MATHEMATICS 101
TUTORIAL 3 (SOLUTIONS TO DISCUSSION PROBLEMS)
MODULE 5 – LINEAR AND MATRIX ALGEBRA 1
Exercise 13.3 (page 849)
Applications of dot product
Question 43
If a = 3 , 0 ,

1 , find a vector b such that comp a b = 2 .
Solution
From definition, comp a b is the scalar projection of b onto a . Hence,
comp a
b
 a.b
a
 a a
.b
= 2



3 , 0 ,

1
 b = 2
3
( 3 )
2
,

( 0 )
2
0 ,

1


(

1 )
2 b = 2
3 ,
10
0 ,

1
 b = 2 10
One possible vector would be 0 , 0 ,

2 10 .
Courtesy Mr. Rajesh Gunesh

Question 49
Use scalar projection to show that the distance from a point ax
 by
 c

0 is ax
1
 a
2 by
1
 b
2
 c
P
1
( x
1
, y
1
) to the line
Use this formula to find the distance from the point (–2, 3) to the line 3 x

4 y

5

0 .
Solution
P
1
( x
1
, y
1
) ax
 by
 c

0 d
A ( x
1
, y
2
)
F
Fig. 1
Choose any point A on line ax
 by
 c

0 such that its x -coordinate is also x (definitely
1 possible, think about it!) and let F be the foot of the perpendicular from P
1
to the line. The equation of the line can be re-arranged in the y = mx + c form as y
  a b x
 c b
. Since its gradient is
 a b
, the direction vector of the line is b ,
 a . Also, when x
 x
1
, y
2
 
1 b
( ax
1
 c ) (see Fig. 1 ). Therefore, A has co-ordinates x
1
,

1 b
( ax
1
 c ) .
Now, FA is the projection of PA onto line ax
 by
 c

0 , that is, FA is given by
PA

unit vector in the direction of line ax
 by
 c

0
Since PA
 x
1
,

1 b
( ax
1
 c )
 x
1
, y
1

0 ,

1 b
( ax
1
 c )
 y
1

0 ,

1 b
( ax
1
 by
1
 c ) ,
FA =
0 ,

1 b
( ax
1
 by
1
 c ) b
2 
(
 a )
2
 b ,
 a
 a b
( ax
1
 by
1 a
2  b
2
 c )
 a ( ax
1 b
 by
1
 c )
. a
2  b
2
Courtesy Mr. Rajesh Gunesh

Using
Pythagoras’ Theorem
in triangle P
1
AF , d

Clearly, PA =

1 b
( ax
1
 by
1
 c )
Hence, d

( ax
1
 by
1 b
2
 c )
2

1 b
( ax
1
 by
1
 c )
 a
2
( ax b
2
1
( a

2 by
1
 b
2

) c )
2
PA
2 
FA
2
.
(is this obvious to you?)



( ax
1
 by
1 b
2
 c )
2



 1

( a
2 a
2
 b
2
)





( ax
1
 by
1 b
2
 c )
2




( a
2 b
2
 b
2
)



( ax
1
 a
2 by
1
 b
2
 c )
2

( ax
1
 by
1 a
2  b
2
 c )
2
 ax
1
 a
2 by
1
 b
2

The distance from point (–2, 3) to the line 3 x

4 y

5

0 is thus c
( 3 )(

2 )

(

4 )( 3 )

5


6

12

5


13
( 3 )
2 
(

4 )
2 25 5
.

13
5
.
Courtesy Mr. Rajesh Gunesh

Exercise 13.4 (page 856)
The cross product
Question 9
State whether each expression is meaningful. If not, explain why. If so, state whether it is a vector or a scalar.
(a) a

( b
 c )
(c) a

( b
 c )
(b)
(d) ( a a


( b b )

 c c
)
(e) ( a
 b )

( c
 d ) (f) ( a
 b )

( c
 d )
Solution
(a) a

( b
 c )
Meaningful. The result is a scalar.
(b) a

( b
 c )
Meaningless since we cannot perform a and the scalar ( b
 c ). cross-product between the vector a
(c)
(d) ( a

( b
 c )
Meaningful. The result is a vector. a
 b )
 c
Meaningless since we cannot perform a and the scalar ( a
 b ).
cross-product between the vector c
(e) ( a
 b )

( c
 d )
Meaningless since we cannot perform a cross-product between scalars ( a
 b ) and ( c
 d ).
(f) ( a
 b )

