. λ(u + v ) = (λu) + (λv ). Definition 3.1.11: A set V equipped with a map taking pairs of elements from V to an element of V (which we call addition and a map S3 (Distributivity II) For all vectors v and scalars λ, µ, we have taking a scalar and an element of V to an element of V (which we call scalar multiplication) satisfying (A1-A4) and (S1-S4) is called a vector (λ + µ)v = (λv ) + (µv ). space. S4 (Special scalars) For all vectors v we have The theory of vector spaces will be developed in the Linear Algebra module in the spring term. 1v = v and 0v = 0 and (−1)v = −v. Joe Chuang (City St George’s) MA1623 Lecture 13 3/9 Joe Chuang (City St George’s) MA1623 Lecture 13 4/9 We can give an algebraic definition of the notion of dimension for a general vector space which matches the geometric definition that we Example 3.1.13: (a) The set V (Rn ) with addition and scalar already know for ordinary vectors. multiplication defined as before forms an n dimensional vector space. Definition 3.1.12: Vectors v 1 ,... , v n in a vector space V form a basis (b) The set of solutions of a second order homogeneous linear of V if every v ∈ V can be uniquely written in the form differential equation is a two-dimensional vector space. However, the set of solutions of a non-homogeneous differential equation do not v = λ1 v1 + · · · + λn v n form a vector space. (c) The set of all functions f : R −→ R forms an infinite dimensional for some scalars λ1 ,... , λn. The coefficients λi are called the vector space, where for x ∈ R we define components of v with respect to the given basis. The dimension of a vector space V is the number of vectors in a basis (f + g)(x) = f (x) + g(x) of V. This definition requires us to know that the number of vectors in a and basis for a given vector space does not depend on the choice of basis. (λf )(x) = λ(f (x)) This and other results about vector spaces will be studied in the Linear. Algebra module next year. Joe Chuang (City St George’s) MA1623 Lecture 13 5/9 Joe Chuang (City St George’s) MA1623 Lecture 13 6/9 It is convenient to have a standard set of basis vectors for V (Rn ). Example 3.1.16: In V (R3 ) it is traditional to denote e1 by i, e2 by j, Definition 3.1.14: For 1 ≤ i ≤ n let ei be the vector corresponding to and e3 by k. We have shifting the ith coordinate by 1 and leaving the rest unchanged. We call the set of such vectors the standard basis of V (Rn ) because of (vi i + vj j + vk k ) + (wi i + wj j + wk k ) = ((vi + wi )i + (vj + wj )j + (vk + wk )k ) Proposition 3.1.15: Every vector in V (Rn ) can be uniquely expressed in the form X n λ(vi i + vj j + vk k ) = (λvi i + λvj j + λvk k ) v= vi ei. i=1 0 = (0i + 0j + 0k ) Proof: If v is the vector which shifts by (v1 ,... , vn ) then n X There are other possible choices for a basis of V (Rn ). We will learn v= vi ei more about how to find bases for a vector space in the Linear Algebra i=1 module. and vice versa. Joe Chuang (City St George’s) MA1623 Lecture 13 7/9 Joe Chuang (City St George’s) MA1623 Lecture 13 8/9 Example 3.1.17: The set of solutions of the equation y 00 − 3y 0 + 2y = 0 consists of the vector space made up of all functions of the form Aex + Be2x. This space has a basis given by the elements ex and e2x. Remark 3.1.18: If we identify the vector v1 e1 + v2 e2 + · · · + vn en in V (Rn ) with the point (v1 , v2 ,... , vn ) of Rn , then Rn becomes a vector space under componentwise addition and scalar multiplication. It is very common to identify the two vector spaces V (Rn ) and Rn ; we have kept them separate to emphasise the difference between a point in space and a vector. Joe Chuang (City St George’s) MA1623 Lecture 13 9/9  3.2 Dot products The definition of length for a vector is the natural one, as it matches up Definition 3.2.1: The length of a vector v = v1 e1 + · · · + vn en ∈ V (Rn ) with the notion of distance between points that we saw earlier, by the is defined to be