Math 311-504Topics in Applied MathematicsLecture 2:Orthogonal projection.Lines and planes.Vectorsn-dimensional vector is an element of Rn, that is,an or der ed n-tuple ( x1, x2, . . . , xn) of real numbers.Elements of Rnare regarded either as vectors or aspoints. If a, b ∈ Rnare points, then the directedsegment−→ab represents the vector b − a.In particular, each point a ∈ Rnhas the samecoordinates as its position vector−→0a.ba0b − aa = (2, 1), b = (−3, 2), b − a = (−5, 1)Let x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) ben-dimensional vectors.Addition: x + y = (x1+ y1, x2+ y2, . . . , xn+ yn).Scalar multiplication: rx = (rx1, rx2, . . . , rxn).Length: |x| =px21+ x22+ ···+ x2n.Dot product: x · y = x1y1+ x2y2+ · + xnyn.Angle: ∠(x, y) = ar ccosx · y|x||y|.Problem. Find the angle θ between v ectorsx = (2, −1) and y = (3, 1).x · y = 5, |x| =√5, |y| =√10.cos θ =x · y|x||y|=5√5√10=1√2=⇒ θ = 45oProblem. Find the angle φ between vectorsv = (−2, 1, 3) and w = (4, 5, 1) .v · w = 0 =⇒ v ⊥ w =⇒ φ = 90oOrthogonal projectionLet x, y ∈ Rn, with y 6= 0.Then there exists a unique decomposition x = p + osuch that p is parallel to y and o is orthogonal to y.ypxop = orthogonal projection of x onto yOrthogonal projectionLet x, y ∈ Rn, with y 6= 0.Then there exists a unique decomposition x = p + osuch that p is parallel to y and o is orthogonal to y.Namely, p = αu, where u is the unit vector of thesame direction as y, and α = x · u.Indeed, p · u = (αu) · u = α(u · u) = α|u|2= α = x ·u.Hence o · u = (x − p) · u = x ·u −p · u = 0 =⇒ o ⊥ u=⇒ o ⊥ y.p is called the vector projection of x onto y,α = ±|p| is called the scalar projection of x onto y.u =y|y|, α =x · y|y|, p =x · yy · yy.LinesA line is specified by one poi nt and a direction.The direction is specified by a nonzero vector.Definition. A line is a set of all points tu + v,where u 6= 0 and v are fixed vectors while t rangesover all real numbers.Here v is a point on the line, u is the directio n.tu + v is a parametric representation of the line.Example. t(1, 3, 1) + (−2, 0, 3) is a line in R3.If (x, y , z) is a point on the line, thenx = t − 2,y = 3t,z = t + 3for some t ∈ R.0uvu + vtu + vLine tu + vProblem. Let ℓ denote a line tu + v.(i) Find the distance from a point x to ℓ.(ii) Find the point on the li ne ℓ that is closest to x.0 uvv + pxv + op = orthogonal projection of x − v onto u.The distance equals |o|. The closest point is v + p.Alternatively, a line is specified by two distinctpoints a and b. Then the vector b − a is parallel tothe line, hence t(b − a) + a is a parametricrepres entation.Let x = t(b − a) + a.Then x lies between a and b if 0 < t < 1.If t > 1 then b lies between a and x.If t < 0 then a li es between x and b.Definition. The segment joining points a and b isthe set of all points t(b − a) + a, where 0 ≤ t ≤ 1.Note that t(b − a) + a = (1 − t)a + tb.0b − aabt(b − a) + aLine through a and b0xyx − x0x0pxIn R2, a line can also be specified by one po int andan or thogonal di rection.0xyx − x0x0pxLine through x0orthogonal to px is on line ⇐⇒ p ·(x − x0) = 0Proposition Let ℓ ⊂ R2be the line passingthrough a poi nt x0and or tho gonal to a vectorp 6= 0. Then a po int x ∈ R2is on ℓ if and only ifp · (x − x0) = 0.Suppose p = (a, b), x = (x, y ), and x0= (x0, y0).Then the equation of the line ℓ becomesa(x − x0) + b(y − y0) = 0orax + by = c, where c = ax0+ by0.Distance to a line in a pl aneProposition Suppose ℓ is a line in R2given by theequation ax + by = c. T hen(i) the distance from a point (x1, y1) to the line ℓequals|ax1+ by1− c|√a2+ b2;(ii) two poi nts ( x1, y1) and (x2, y2) are on the sameside of ℓ if and only if the numbers ax1+ by1− cand ax2+ by2− c have the same sign.0xyax + by = cx0p = (a, b)x1Distance from x1to ℓ is equal to |x1− x0|Vector x1− x0is parall el to pProof o f (i)The vector p = (a, b) is orthogonal to the li ne ℓ.The equation ax + by = c can be rewritten asp · x = c, where x = (x, y ).Given a poi nt x1= (x1, y1), let x0be its orthogonalprojection on ℓ. Then the distance dist(x1, ℓ) isequal to |x1− x0|.Since vectors x1− x0and p are parallel,p · (x1− x0) = ±|p||x1− x0|.dist =|p ·(x1− x0)||p|=|p · x1− p · x0||p|=|ax1+ by1− c|√a2+ b2PlanesA plane is specified by two intersecting lines.Definition. A plane is a set of all pointstu + sw + v, where u, w, and v are fixed vectorssuch that u and w are not parallel, while t and srange over all real numbers.The plane tu + sw + v contains lines tu + v andsw + v that intersect at the point v.tu + sw + v is a parametric representation.Planes0uwvv + tuv + tu + swv + swAlternatively, a plane is specified by a line tu + vand a point a outside it. Then a parametricrepres entation is tu + s(a − v) + v.Alternatively, a plane is specified by three points a,b, and c that are not on the same l ine. Then aparam etric representation ist(b −a) + s(c − a) + a= (1 − t −s)a + tb + sc.In R3, a plane can also be specified by one point x0and an ortho gonal dir ecti on p 6= 0. Then the planeis given by the equation p · (x − x0) = 0.Let p = (a, b, c), x = (x, y , z), and x0= (x0, y0, z0).Then the equation of the plane becomesa(x − x0) + b(y − y0) + c(z − z0) = 0orax + by + cz = d, where d = ax0+ by0+ xz0.Distance to a plane in spaceProposition Suppose Π is a plane in R3given bythe equation ax + by + cz = d. Then(i) the dis tance from a point (x1, y1, z1) to the planeΠ equals|ax1+ by1+ cz1− d|√a2+ b2+ c2;(ii) two poi nts ( x1, y1, z1) and (x2, y2, z2) are on thesame side …