Math 251 ASSIGNMENT #2A ANSWERS
Instructions: Show all solution steps and use your calculator to check your answers and to
compute determinants and RREF.

1) ( 15 p ) Let



a) w v ,

v  (1,  3, 7) and w  (1,5 ,1) .Find the following:

w  v,






v , w , w v , and w v .
Solution




w  v  (1, 5, 1)  (1,  3, 7)  (0, 2, 8)
w  v  (1, 5, 1)  (1,  3, 7)  (2, 8,  6)

v  12  (3) 2  7 2  59

w  (1) 2  5 2  12  3 3


b) The scalar (or Dot) product w  v and the angle between the two vectors.
Solution


w  v  (1, 5, 1)  (1,  3, 7)  1  15  7  9




w
v
w v
cos  

9
59  3 3

3
177
   cos 1 (
3
177
)  103


c) The unit vector in the direction of v and the unit vector in the direction of w .
Solution



u v = the unit vector in the direction of v =
v
1


59
v
1
59
,
3
59
7
,
59
)



(1,  3, 7)  (
w
u w = the unit vector in the direction of w =

w


1
3 3
(1, 5, 1)  (
1
,
5
,
1
3 3 3 3 3 3
)

d) The vector component (or projection) of w along v and the vector component (or projection)


of w orthogonal to v .
Solution



proj  w  ( w cos  ) u v  [3 3 (
v
3
177
)]
1
59
(1,  3, 7)  
9
59

(1,3,7)  w
\\

w


 w  w ||  (1,5,1) 

9
59
(1,3,7)  (
9  59
59

,5 
27
59
,1 
63
59
)

e) The vector component (or projection) of v along w and the vector component (or projection)


of v orthogonal to w .
Solution




1
1 5 1
(1,5, 1)   (1,5, 1)  ( , , )  v \ \
3
3 3 3
177 3 3


1 5 1
2 4 22
 v  v ||  (1,3,7)  ( , , )  ( , , )
3 3 3
3 3 3
proj w v  ( v cos  ) u w  59 (

v

3
)
1

2) (5p) Find two unit vectors perpendicular on the nonzero vector v  (a, b)
Solution



X  ( x, y )  ? such that v  X  0 i.e. ax+by = 0,assuming b nonzero  y  

X  ( x, y )  ( x, 

X  (b, a ) 
ax
) ,with x a parameter, choose x=b for simplicity
b

X

X
ax

b

1
a2  b2

(b, a ) is a unit vectors perpendicular on v  (a, b)

Another unit vectors perpendicular on v is 
1
a b
2
2
(b,a) 
1
a  b2
2
(b, a)
Note
By simple geometric considerations it can be deduce that both (a, -b) and (-a , b ) are
perpendicular on (a, b)
3) (10p) Find the distance between the line 3x+2y = 6 and the point P (5, 10).
Solution
Method 1
either learn to deduce or memorize
| ax0  by 0  c |
D
with P( x0  5, y0  10) and the line 3x  2 y  6  0
a2  b2
a =3, b = 2, c = -6
| 3  5  2  10  6 | 29
D

13
32  2 2
Method 2 No formula to memorize!!!
3
Consider the line (l) of equation 3x+2y = 6 or y   x  2 .
2



The vector v  ( 2,3) is parallel to (l) and u  (3,2)  v .Take any point on (l) ,e.g. Q(2,0)




QP0  (5,10)  (2,0)  (3,10)  D  Pr oj  QP0  QP0 cos   QP0
u

D

QP0  u


(3,10)  (3,2)
3 2
2
u






QP0  u
QP0 u
29
13

4) (5p) Let v  (1, 9, 7) and w  (1 ,5 , 1) . Find:


a) A unit vector perpendicular on both v and w .
Solution
i j k


v  w  1 9 7  (44, 8,  4)  4(11,2,1) 
1 5 1


v  w  4 (11) 2  2 2  (1) 2  12 14


 u1, 2  

v w


v w

1
3 14
(11, 2,1)


b) The area of the parallelogram formed by u and v .
Solution


A  v  w  (11) 2  2 2  (1) 2  12 14
5) (10p) Find the area of the triangle PQR if P (2, 6,-1), Q (1, 1, 1) and R (4, 6, 2).
Solution


PQ  (1,1,1)  (2,6,1)  (1,5,2) and PR  (4,6,2)  (2,6,1)  (2,0,3)
i j
k


PQ PR   1  5 2  (15,7,10) 
2 0
3
ATRIANGLE 
1  
1
374
PQ PR 
(15) 2  7 2  (10) 2 
2
2
2
6) (5p) Find all unit vectors parallel to yz-plane and perpendicular to the vector ( 3,-1, 2 ).
Solution


v  (0, y, z )  (3,1,2)   y  2 z  0  y  2 z  v  z (0,2,1) with z a parameter
(0,2,1)
(0,2,1)
u