q following easy result. |v | = v12 + v22 + · · · + vn2. Proposition 3.2.3: Let P and Q be points in Rn. Then A vector v is a unit vector if |v | = 1. −→ |PQ| = |PQ|. Example 3.2.2: Show that the vector −→ Proof: Let P = (x1 , x2 ,... , xn ) and Q = (y1 , y2 ,... , yn ). Then PQ is u = i + λj + 2k the vector which shifts by (y1 − x1 , y2 − x2 ,... , yn − xn ). Hence, √ has length at least 5. −→ q |PQ| = (y1 − x1 )2 + (y2 − x2 )2 + · · · (yn − xn )2 = |PQ|. p p √ |u| = 12 + λ2 + 22 = 5 + λ2 ≥ 5. Joe Chuang (City St George’s) MA1623 Lecture 14 1/6 Joe Chuang (City St George’s) MA1623 Lecture 14 2/6 The definition of the length of a vector is a special case of the following Here are some easily deduced properties of the lengths. important definition. Proposition 3.2.4: Let v ∈ Rn and λ ∈ R. Then Definition 3.2.5: Let 1 |λv | = |λ||v | u = u1 e 1 + · · · + un e n and v = v1 e1 + · · · + vn en 2 |v | = 0 if and only if v = 0. 3 If v 6= 0, then be vectors in Rn. The dot product, or (standard) inner product of u and 1 v is the real number v = 1. |v | u · v = u1 v1 + · · · + un vn. Proof completed by hand in the lecture. Taking u = v we see that v · v = |v |2. Joe Chuang (City St George’s) MA1623 Lecture 14 3/6 Joe Chuang (City St George’s) MA1623 Lecture 14 4/6 Example 3.2.6: Let u = 2i − 3j + 7k and v = 4j − 2k ∈ V (R3 ). Then u.v = 2 × 0 + (−3) × 4 + 7 × (−2) = −26. We have seen that the length of a vector can be calculated using the The dot product satisfies the following properties. dot product. In fact, we can reverse this process and calculate dot Proposition 3.2.7: For any vectors u, v , w ∈ V (Rn ) and any scalars products using only the length function: λ, γ ∈ R, we have Proposition 3.2.8: (Polarisation formula) D1 (Linearity) For any vectors u, v ∈ V (Rn ) we have (λu + γv ) · w = λu · w + γv · w. 1 D2 (Symmetry) u·v = (|u|2 + |v |2 − |u − v |2 ). 2 u · v = v · u. D3 (Positive definiteness) Proof completed by hand in the lecture. v ·v ≥0 and v · v = 0 if and only if v = 0. Proof completed by hand in the lecture. Joe Chuang (City St George’s) MA1623 Lecture 14 5/6 Joe Chuang (City St George’s) MA1623 Lecture 14 6/6 Proposition 3.2.9: (The Cauchy-Schwarz inequality) For any two vectors u and v we have |u · v | ≤ |u||v | We have defined the dot product by a simple algebraic formula. However, the importance of the dot product is that it also has a purely with equality if and only if u is a scalar multiple of v , or vice versa. geometric interpretation. To see this we will define a second, geometric Proof: If u or v is zero then the result clearly holds, so we assume that product of vectors, and show that it agrees with the dot product. they are both non-zero. For any real number λ ∈ R consider the vector Definition 3.2.10: We will define the geometric dot product u ◦ v of u − λv. We have two vectors u and v by 0 ≤ |u − λv |2 = (u − λv ) · (u − λv ) u ◦ v = |u||v | cos θ = |u|2 + λ2 |v |2 − 2λu · v. u·v where θ is the angle between the two vectors u and v if both vectors Taking λ = |v |2 and rearranging we get are non-zero, or 0 otherwise. (u · v )2 0 ≤ |u|2 − ie |u · v | ≤ |u||v |. |v |2 Equality holds when u − λv = 0, ie when u = λv. Joe Chuang (City St George’s) MA1623 Lecture 16 1/8 Joe Chuang (City St George’s) MA1623 Lecture 16 2/8 Now consider the sum of vectors u + v and w as shown below. Rearranging this equation, and recalling the geometric definition of v cosine, we see that the length of the projection of u onto v (which is u τ |u| cos θ) is given by u◦v |u| cos θ =. θ γ |v | u |u|cos θ |v|cos τ w |u+v|cos γ θ By the definition of ◦ we have |u|cos θ v u◦w v ◦w |u| cos θ = and |v | cos τ = Replacing u by λu we see that |w| |w| (λu) ◦ v = λ(u ◦ v ). and (u + v ) ◦ w |u + v | cos γ =. |w| Joe Chuang (City St George’s) MA1623 Lecture 16 3/8 Joe Chuang (City St George’s) MA1623 Lecture 16 4/8  Now consider our standard basis vectors ei with 1 ≤ i ≤ n. As these are at right angles to each other, and all have length one, we see that From the diagram it is clear that ei ◦ ej = δij |u| cos θ + |v | cos τ = |u + v | cos γ where δij is the Kronecker delta defined by and so u◦w v ◦w (u + v ) ◦ w + =.  