# NAME: MATH 151 September 17, 2014 QUIZ 2 ```NAME:
MATH 151
September 17, 2014
QUIZ 2
• Calculators are NOT allowed!
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| &middot; |b| cos θ where θ is the angle between a and b. So
1
a•b
4
= arccos √ .
θ = arccos
= arccos √ √
|a| &middot; |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 &middot; 4 + 1 &middot; 11 3
=p
d = |compa⊥ b| = ⊥ = =√ .
(−2)2 + 12 ka k
kh−2, 1ik
5
d = compa⊥ b
b
a
0
y = 2x
a⊥
```