( c
 d )
Meaningful. The result is a scalar.
Courtesy Mr. Rajesh Gunesh

(
Question 33
Use the scalar triple product to verify that the vectors a = 2 i + 3 j + k , b = i – j and c = 7 i + 3 j + 2 k are coplanar.
Solution
The scalar triple product of three vectors a , b and c is given by a

( b
 c ).
Note If a

( b
 c ) = 0, then a , b and c are coplanar. b
 c ) = i
1

1 j
7 3 k
0
2
= i


1
3
0
2
– j

1
7
0
2
+ k

1

1
7 3 a

( b
 c ) =
= –2 i – 2 j + 10 k .
2 , 3 , 1
 
2 ,

2 , 10

( 2 )(

2 )

( 3 )(

2 )

( 1 )( 10 )
 
4

6

10

0 .
Therefore, a , b and c are coplanar.
Question 41
Prove that ( a – b )

( a + b ) = 2( a
 b ).
Solution
( a – b )

( a + b ) = [ a

( a + b )] – [ b

( a + b )]
= [( a
 a ) + ( a
 b )] – [( b
 a ) + ( b
 b )]
= ( a
 b ) – ( b
 a ) since ( a
 a ) = ( b
 b ) = 0
= 2( a
 b ) since ( a
 b ) = – ( b
 a ).
(distributive property)
Courtesy Mr. Rajesh Gunesh

Exercise 13.5 (page 865)
Equation of a line
Question 4
Find a vector equation and parametric equations for the line through the origin and parallel to the line x = 2 t , y = 1 – t , z = 4 + 3 t .
Solution
The line with parametric equations x = 2 t , y = 1 – t , z = 4 + 3 t has vector equation r = 0 , 1 , 4
 t 2 ,

1 , 3
Since the required line has the same direction vector and passes through (0, 0, 0), it has vector equation r = 0 , 0 , 0
 t 2 ,

1 , 3 or r = t 2 ,

1 , 3 .
Its parametric equations are x = 2 t , y = – t , z = 3 t .
Question 17
Find a vector equation for the line segment from (10, 3, 1) to (5, 6, –3).
Solution
The direction vector of the line segment is 5 , 6 ,

3

10 , 3 , 1
 
5 , 3 ,

4 .
The equation of the line segment is therefore r = 10 , 3 , 1
 t for 0
 t

1 .

5 , 3 ,

4
Courtesy Mr. Rajesh Gunesh

Exercise 2.3 (page 90)
MODULE 2 – INTRODUCTORY CALCULUS
The Squeeze Theorem
Question 37
Prove that lim x

0 x
4
 cos

2 x

0 .
Solution
From definition,

1
 cos
 
1
   
.
This result is therefore also true for
 
2
, x x

0 .
Hence,

1
 cos

2 x

1
 x
 
, x

0
Multiplying by x
4
, we have
 x
4  x
4 cos

2 x
Taking limits as x approaches 0, we obtain
 x
4  lim x

0
(
 x
4
)
 lim x

0 x
4
 cos

2 x x
 
, x

0 .
 x lim

0 x
4
Since lim x

0
(
 x
4
) = lim x

0 x
4
= 0, by the Squeeze Theorem , lim x

0 x
4 cos


2 x

0 .
Courtesy Mr. Rajesh Gunesh

Exercise 4.4 (page 261)
Limits at infinity
Question 19
Evaluate lim x
   x x
2 
4
.
Solution lim x
   x x
2 
4
= x lim
   x x
2

1 
4 x 2
=
= x lim x
   lim x
  
(
 x )

1

1 
4 x 2
=

1  x
4
2
= x lim
  
since


1 
1
0
 x
2
 
1 .
x

1 
4 x 2
= lim x
   x
  x for x < 0 x x

1 
4 x 2
Courtesy Mr. Rajesh Gunesh

Question 20
Evaluate lim x
 
 x
4 
6 x
2  x
2

.
Solution lim x
 
 x
4 
6 x
2  x
2

= lim x
 
 x
4 
6 x
2 x
4
 x
2


6 x
2 x
4


6 x
2 x
2
 x
2

(rationalising the numerator)
= lim x
 
 x
4  x
4
6 x
2

6 x
2
2



  x
2
2
= lim x
 

 x
4 
6 x
2  x
4 x
4 
6 x
2  x
2


=
= x lim
 

 lim x
 



 x
2 x
4
6 x
2

6 x
2  x
2

 =

 1

6 x
2
6 x
2


 x
2




= lim x
 



 x lim
 



 x
4

 1

6 x
2
6 x
2


 x
2





 1

6
6 x
2



1




(divide by
2 x )
=

6

1 0


1

6
2

3 .
Courtesy Mr. Rajesh Gunesh