1 if i = j |w| |w| |w| δij = 0 otherwise. In summary, we have shown Combining this observation P with Lemma 3.2.11 we deduce that if Lemma 3.2.11: For any vectors u, v , w and scalar λ we have u = ni=1 ui ei and v = ni=1 vi ei then P (λu) ◦ v = λ(u ◦ v ) and (u + v ) ◦ w = u ◦ w + v ◦ w. n X u◦v = ui vi = u · v. i=1 Joe Chuang (City St George’s) MA1623 Lecture 16 5/8 Joe Chuang (City St George’s) MA1623 Lecture 16 6/8 In summary, we have shown that the two different definitions of dot products coincide: Example 3.2.14: Consider the vectors Theorem 3.2.12: For any (non-zero) vectors u and v we have u = 2i − j + 3k v =i +j w = i + λj u · v = |u||v | cos θ in V (R3 ) where λ ∈ R. (i) Calculate the angle between u and v. where θ is the angle between the vectors u and v. (ii) For which values of λ is the angle between v and w strictly greater Corollary 3.2.13: For any non-zero vectors u and v , the angle than π/2? between them is Example completed by hand in the lecture.   −1 u·v θ = cos. |u||v | Joe Chuang (City St George’s) MA1623 Lecture 16 7/8 Joe Chuang (City St George’s) MA1623 Lecture 16 8/8 The Cauchy-Schwarz inequality implies the well-known triangle inequality: the sum of the lengths of any two sides of a triangle is Definition 3.2.16: Given three distinct points A, B, C in Rn , denote by −→ −→ always at least the length of the remaining side. ∠BAC the angle between the vectors AB and AC. Proposition 3.2.15: (The triangle inequality) Example 3.2.17: Let A = (1, 1, −1), B = (3, 0, 0), C = (0, −1, 0). For any two vectors u and v in V (Rn ) we have Calculate ∠BAC. |u + v | ≤ |u| + |v |. Example completed by hand in the lecture. The next definition is very important when dealing with problems Proof: We have involving vectors. |u + v |2 = (u + v ) · (u + v ) Definition 3.2.18: Two non-zero vectors u and v are said to be = |u|2 + |v |2 + 2u · v perpendicular or orthogonal if ≤ |u|2 + |v |2 + 2|u||v | (by Cauchy-Schwarz) = (|u| + |v |)2. u · v = 0. The result now follows by taking the square root of each side. Joe Chuang (City St George’s) MA1623 Lecture 17 1/7 Joe Chuang (City St George’s) MA1623 Lecture 17 2/7 Example 3.2.19: Consider the vectors u = 2i − j and v = 3i + 6j. We have |u + v | = |u − v | u · v = 2 × 3 + (−1) × 6 = 0 if and only if and so these vectors are orthogonal. |u + v |2 = |u − v |2. Proposition 3.2.20: Let u and v be non-zero vectors in V (Rn ). Then Expanding both sides we see that this is equivalent to the following statements are equivalent. 1 u and v are orthogonal. |u|2 + |v |2 + 2u · v = |u|2 + |v |2 − 2u · v. 2 The angle between u and v is π/2. Clearly this is equivalent to u · v = 0 which is the definition of 3 We have |u + v | = |u − v |. orthogonality. Proof: The first two statements are clearly equivalent using the Definition 3.2.21: The projection of a vector u onto a non-zero vector geometric description of the dot product. To show that the first and v is the unique scalar multiple λv of v such that u − λv and v are third statements are equivalent, we have perpendicular. Joe Chuang (City St George’s) MA1623 Lecture 17 3/7 Joe Chuang (City St George’s) MA1623 Lecture 17 4/7 Using the formula for perpendicular vectors we see that (u − λv ).v = 0 Example 3.2.22: Consider the vectors and so u·v λ=. u = i + λj + k and v = λi + (λ + 2)j + 2k v ·v Thus the projection of u onto v is where λ ∈ R.  u·v  (i) Let λ = 3 and determine the projection of u onto v. v (ii) Let λ = 5 and construct all unit vectors perpendicular to both u and |v |2 v. and the length of this projection is Example completed by hand in the lecture. |u · v |. Later in this module we will learn a more efficient way of solving the |v | second part of this example. Joe Chuang (City St George’s) MA1623 Lecture 17 5/7 Joe Chuang (City St George’s) MA1623 Lecture 17 6/7 Definition 3.2.23: Two non-zero vectors are parallel if one is a scalar multiple of the other. Proposition 3.2.24: Let u and v be non-zero vectors in V (Rn ). Then the following statements are equivalent. 