Precalculus H
Sections 6/1 – 6/3
Review Worksheet answers
For the given initial and
terminal points, find the
vector u in the form
requested.
6.
=
=
=
u–v
(2i + 3j) - (-4i + j)
2i + 3j + 4i - j
6i + 2j
1. I: (2,7) T: (9,4)
in component form
u = < 9-2, 4-7 >
= < 7, -3 >
7.
=
=
=
2u – 5v
2(2i + 3j) - 5(-4i + j)
4i + 6j + 20i – 5j
24i + j
2. I: (-1,0) T: (3,-2)
in linear combination form
Find a unit vector in the
same direction as the
given vector.
u = (3- -1)i + (-2-0)j
= 4i – 2j
Find the magnitude and
angle direction of the
vector.
3. u = < -4, 3 >
Magnitude
║u║ = (-4) 2  3 2 = 5
Direction angle
tan θ = |3/-4| = 0.75
ref  : 36.9
< -4, 3 > is in Quad II
So, θ = 180 - 36.9
= 143.1
4. u = < 0, 3 >
Draw a graph.
Magnitude
║u║= 3
Direction angle
θ = 90
u = 2i + 3j and v = -4i + j
Find :
5.
=
=
=
u+v
(2i + 3j) + (-4i + j)
2i + 3j + -4i + j
-2i + 4j
8. v = < -8, -6 >
unit vector = v_
║v║
 8,6 
=
(-8) 2  (6) 2
= < -8, -6 >
10
= < -0.8, -0.6 >
9. v = < 1, 2 >
unit vector = v_
║v║
 1,2 
=
(1) 2  (2) 2
 1,2 
=
5
= < 1/ 5 , 2/ 5 >
Find the dot product of
the vectors.
10. < 3, 7 > < -1, 0 >
(3)(-1) + (7)(0)
= -3 + 0 = -3
11. < 4, -1 > < 2, 3 >
(4)(2) + (-1)(3)
= 8 + -3 = 5
Find the angle between
the two vectors.
12. < -1, 3 > and < 2, 5 >
u v
cos θ =
u v
 2  15
cos θ =
10  29
13
cos θ =
290
cos θ = 0.7933862854
θ = 40.2
13. < 3, 2 > and < -4, 0 >
u v
cos θ =
u v
 12  0
=
13  16
 12
=
208
cos θ = -0.8320502943
θ = 146.3
Find the dot product of
the vectors.
14. ║u║ = 8 ║v║ = 10
 between u and v : 60
u ∙ v = ║u║∙║v║∙ cos θ
= 8 ∙ 10 ∙ cos 60
= 40
15. ║u║ = 20 ║v║ = 50
 between u and v : 135
u ∙ v = ║u║∙║v║∙ cos θ
= 20 ∙ 50 ∙ cos 135
= 500 2
Are u and v orthogonal,
parallel, or neither?
16. u = < 4, 2 >
v = < -3, -6 >
Dot product
Precalculus H
Sections 6/1 – 6/3
Review Worksheet answers
(4)(-3) + (2)(6) = 0
So, orthogonal.
17. u = < 5, -2 >
v = < -10, 4 >
Dot product
(5)(-10) + (-2)(4) = -58
So, NOT orthogonal.
Cross-product
(5)(4) = (-2)(-10) True
So, parallel
18. u = < 4, 7 >
v = < -11, -6 >
Dot product
(4)(-11) + (7)(-6) = -86
So, NOT orthogonal.
Cross-product
(4)(-6) = (7)(-11) False
So, NOT parallel
Neither
Graph the parametric
system for the given
domain of t.
19. x = 2t
y = t+1
-2 < t < 2
t
-2
-1
0
1
2
x
-4
-2
0
2
4
y
-1
0
1
2
3
20. x = 2cos(t)
y = -4sin(t)
0<t<π
t
0

4

2
3
4
π
Eliminate the parameter.
22. x = t - 3
y = t2
x
4
y
0
Since x = t – 3, t = x + 3
2 2
0
2
4
Since y = t2,
y = (x + 3)2
y = x2 + 6x + 9
-2 2
4
2
0
23. x = 2cos(t)
y = -3sin(t)
Since x = 2cos(t),
cos(t) = 2x
Since y = -3sin(t),
cos(t) = y3
21. x = t
y = t3/8
-3 < t < 3
t
-3
-2
-1
0
1
2
3
x
y
3 -3.375
2
-1
1 -0.125
0
0
1 0.125
2
1
3 3.375
Since sin2(t) + cos2(t) = 1,
2
2
 x  y 
  
 1
 2  3
x2 y2

1
4
9
or 9x2 + 4y2 = 36
24. x =
y =
1
t
t
t 1
Since x =
1
t
,t =
t
Since y =
,
t 1
y =
y =
y =
1
x
1
x
1
1
x
1
x
x
1 x
1
1 x

1
x
Precalculus H
Sections 6/1 – 6/3
Review Worksheet answers
25. x = ln(t)
y = t
28. y =
ex
or y =
e
x
2
29. (y+3)2 + 1 = (x-2)2
Find a parametric system
for the equation.
(There may be more than
one correct system.)
26. y = 2x + 3
Let t = 2x. (So x =
Then y = t + 3
System:
x = 2t
y = t+3
x4
Let t = x – 4 (So x = t + 4)
Then y = 3 t
System:
x=t+4
y = 3t
Since x = ln(t),
t = ex
Since y = t ,
y =
3
t
2
)
27. 4y2 + 25x2 = 100
4 y 2 25 x 2 100


100 100 100
y2 x2

1
25 4
2
2
 y  x
    1
 5 2
This fits the form of
sin2(t) + cos2(t) = 1
y
 sin( t )
Let
5
x
and  cos(t )
2
System:
x = 2cos(t)
y = 5sin(t)
This fits the form of
tan2(t) + 1 = sec2(t)
x - 2 = sec(t)
y +1 = tan(t)
System:
x = 2 + sec(t)
y = -1 + tan(t)