5
2 2  12
7) (10p) Are the points A(1,2,6), B(0,2,3) and C(-1,4,5) collinear? Why?
Solution
Method 1 :



If the points are collinear then the vectors OA, OB, OC , are in the same plane.
1 2 6
OA (OB  OC )  0 2 3  4  0  A,B,C are NOT collinear
1 4 5
Method 2





AB  (0.2,3)  (1,2,6)  (1,0,3) , AC  (1,4,5)  (1,2,6)  (2,2,1)
Because the 2 vectors are not proportional A,B,C are NOT collinear
8) (10p)Find an equation for the plane thought (2,-1, 4) that is perpendicular to the line of
intersection of the planes –x+ 2 y+ z +1= 0 and 3x+6 y+3 z = 7.
Solution (see also problem #28 pg 164)
Method 1

The plane –x+ 2 y+ z +1= 0 has n1 (1,2,1)

The plane 3x+6 y+3 z = 7 has n2  (3,6,3) .The line of intersection of the two planes is

perpendicular on both



i
j


n1 and n2 and therefore has the direction of n


3  n1  n 2
k
n3  n1  n2   1 2
1  (0,2,4)  2(0, 1,2)
1
2
1
Now we have to write the equation for the plane thought (2,-1, 4) and having the normal (0,1,-2)
(0,2,2)  ( x  2, y  1, z  4)  0 gives y  2 z  9
Method 2
The line of intersection of the two planes has to satisfy both equations
–x+ 2 y+ z +1= 0 and 3x+6 y+3 z = 7
5
5 / 3
 1 2 1  1 1 0 0
3
i.e. the solution of the system
RREF 


 that gives
1
1
3
6
3
7
0
1
1
/
2
1
/
3

 

y z
2
3
5
1
1
is x  , y   t  , z  t and therefore the line of intersection of the two planes has the
3
2
3
x

direction of v  (0,  1 / 2, 1) or (0, -1 2)
Now, the equation for the plane thought (2,-1, 4) and having the normal (0,1,-2) is
(0,2,2)  ( x  2, y  1, z  4)  0 or y  2 z  9
9) (10p) (#30 pg 164 Text) Show that the lines:
x=3 - 2t, y=4+t, z=1-t and x =5+2t
are parallel, and find the plane they contain.
Solution: see the Student Solution Manual
y=1- t z = 7+ t (where t is a parameter)
10) (10 p) Show that the plane whose intercepts with the coordinate axes are x =a, y=b and z =c
x y z
has the equation    1 , provided that abc  0.
a b c
Solution
Let A(a,0,0) ,B(0,b,0) and C(0,0,c) be the given intercepts


AB  (0, b,0)  (a,0,0)  (a, b,0)
AC  (0,0, c)  (a,0,0)  (a,0, c)
i j
k
n  AB  AC   a b 0  (bc, ac, ab) is the normal to the plane
a 0 c
Use one of the intercepts, for e.g. A (a, 0, 0), and P(x, y, z) to write the equation of the plane:





bc(x-a)+acy+abz=0  bcx  acy  abz  abc ,divide by abc  0 
n  PA  0 
x y z
  1
a b c
11) ( 10 p )Using the properties of the dot (or Scalar) product show that

 2
 2
a) v  w  v
 2


 w  2 v w cos 
c 2  a 2  b 2  2ab cos  )
(Compare to Law of Cosine
Solution

 2



 2

v  w  (v  w)  (v  w)  v


 2


 2


 w  2 v w cos 
 2






because v  v  v , w  w  w and v  w  w  v  v w cos 

 2
 2
 2


 w  2 v w cos 
b) v  w  v
Solution
 2




 2

v  w  (v  w)  (v  w)  v

 2


 2
 2



 w  2 v w cos 






because v  v  v , w  w  w and v  w  w  v  v w cos 
Then, using the above relations show that:
2
 