1 u and v are parallel. 2 The angle between u and v is 0 or π. 3 We have |u · v | = |u||v |. Proof: Proof completed by hand in the lecture. Joe Chuang (City St George’s) MA1623 Lecture 17 7/7  3.3. Linear geometry in three dimensions Definition 3.3.1: The parametric form of the line through P in the direction a is given by the set of points of the form For the remainder of this chapter we will focus on the special case of three dimensional space. We start by considering lines. Recall that a Q(λ) = (x0 + λa1 , y0 + λa2 , z0 + λa3 ) line (in any space) is determined by a point and a direction vector. with λ ∈ R. Consider the line through the point P = (x0 , y0 , z0 ) ∈ R3 in the direction The Cartesian form of the equation of the same line is of the vector a = a1 i + a2 j + a3 k. Then a point Q = (x, y , z) lies on this −→ x − x0 y − y0 z − z0 line if and only if PQ = λa for some λ ∈ R. Since = = = λ. a1 a2 a3 −→ PQ = (x − x0 )i + (y − y0 )j + (z − z0 )k This does not make sense if ai = 0; in that case we replace the we must have corresponding fraction by the appropriate constant function. For example if a1 = 0 then we replace x − x0 = λa1 y − y0 = λa2 z − z0 = λa3. x − x0 =λ by x = x0. From this equation we obtain two forms for the equation of a line. a1 Joe Chuang (City St George’s) MA1623 Lecture 18 1/6 Joe Chuang (City St George’s) MA1623 Lecture 18 2/6 Example 3.3.2: Find the equation in Cartesian form of the line through the point P = (1, 3, −1) in the direction of a = 2i + 3j + 4k , and The two lines x − x1 y − y1 z − z1 parametrise the points on this line. L1 : = = a1 a2 a3 Example completed by hand in the lecture. and x − x2 y − y2 z − z2 L2 : = = Consider two lines in three dimensions. These lines either b1 b2 b3 1 are equal; intersect if and only if 2 intersect in a point; Q1 (λ) = (x1 + λa1 , y1 + λa2 , z1 + λa3 ) 3 are parallel; 4 are skew. and Q2 (µ) = (x2 + µb1 , y2 + µb2 , z2 + µb3 ) Here we say that two lines are parallel if they belong to a common plane but do not intersect, and are skew if they do not belong to a coincide for some λ and µ in R. common plane. Joe Chuang (City St George’s) MA1623 Lecture 18 3/6 Joe Chuang (City St George’s) MA1623 Lecture 18 4/6  Example 3.3.3: Does the line through the points P1 = (7, 4, 3) and Thus the two lines intersect if and only if the system of equations P2 = (9, 5, 5) intersect the line from Example 3.3.2? If it does then describe the intersection. x1 + λa1 = x2 + µb1 y1 + λa2 = y2 + µb2 Example completed by hand in the lecture. z1 + λa3 = z2 + µb3 Example 3.3.4: Given the point A = (4, 1, 7) and the line has a solution in λ and µ. x +2 y −6 z −9 L: = = If there are many solutions then L1 = L2. If there is exactly one 1 −2 −1 solution then L1 and L2 intersect in a point. find all points P on L such that the line M through A and P is If there are no solutions then either a = λb for some λ ∈ R and so L1 perpendicular to L. and L2 are distinct parallel lines, or L1 and L2 are skew. Example completed by hand in the lecture. Joe Chuang (City St George’s) MA1623 Lecture 18 5/6 Joe Chuang (City St George’s) MA1623 Lecture 18 6/6 Many problems in geometry reduce to finding a set of perpendicular vectors. Dot products are very useful for showing that two vectors are perpendicular, but it would be very useful to have a method for Remark 3.3.6: The cross product of two vectors can be expressed as constructing perpendicular vectors. The cross product will enable us to a determinant: do this. i j k u × v = u1 u2 u3. Definition 3.3.5: Let v1 v2 v3 u = u1 i + u2 j + u3 k and v = v1 i + v2 j + v3 k Example 3.3.7: Let u = i + 5j + k and v = 5i + 7j + 2k. Find be vectors in V (R3 ). The cross-product (or vector product) of u and v u×v v ×u v × v. is a vector, denoted by u × v and defined by u × v = (u2 v3 − u3 v2 )i + (u3 v1 − u1 v3 )j + (u1 v2 − u2 v1 )k. Example completed by hand in the lecture. At first sight this seems like a rather arbitrary definition. however we will see that it is a very natural construction geometrically. Our first hint that this may be the case is Joe Chuang (City St George’s) MA1623 Lecture 19 1/6 Joe Chuang (City St George’s) MA1623 Lecture 19 2/6 Proof: All of these properties can be shown by writing out explict Note that in the above example changing the order of the vectors u formulas in terms of expressions for u, v and w in terms of i, j and k. and v in the cross product changes the cross product by a sign, and This is quite lengthy and so will be omitted here, but is a useful the cross product of the vector v with itself is zero. The next exercise for you to undertake. proposition, shows, amongst other things, that this always holds. All of the following properties should be known. Remark 3.3.9: The definition of cross-product is based on a “right-handed” system, i.e. a system where Proposition 3.3.8: For all u, v , w in V (R3 ) and λ, µ in R we have 1 (λu + µv ) × w = λu × w + µv × w and i × j = k. w × (λu + µv ) = λw × u + µw × v. 2 (u × v ) · u = 0 and (u × v ) · v = 0. If instead we work with a left-handed system, where i × j = −k then we obtain the left-handed cross product which also satisfies the 3 v × v = 0. properties of Proposition 3.3.8. In fact, it can be shown that any 4 u × v = −v × u. operation V (R3 ) × V (R3 ) → V (R3 ) which satisfies Proposition 3.3.8 is 5 If |u| = 1, |v | = 1 and u · v = 0, then |u × v | = 1. one of these two cross products. We will only working with the standard (right-handed) definition of the cross product. Joe Chuang (City St George’s) MA1623 Lecture 19 3/6 Joe Chuang (City St George’s) MA1623 Lecture 19 4/6 One of the reasons that the cross product is so important is Proposition 3.3.10: Let u and v be non-zero vectors in V (R3 ). Then We have already seen how the dot product can be used to calculate any non-zero vector of the form angles. So far we have seen that the direction of the cross product of two vectors has a special meaning, but we can also give a geometric λ(u × v ) interpretation to the length of this product. is orthogonal to both u and v. Further, if u × v 6= 0 then any vector Proposition 3.3.12: Let u, v be non-zero vectors in V (R3 ). Then orthogonal to u and v is of this form. |u × v | = |u||v | sin θ Example 3.3.11: Let u = i + 5j + k and v = 5i + 7j + 2k. Find all unit vectors perpendicular to both u and v. where θ is the angle between u and v. Calculate Another way of stating this result is that the length of the cross product (u × v ) · (17u − 78v + (u × v )). of two vectors is equal to the area of the parallelogram defined by those vectors. Example completed by hand in the lecture. Joe Chuang (City St George’s) MA1623 Lecture 19 5/6 Joe Chuang (City St George’s) MA1623 Lecture 19 6/6  The scalar triple product of vectors u, v , w is defined to be Given three vectors u, v and w of R3 , how can we combine them using dot and cross products? In answering this question, we should keep in u · (v × w). mind that the dot product u · v is a scalar, while the cross product u × v is a vector. For this reason dot and cross products are also called Note that the scalar triple product is a number, not a vector. Here is a scalar and vector products, respectively. useful formula for calculating the scalar triple product If we take the dot product of two of the vectors, the resulting real u1 u2 u3 number may be used to scale the third vector: u · (v × w) = v1 v2 v3 w1 w2 w3 (u · v )w. Standard properties of determinants now translate into properties of the scalar product. For