**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 c_{1}, c_{2} and c_{3}
such that

-3c_{1}
+ 4c_{2} + 6c_{3} = 2

c_{1}
- c_{3} = 0

2c_{1}
- 8c_{2} -4c_{3} = 4

This
system has solution c_{1 }= 2, c_{2} = -1 and c_{3}
= 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 ||-2**u**|| = 2 ||**u**|| then ||-2**u**|| + 2 ||**u**||
= 4 ||**u**|| = 4[17]^{1/2}

(d)
3**u** - 5**v** + **w** = (4, 15, -15) then || 3**u** -
5**v** + **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.
proj** _{a}u** = (

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

(b)
**u**.**a** = -4,
||**a**||^{2} =
13, proj** _{a}u
**= (8/13, -12/13)

(c)
**u**.**a** = -32, ||**a**||^{2}
= 26, proj** _{a}u
**= (-16/13, 0, -80/13)

(d)
**u**.**a** = 4, ||**a**||^{2}
= 89, proj** _{a}u
**= (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

p^{2}
+ q^{2} + r^{2} = 1 = 3t^{2}. Thus t =
1/3^{1/2} and **w**
= ( 1/3^{1/2}, 1/3^{1/2}, - 1/3^{1/2})
or -( 1/3^{1/2}, 1/3^{1/2}, - 1/3^{1/2}).

14.
(a) If **p** and **q** are parallel then **p** = c**q**.
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 + k^{2}]^{1/2} 34^{1/2}. If we
square this equation we get 100k^{2} + 240k + 144 = 34k^{2}
+ 136 or 33k^{2} + 120k + 4 = 0 or k =

(d)
As in part (c) we have cos(pi/4) = 2^{-1/2} = (6 + 5k)/[4 +
k^{2}]^{1/2} 34^{1/2 } or k = -8 or 1/2.

Section 3.4

1. See answers in text.

10.
Since the volume of the parallelepiped is |**u**.(**v**x**w**)|
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** **= PQ**x**PR**
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 **n**x**p** = (-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**|| = 26^{1/2} and the distance is |1 - 3/2| 26^{-1/2}
= 1/(2x26^{1/2})

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

(c)
**n** = (2, -1, 1) and
the distance is 2/6^{1/2}.