MATH 102 - ASSIGNMENT 5 (1) Let 

(a) (b) (c) (d) (e) (f) (g)

     1 0 1 u= 2  v =  −1  w= 0  1 6 1 Find a vector that is perpendicular to both v and w. Find the angle between u and v. Find dist(u, v). Find proju v. Find the area of the parallelogram spanned by v and w. Find the volume of the parallelepiped defined by u, v, w. Do the three vectors lie in a plane? Explain. 

 −1 (a) v × w =  6  is perpendicular to both v and w.  1 √ √ u·v = arccos(4/ 6 37) = 1.299 rad. (b) θ = arccos kukkvk

1

p

√ 2 2 2

(c) dist(u, v) = ku − vk =

3 = 1 + 3 + (−5) = 35

−5     2/3 1 u·v (d) proju v = u·u u = 46  2  =  4/3  2/3 1 (e) The parallelogram spanned by v and w has area p √ kv × wk = (−1)2 + 62 + 12 = 38 (f) The volume of the parallelepiped defined by u, v, w is u · (v × w) = 1(−1) + 2(6) + 1(1) = 12 (g) No, if the three vectors did lie in one plane then they would form a parallelepiped of volume 0 (i.e. the shape would have no height). (2) (a) Find the distance from the line y = 2x + 3 to the point (1, 2). (b) Find the distance from the point (2, 0, 2) to the plane x − 2y + 2z = 5. Both of these problems can be solved using projections, but we will use the distance formulas here. (a) We write the line as 2x − y + 3 = 0, to get the distance √ |2(1) − 1(2) + 3| = 3/ 5 d= p 2 2 2 + (−1) (b) The distance is d=

|1(2) − 2(0) + 2(2) − 5| p = 1/3 12 + (−2)2 + 22

1

2

   0 1  3   2     (3) Let u =   −1  and v =  −2 . 2 −2 (a) Find the projection of u onto v. (b) Find the portion of u that is perpendicular to v. (c) What is the distance from u to v? (d) What is the distance from the point given by u to the line passing through the origin and the point given by v?   0   4  3  (a) projv u = u·v v·v v = 17  −2  = ukv the portion of u that is parallel 2 to v. (b) The portion of u that is perpendicular to v is       1 0 1  2   12/17   22/17       u⊥v = u − ukv =   −1  −  −8/17  =  −9/17  −42/17 8/17 −2

1

−1 √

(c) dist(u, v) = ku − vk =

1 = 19 = 4.359

−4 √ 1 | 2618 = 3.01. Notice that this (d) This distance is given by ku⊥v k = | 17 is shorter than dist(u, v). 

(4) Consider the lines in R3 y(x) = (1, 2, 3)x + (1, 0, 1) and y(x) = (1, −1, 1)x + (0, 1, 1) (a) Are the lines parallel? Explain. (b) Find a vector n, that is perpendicular to both lines (hint: you may find the cross product useful for this). (c) Find the minimum distance between these two lines (hint: start by finding some vector that starts on one line and ends on the other; then think about projections). (d) Do theses lines intersect? Explain. (a) No, their slope vectors are not parallel (i.e. (1, 2, 3) 6= α(1, −1, 1)) (b) n = (1, 2, 3) × (1, −1, 1) = (5, 2, −3) is perpendicular to both lines. (c) First we choose a point on each line, the most obvious point on a line is the point for when x = 0. We will let P = (1, 0, 1) and Q = (0, 1, 1). w = P − Q = (1, −1, 0) joins the two lines. The projection projn w =

3 n 38

3

represents the shortest line segment joining the two lines (it is perp to both lines). So the distance between the two lines is √ 3 k projn wk = knk = 3/ 38 38 . (5) (a) Find the equation of the plane that passes the point   through  0 1 P =  1  and is perpendicular to n =  2 . 1  1   0 1 (b) Find the distance from the line y =  1  + t  −1  to the plane 2 1 from part (a). (a) By n, we know that the the plane is of the form 1x + 2y + 1z = d, and by plugging P into this equation, we are able to solve for d; d = 1(0) + 2(1) + 1(1) = 3. So the plane has the equation x + 2y + z = 3 −−→ (b) Let Q = (0, 1, 2), a point on the line. So w = P Q = Q − P = (0, 0, 1) joins the line to the plane. Notice that n is perpendicular to both the plane and the slope of the line, so 1 projn w = (1, 2, 1) 6 represents the shortest line segment joining the line to the plane. So √ 1 k projn wk = | |k(1, 2, 1)k = 1/ 6 6 is the distance between the line and the plane. (6) Find the distance from the point P to the    1 1  2   −1    P = L(x) =  1  1 −1 −1

line L.  

 1    x +  0   1   0

• First join the point to the line: Let Q = (1, 0, 1, 0), this is a point on −−→ the line. Let w = P Q = Q − P = (0, −2, 0, 1), this joins the point to the line. • Next project w onto the slope of the line: Let v = (1, −1, 1, −1) the slope of the line. 1 projv w = (1, −1, 1, −1) = wkv 4 This is the part of w that is parallel to the line.