example swapping two rows changes the sign On the other hand the expression u(v × w) is not valid, as both u and of the determinant, so v × w are vectors. We must either take the dot or cross product of the two. u · (v × w) = −v · (u × w). Joe Chuang (City St George’s) MA1623 Lecture 20 1/7 Joe Chuang (City St George’s) MA1623 Lecture 20 2/7 The vector triple product of u, v and w is defined to be Neither of these approaches is particularly enlightening. Instead we u × (v × w). just try to convince that the identity is reasonable. Suppose u = v × w. Then the LHS of the given equation is just We claim that u × u = 0. On the other hand, the RHS is also 0, as each of the u × (v × w) = (u · w)v − (u · v )w. coefficients u · w and u · v is zero. If the given equation is true, then the RHS should be This can be proven by writing everything in terms of the standard basis perpendicular to both u and to v × w. We can check this. and calculating explicitly. Alternatively, one can note that both sides of Example completed by hand in the lecture. the equation are linear in each of the arguments u, v and w; this In fact these verifications are enough to prove that the two sides of the implies that it is suffices to check the identity when u, v , w ∈ {i, j, k }, given equation are scalar multiples of each other. which is easy to do. Joe Chuang (City St George’s) MA1623 Lecture 20 3/7 Joe Chuang (City St George’s) MA1623 Lecture 20 4/7 We now consider combinations of four vectors. For vectors u, v , w and x, the scalar quadruple product is defined to be The alert reader will have noticed that we never proved Proposition 3.3.12. From our identity for the scalar quadruple product we can now (u × v ) · (w × x). show this result as follows. Proof: Let θ be the angle between nonzero vectors u and v. Then We have the following identity. |u × v |2 = (u × v ) · (u × v ) (u × v ) · (w × x) = (u · w)(v · x) − (u · x)(v · w). = (u · u)(v · v ) − (u · v )(v · u) To see this, we first use the triple scalar product property to swap rows = |u|2 |v |2 − |u|2 |v |2 cos2 θ 1 and 2 and then 2 and 3 of the determinant: = |u|2 |v |2 sin2 θ. (u × v ) · (w × x) = w · (x × (u × v )) Here we have used the scalar quadruple product, and then the = w · ((x · v )u − (x · u)v ) geometric description of the dot product. = (x · v )(w · u) − (x · u)(w · v ). Joe Chuang (City St George’s) MA1623 Lecture 20 5/7 Joe Chuang (City St George’s) MA1623 Lecture 20 6/7 For vectors u, v , w and x, the vector quadruple product is defined to be (u × v ) × (w × x). We have the following identity. (u × v ) × (w × x) = (u · (v × x)) w − (u · (v × w)) x. The proof of this is routine but somewhat tedious, and is left as an exercise. Joe Chuang (City St George’s) MA1623 Lecture 20 7/7 To determine the equation of a line, we need a point on the line and a direction vector. What do we need to define a plane in three dimensions? Simplifying this becomes Definition 3.3.13: A plane P is determined by a point A = (x0 , y0 , z0 ) it a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0 contains, together with a normal vector n to the plane. That is, n is a vector such that n · v = 0 for all v ∈ P. or just We would like a single simple equation to define a plane. ax + by + cz − (x0 a + y0 b + z0 c) = 0. Let B = (x, y , z) be a point in R3. Then B lies in the plane through Thus the equation of the plane through A = (x0 , y0 , z0 ) with normal A = (x0 , y0 , z0 ) with normal vector n = ai + bj + ck if and only if vector n = ai + bj + ck is −→ ax + by + cz + d = 0 AB · n = 0 that is if and only if with d = −(x0 a + y0 b + z0 c). (ai + bj + ck) · ((x − x0 )i + (y − y0 )j + (z − z0 )k ) = 0. Joe Chuang (City St George’s) MA1623 Lecture 21 1/7 Joe Chuang (City St George’s) MA1623 Lecture 21 2/7 Example 3.3.14: Find the equation of the plane passing through We claim that the vector P = (2, 4, −6) that is normal to the vector n = 4i − j + k. −→ −→ n = PQ × PR Example completed by hand in the lecture. is a normal vector to the plane. First, since P, Q and R do not lie on a There are various combinations of data which can be used to define a common line we have n 6= 0. plane, and we need to know how to determine the equation of the plane from this data. Now consider the plane P through P and with normal vector n. By Property 2 of Proposition 3.3.8 for cross-products, we have For our first case, suppose that we are given three points −→ −→ −→ −→ n · PQ = (PQ × PR) · PQ = 0. P = (x1 , y1 , z1 ) Q = (x2 , y2 , z2 ) R = (x3 , y3 , z3 ) Thus, by the definition of P, Q lies in P. Similarly R lies in P. Thus, P which do not lie on a common line. Then there is a unique plane is a plane containing the three points P, Q and R. containing the given points, which can be determined as follows. Joe Chuang (City St George’s) MA1623 Lecture 21 3/7 Joe Chuang (City St George’s) MA1623 Lecture 21 4/7  Another variation: Suppose that we are given a line L and a point P not Example 3.3.15: Find the equation of the plane P through the three on L. To find the plane containing L and P we reduce to the preceding −→ points P = (1, 3, 2), Q = (3, 4, 2) and R = (5, 2, 4). case by finding a point Q on L and determining the vector PQ. Then we can use this and the direction vector for L and proceed as before. Example completed by hand in the lecture. Example 3.3.16: Find the equation of the plane P containing the line Now suppose we are given two lines L and M that intersect in some point and do not coincide. Then by repeating the arguments above, the x −2 y z +1 L: = = vector n obtained by taking the cross product of the direction vectors 3 2 −1 for L and M will be a normal to the plane, and by picking a point on and the point Q = (3, 1, 2). the plane we can work out the value of d in our equation for the plane. Example completed by hand in the lecture. Joe Chuang (City St George’s) MA1623 Lecture 21 5/7 Joe Chuang (City St George’s) MA1623 Lecture 21 6/7 One feature of questions about planes and lines in three dimensions is that you can be given the necessary data in a wide variety of ways. It is then up to you to work out how to manipulate it into the necessary form. By finding appropriate points and lines, and perpendicular vectors, you can usually solve all of these quite easily. For an (easy) example, how would you find the equation of a plane containing two parallel but distinct lines? Taking the cross product of these lines will not work, as they are parallel. But you can choose a point on each line and construct a third line which is not parallel to either, and then use this with one of the given lines to determine the plane. Joe Chuang (City St George’s) MA1623 Lecture 21 7/7 Next we will consider the problem of finding the intersection of a plane The points on L have the form and a line. Consider a line L and a plane P. Then one of the following holds: Q(λ) = (x0 + a1 λ, y0 + a2 λ, z0 + a3 λ). L and P intersect in exactly one point P So Q(λ) ∈ P if and only if L and P do not intersect a(x0 + a1 λ) + b(y0 + a2 λ) + c(z0 + a3 λ) + d = 0. L is contained in P Let P and L have equations: Solving for λ we deduce that the intersection of L and P is the point P: ax + by + cz + d = 0 (x0 + a1 λ̃, y0 + a2 λ̃, z0 + a3 λ̃) x − x0 y − y0 z − z0 where L: = = =λ ax0 + by0 + cz0 + d a1 a2 a3 λ̃ = −. aa1 + ba2 + ca3 Joe Chuang (City St George’s) MA1623 Lecture 22 1/7 Joe Chuang (City St George’s) MA1623 Lecture 22 2/7 Now of course this answer is valid only when the denominator in the expression for λ̃ is non-zero. If instead we have Example 3.3.18: Find the equation of the plane P through the point aa1 + ba2 + ca3 = 0 A = (4, 1, 2) that intersects neither the line then either x −3 y −1 z −2 ax0 + by0 + cz0 + d = 0 L1 : = = 5 2 1 which implies that L ⊂ P or nor the line x y −2 z −3 ax0 + by0 + cz0 + d 6= 0 L2 : = =. 