MATH1025 AF


FW00F


Answers to Tutorial 5


Section 3.1


5. (a) Let the initial point of u be P = (x, y, z). Then PQ = (3-x, -y, -5-z). If this is parallel to v thenThus one possibility is x = -1, y = 2 and z = -4.

(b) As in part (a) if PQ is parallel to -v thenThus one possibility is

x = 7, y = -2 and z = -6.


6. See answers in text.


8. We must find scalars c1, c2 and c3 such that

-3c1 + 4c2 + 6c3 = 2

c1 - c3 = 0

2c1 - 8c2 -4c3 = 4

This system has solution c1 = 2, c2 = -1 and c3 = 2.



Section 3.2


3. (a) Since u + v = (3, -5, 7), ||u|| = [9 + 25 + 49}1/2 = [83]1/2

(b) ||u|| = [4 + 4 + 9]1/2 = [17]1/2 and ||v|| = [1 + 9 + 16}1/2 = [26]1/2 then ||u|| + ||v|| = [17]1/2 + [26]1/2.

(c) Since ||-2u|| = 2 ||u|| then ||-2u|| + 2 ||u|| = 4 ||u|| = 4[17]1/2

(d) 3u - 5v + w = (4, 15, -15) then || 3u - 5v + w|| = [16 + 225 + 225]1/2 = [466]1/2.

(e) Since ||w|| = [9 + 36 + 16]1/2 = [61]1/2 then (1/||w||) w = ( 3/[61]1/2, 6/[61]1/2, -4/[61]1/2)

(f) Since (1/||w||) w is a unit vector,its length is 1.


6 (b) Since ||v|| = [9 + 16]1/2 = 5 then the unit vector is (3/5, 4/5)

(c) Since ||v|| = [4 + 9 + 36]1/2 = 7 then the unit vector is (1/||v||) (-v) = (2/7, -3/7, 6/7)



Section 3.3


Questions 1, 2 - see answers in text.


4. projau = (u.a/||a||2) a

(a) since u.a = 0 the projection is zero

(b) u.a = -4, ||a||2 = 13, projau = (8/13, -12/13)

(c) u.a = -32, ||a||2 = 26, projau = (-16/13, 0, -80/13)

(d) u.a = 4, ||a||2 = 89, projau = (16/89, 12/89, 32/89)


13. Let the vector be w = (p, q, r). Then u.w = p + r = 0 and v.w = q + r = 0. These equations have an infinite number of solutions p = q = -r = t. But since w is a unit vector we must have

p2 + q2 + r2 = 1 = 3t2. Thus t = 1/31/2 and w = ( 1/31/2, 1/31/2, - 1/31/2) or -( 1/31/2, 1/31/2, - 1/31/2).


14. (a) If p and q are parallel then p = cq. Thus 2 = 3c and k = 5c so that c = 2/3 and k = 10/3

(b) If p and q are orthogonal then p.q = 0 = 6 + 5k. Thus k = -6/5.

(c) We must have cos(pi/3) = 1/2 = p.q/||p|| ||q|| = (6 + 5k)/[4 + k2]1/2 341/2. If we square this equation we get 100k2 + 240k + 144 = 34k2 + 136 or 33k2 + 120k + 4 = 0 or k =

Only the plus sign yields the correct result. The minus sign gives an angle 2pi/3.

(d) As in part (c) we have cos(pi/4) = 2-1/2 = (6 + 5k)/[4 + k2]1/2 341/2 or k = -8 or 1/2.


Section 3.4


1. See answers in text.


10. Since the volume of the parallelepiped is |u.(vxw)| we get the results:

(a) and we take the absolute value to get the volume = 16

(b)so the volume is 45.


Section 3.5


1. See answers in text.


4. (a) The normal to the plane is given by n = PQxPR where PQ = (2, 1, 2) and PR = (1, -1, -2) so that n = (0, 6, -3) and the plane is 6(y + 1) - 3(z + 1) = 0 or 2y - z + 1 = 0.

(b) As in part (a) PQ = (-1, -1, -2) and PR = (-4, 1, 1) so that n = (1, 9, -5) and the plane is

(x - 5) + 9(y - 4) - 5(z - 3) = 0.


5. (a) The normal vectors to the planes are (4, -1, 2) and (7, -3, 4) and these are not parallel so the planes are not parallel.

(a) The normal vectors to the planes are (1, -4, -3) and (3, -12, -9) and these are parallel so the planes are parallel.

(a) The normal vectors to the planes are (8,-2, -4) and (1, -1/4, -1/2) and these are parallel so the planes are parallel.


9. See answers in text.


10. (a)If we call the points P and Q the line is parallel to PQ = (2, 4, -8) and its parametric equations are x = 5 + 2t, y = -2 + 4t, z = 4 - 8t.

(b) As in part (a) PQ = (2, -1, -3) and the equations are x = 2t, y = -t and z = -3t.


11. (a) Solve the equations 7x - 2y + 3z = -2 and -3x + y + 2z = -5 to get x = -12 -7t,

y = -41 -23t and z = t.

(b) Here the solution is x = 5t/2, y = 0 and z = t.


17. The line is parallel to (2, 3, -5) so the equation of the plane is 2(x+2) + 3(y-1) - 5(z - 7) = 0.


18. See answers in text.


22. Substituting the equations for the line into the equation for the plane gives 2(9-5t) - 3(-1-t) + 4(3+t) + 7 = 0 or -3t + 40 = 0 or t = 40/3 so the point of intersection is (-173/3, -43/3, 49/3).


28. The normal to the first plane is n = (4, 2, 2) and to the second is p = (3, 6, 3) so that the line of intersection of the plane is parallel to nxp = (-6, -6, 18) so that the equation of the plane is

-6(x-2) - 6(y+1) + 18(z -4) = 0 0r x + y - 3z + 11 = 0.


39. Use theorem 3.5.2 - see answers in text.


40.If we write the parallel planes as n.x = d and n.x = f then the distance between them is

|d-f|/||n||.

(a) Writing the planes as 3x - 4y + z = 1 and 3x - 4y + z = 3/2 we have ||n|| = 261/2 and the distance is |1 - 3/2| 26-1/2 = 1/(2x261/2)

(b) The distance is zero (both planes are the same one).

(c) n = (2, -1, 1) and the distance is 2/61/2.