4

• Find kw⊥v k: w⊥v = w − wkv = (−1/4, −7/4, −1/4, 5/4) represents the shortest line joining the point to the line. So 1√ kw⊥v k = k(−1/4, −7/4, −1/4, 5/4)k = 76 4 is the distance from P to L. (7) Let T (x1 , x2 ) = (x1 − 2x2 , −x1 + 3x2 , 3x1 − 2x2 ) (a) Find the domain and codomain of T (b) Find the standard matrix for T (c) Determine if T is a linear transformation (d) Find a vector ~x such that T (~x) = (−1, 4, 9) (a) T defines a transformation from R2 to R3 . That is, T : R2 → R3 so the domain is R2 and the codomain is R3 . (b) Rewriting the transformation in matrix notation,      w1 x1 1 −2  w2  =  −1 3   x2  . 3 −2 w3 x3   1 −2 3  Then the standard matrix is A =  −1 3 −2 (c) For ~u = (u1 , u2 ), ~v = (v1 , v2 ) ∈ R2 , T (~u + ~v ) = ((u1 + v1 ) − 2(u2 + v2 ), −(u1 + v1 ) + 3(u2 + v2 ), 3(u1 + v1 ) − 2(u2 + v2 ) = (u1 − 2u2 , −u1 + 3u2 , 3u1 − 2u2 ) + (v1 − 2v2 , −v1 + 3v2 , 3v1 − 2v2 ) = T (~u) + T (~v ). And for c a scalar, T (c~u) = ((cu1 ) − 2(cu2 ), −(cu1 ) + 3(cu2 ), 3(cu1 ) − 2(cu2 )) = (c(u1 − 2u2 ), c(−u1 + 3u2 ), c(3u1 − 2u2 )) = c(x1 − 2x2 , −x1 + 3x2 , 3x1 − 2x2 ) = cT (~u) Thus T is a linear transformation. (d) For ~x = (5, 3) then T (~x) = (−1, 4, 9) (8) Given the vectors ~u = (−7, 6, 10, −3, −3), ~v = (5, −9, −2, 4, −3) (a) Find the Euclidean inner product ~u · ~v (b) Find the Euclidean distance between ~u and ~v (c) Verify the Cauchy-Schwarz inequality holds (a) ~u · ~v = (−7)(5) + (6)(−9) + (10)(−2) + (−3)(4) + (−3)(−3) = −112 p (b) √ d(~u, ~v ) = (−7 − 5)2 + (6 + 9)2 + (10 + 2)2 + (−3 − 4)2 + (−3 + 4)2 = 562

5

(c) |~u · ~v | = p112 √ k~uk = p (−7)2 + 62 + 102 + (−3)2 + (−3)2 = 203 √ k~v k = 52 + (−9)2 + (−2)2 + 42 + (−3)2 = 3 15 Then k~ukk~v k ≈ 165.54 ≥ |~u · ~v | (9) Find the standard matrix of the linear transformation T : R3 → R3 given by orthogonal projection onto the plane x − 2y + z = 0. (It is helpful to note that this plane contains the origin). 

 a1 Let ~a =  a2  be a point in R3 . We want to find the standard matrix a3 corresponding to T (~a). First we find a vector from ~a to point in the plane. Since (0, 0, 0) is a point in the plane, we have the vector       a1 0 a1 ~v =  a2  −  0  =  a2  a3 0 a3   1 Now, the plane has the normal vector ~n =  −2 , so the projection of ~v 1 along ~n is given by   1 a1 − 2a2 + a3  −2  . proj~n~v = 6 1 Then the orthogonal projection of ~v onto the plane (the transformation we’re looking for) is given by ~v −proj~n~v   1 a1 =  a2  − a1 −2a62 +a3  −2  1 a3  5a  1 +2a2 −a3 = 

6 a1 −5a2 +a3 3 −a1 +2a2 +5a3 6



  5/6 1/3 −1/6 a1 1/3   a2  =  1/3 −5/3 −1/6 1/3 5/6 a3 Thus the standard matrix of   a1 T  a2  a3 is



5/6  1/3 −1/6

1/3 −5/3 1/3

 −1/6 1/3  5/6

6

(10) Determine the standard matrix for the linear transformation T : R2 → R2 that first dilates a vector by a factor k = 3, then reflects about the line y = x, then projects onto the y-axis. Dilation by a factor 3 can be given as multiplication by the vector   3 0 0 3 Reflection about the line y = x can be given as multiplication by the vector   0 1 1 0 And projection onto the y-axis can be given as multiplication by the vector   0 0 0 1 Then the transformation T (~u) for ~u = (u1 , u2 ) ∈ R2 is given by         0 0 0 1 3 0 u1 0 0 u1 T (~u) = = 0 1 1 0 0 3 u2 3 0 u2 and the standard matrix is 

0 3

0 0