2 1 4 which implies that L ∩ P = ∅. Example completed by hand in the lecture. Example 3.3.17: Find the point of intersection of the line L through P = (1, 3, 2) and Q = (2, 1, 1) with the plane P : 2x + y − z = 11. Example completed by hand in the lecture. Joe Chuang (City St George’s) MA1623 Lecture 22 3/7 Joe Chuang (City St George’s) MA1623 Lecture 22 4/7 The intersection of two distinct planes is either There are two possibilities: empty or If n1 × n2 6= 0, then P1 and P2 intersect in a line. a line. The line L of intersection is perpendicular to both n1 and n2 , and thus Suppose that the planes are given by the equations parallel to n1 × n2. Then to determine L it suffices to find the coordinates of one of its points; one way to do so is to fix the value of P1 : a1 x + b1 y + c1 z + d1 = 0 one coordinate (e.g. z = 0) and solve the remaining system of two and equations in two variables. P2 : a2 x + b2 y + c2 z + d2 = 0. If n1 × n2 = 0, then the planes are parallel and do not intersect. Then n1 = a1 i + b1 j + c1 k and n2 = a2 i + b2 j + c2 k are normal vectors We demonstrate the method of finding the intersection in the following to P1 and P2 respectively. example. Joe Chuang (City St George’s) MA1623 Lecture 22 5/7 Joe Chuang (City St George’s) MA1623 Lecture 22 6/7 Example 3.3.19: Find the intersection of P1 : x +y +z −5=0 and P2 : 4x + y + 2z − 15 = 0. Example completed by hand in the lecture. Finally we note that the angle between two planes is equal to the angle between their two normals, and so we can determine this angle by finding the angle between the two normals in the usual manner. Joe Chuang (City St George’s) MA1623 Lecture 22 7/7 We shall derive a formula for the distance from a point P = (x0 , y0 , z0 ) Since Q ∈ P we have to the plane P : ax + by + cz + d = 0. ax1 + by1 + cz1 + d = 0. Let Q be the point on P closest to P and let L be the line containing −→ Substituting for x1 , y1 and z1 we obtain both P and Q. Then the distance between P and P is |PQ|. −→ ax0 + by0 + cz0 + d Now PQ is parallel to λ̃ = −. a2 + b 2 + c 2 n = ai + bj + ck We conclude that the distance between P and P is and so any point on the line through P and Q is of the form q R(λ) = (x0 + λa, y0 + λb, z0 + λc). |PQ| = (x1 − x0 )2 + (y1 − y0 )2 + (z1 − z0 )2 q Since Q lies on L we have that = λ̃2 (a2 + b2 + c 2 ) |ax0 + by0 + cz0 + d| Q = (x1 , y1 , z1 ) = (x0 + λa, y0 + λb, z0 + λc) = √. a2 + b 2 + c 2 for some λ ∈ R. Joe Chuang (City St George’s) MA1623 Lecture 23 1/6 Joe Chuang (City St George’s) MA1623 Lecture 23 2/6 Example 3.3.20: Verify that the lines Two planes in R3 either intersect in a line or are parallel. If they are x −4 y −1 z + 12 parallel we can ask what the distance is between them. L1 : = = 1 −1 −4 Suppose P1 and P2 are parallel planes; they can then be described by a common normal vector n = ai + bj + ck , and thus by equations and x +2 y − 11 z L2 : = = P1 : ax + by + cx + d1 = 0 −1 2 1 intersect. Find the equation of the plane P containing both lines, and the distance from the point P = (0, 1, 3) to P. P2 : ax + by + cx + d2 = 0 Example completed by hand in the lecture. for some d1 and d2. Joe Chuang (City St George’s) MA1623 Lecture 23 3/6 Joe Chuang (City St George’s) MA1623 Lecture 23 4/6 The distance between P1 and P2 is equal to the distance from P2 to a To see how this result can be applied, we conclude our Chapter on point A = (x0 y0 , z0 ) on P1 , which we saw above is geometry with |ax0 + by0 + cz0 + d2 | Example 3.3.21: Given that the two lines √. a2 + b 2 + c 2 x −4 y −1 z L1 : = = Since A is on P1 , we have 1 −1 −4 ax0 + by0 + cz0 + d1 = 0. and x +2 y − 11 z L2 : = = We conclude that the distance between P1 and P2 is −1 2 1 are skew, find the shortest distance between them. |d2 − d1 | |d2 − d1 | √ =. Example completed by hand in the lecture. 2 2 a +b +c 2 |n| Joe Chuang (City St George’s) MA1623 Lecture 23 5/6 Joe Chuang (City St George’s) MA1623 Lecture 23 6/6