NAME:
MATH 151
September 17, 2014
QUIZ 2
• Calculators are NOT allowed!
• Show all your work and indicate your final answer clearly. You will be graded not merely
on the final answer, but also on the work leading up to it.
1. (3 points) Find the angle between the vectors:
a = 6i − 2j
b = i + j.
Solution: By definition, a • b = |a| · |b| cos θ where θ is the angle between a and b. So
1
a•b
4
= arccos √ .
θ = arccos
= arccos √ √
|a| · |b|
40 2
5
2. (3 points) Determine a vector equation of the straight line which passes through the point
(1, −1) and is parallel to the vector h−2, 3i.
Solution: The slope of the line is given by the vector h−2, 3i. So the vector equation is
hx0 − 2t, y0 + 3ti
where (x0 , y0 ) is a point on the line. Since the line is given to go through the point (1, −1),
the vector equation of the line is
h1 − 2t, −1 + 3ti.
NAME:
MATH 151
September 17, 2014
3. (3 points) Find the distance from the point (4, 11) to the line y = 2x.
Solution:The points (0, 0) and (1, 2) lie on the line y = 2x, so the displacement vector is
a = h1, 2i − h0, 0i = h1, 2i. Since a is parallel to the line y = 2x, then a⊥ = h−2, 1i is
perpendicular to the line. The displacement vector b from (0, 0) to the point (4, 11) is given
by b = h4, 11i − h0, 0i = h4, 11i. Thus the distance from the point to the line is
⊥
a • b h−2, 1i • h4, 11i −2 · 4 + 1 · 11 3
=p
d = |compa⊥ b| = ⊥ = =√ .
(−2)2 + 12 ka k
kh−2, 1ik
5
d = compa⊥ b
b
a
0
y = 2x
a⊥