   
 

c)  v  w   v  w   v  w 




Solution
2


2




 1   cos  1   2 v w  2 v w cos   2 v w 
 2
v
 2

 2

 w 2 v w  v
2
 2
 

   
 

v

w

v

w

v

w








2
2

2

 2


 2


 w 2 v w 
2
 

   
 

d)  v  w   v  w   v  w 




Solution

 2

 w  2 v w cos   v

2



 1  cos  1   2 v w  2 v w cos   2 v w 
 2
v
 2

 2

 w 2 v w  v
2
 2

 2
 w  2 v w cos   v
 

   
 

 v  w   v w   v  w 




2



 w 2 v w 
2
12) (10p) Knowing that v  5 , w  7 and the angle between the two vectors is   120 find
the following:


a) v  w
Solution
 2

 2
v w  v

 2


 2




 w  2 v w cos   25  49  2  5  7  (0.5)  39  v  w  39

b) v  w
Solution
 2

 2
v w  v



c) v  w
Solution




v  w  v w sin   5  7 sin 120 

35 3
2

d) v  w
Solution


v  w  5  7 cos 120  
35
2


e) The area of the parallelogram formed by u and v .
Solution
 
35 3
A  vw 
2




13) (10p) Show that the vector v  w is perpendicular on both v and w .
Solution
vx
vy
vz
v  ( v  w)  v x
vy
vz









 0 (two identical rows)  v  w  v
wx
wy
wz
wx
wy
wz
w ( v  w)  wx
wy
wz  0 (two identical rows)  v  w  w
similarly




 w  2 v w cos   25  49  2  5  7  (0.5)  109  v  w  109
vx
vy
vz
14) Show that the distance between the a point P0 ( x0 , y0, z0 ) and the plane ax  by  cz  d  0 is
D
| ax0  by0  cz0  d |
a 2  b2  c2
Solution see class notes and/or textbook.
Learn to deduce. DO NOT MEMORIZE!
15) a)Find the volume of the parallelepiped determined by



u  [3,  1, 0] , v  [6,  2, 5] , and w  [8, 5,  5]
Solution
3 1
0
V  6
8
2
5  115  V  115  0
5 5
b) Are the three vectors on the same plane? Why?
The three vectors are not on the same plane once their triple product is nonzero
V  115  0
Solution
16) Find the standard form of the equation of the plane passing through the points
A  (1,  2, 1) , B  (2, 3, 1) , and C  (3,  1, 2)
Solution
Method 1

AB  (2,3,1)  (1,2,1)  (3,5,0);



i

AC  (3,1,2) (1,2,1)  (4,1, 1)
j k
n  AB AC  3 5 0  (5,3,17) is the normal to the plane
4 1 1
The equation of the plane passing through the points is
5( x  1)  3( y  2)  17( z  1)  0 or 5 x  3 y  17 z  16  0
Method 2
The equation of the plane passing through the points is of the form ax  by  cz  d  0
Because A,B,C are on this plane it follows that
 a  2b  c  d  0
5
3
17
2a  3b  c  d  0  a  d , b   d , c  d , d  parameter
16
16
16
3a  b  2c  d  0
5
3
17
dx  dy  dz  d  0 or ,after rearranging
The equation ax  by  cz  d  0 becomes
16
16
16
the terms and dividing by d
5 x  3 y  17 z  16  0
17) Find the parametric and symmetric equation of the line through the points
P  (6,  1, 6) and Q  (3, 7,  1)
Solution


v  PQ  (3,7,1)  (6,1,6)  (3,8,7) 
x  6  3t
y  1  8t
z  6  7t
Parametric equations of the line:
x  6 y 1 z  6


3
8
7
18) Find equation of the line of intersection of the planes 3x- 6y + 2 z = 8 and x+ y+ 3 z = 5.
Solution
Solve the system 3x- 6y + 2 z = 8
x+ y+ 3 z = 5.
3  6 2 8 1 0 20 / 9 38 / 9 
RREF 

1 3 5  0 1 7 / 9
7 / 9
1
Or, after solving for t in each obtain the symmetric equations of the line
38 20
 t
9
9
7 7
y  t
9 9
zt
x


 
19) Let v0  ( x0 , y0 , z0 ) and v  ( x, y, z ) . Describe the set of points for which v v0  9
Solution


Let v  ( x, y, z ) be the coordinates of a any point in 3D and v 0  ( x0 , y 0 , z 0 ) a fixed point




v  v 0  ( x, y , z )  ( x 0 , y 0 , z 0 )  v  v 0
2
 ( x  x0 ) 2  ( y  y 0 ) 2  ( z  z 0 ) 2
( x  x0 ) 2  ( y  y 0 ) 2  ( z  z 0 ) 2  81 is the equation of a circle with centre at ( x0 , y0 , z 0 ) and
radius 9
Note
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
The relation v  v 0  9 can be seen as the set of all points (x,y,z) at 9 units distance of
( x0 , y0 , z 0 ) which is of course … a circle with centre at ( x0 , y0 , z 0 ) and radius 9