Math 0240
Quiz 1
Spring 2014
Solutions
Problem:
Find a unit vector that is orthogonal to both vectors ī − j̄ + k̄ and ī − k̄.
Solution:
Let ā = ī − j̄ + k̄ = h1, −1, 1i and b̄ = ī − k̄ = h1, 0, −1i. The vector
ī
j̄
k̄ 1 = h1, 2, −1i
v̄ = ā × b̄ = 1 −1
1
0 −1 is orthogonal to both ā and b̄. The corresponding unit vector is
1
v̄
1
2 1
1 2
= √ , √ , −√
or
−√ , −√ , √
ū =
|v̄|
6 6
6
6
6 6
Problem: Show that the lines
z−1
x
=y=
and L2 :
L1 :
2
2
are skew lines.
x = s, y = s + 2, z = −s
Solution:
L1 : x = 2t, y = t, z = 2t + 1. Direction vectors are v̄1 = h2, 1, 2i and
v̄2 = h1, 1, −1i .
v̄2 is not a constant multiple of v̄1 . Hence the vectors and the lines are not parallel.
Now, assume that the lines intersect. Then in the point of intersection the following equalities
must hold 2t = s, t = s + 2, 2t + 1 = −s. Taking the second equality and replacing t by
s + 2 in the first and the third equalities we obtain 2s + 4 = s and 2s + 5 = −s or s = −4
and s = −5/3 simultaneously. This condtradiction shows that our assumption was wrong and
the lines do not intersect.
Therefore, L1 and L2 are neither parallel nor they intersect. So, they are skew lines.
√
3, 7i onto
Problem:
Find
the
scalar
(component)
and
vector
projections
of
b̄
=
h5,
−
√
ā = h−2, 3, 3i.
ā · b̄
−10 − 3 + 21
8
= = 2.
= √
|ā|
4+
4+
9
*3 + √
ā
ā
ā
3 3
.
projā b̄ = compā b̄
= 2 · = = −1,
,
|ā|
4
2
2 2
Solution:
compā b̄ =
1
Problem:
Are vectors ā = h1, −2, 3i , b̄ = h4, 1, −2i, and c̄ = h1, 3, 5i coplanar?
Solution:
To answer the question we find the triple product of the vectors:
1 −2
3 1 −2 = 1 · (5 + 6) + 2 · (20 + 2) + 3 · (12 − 1) = 88 6= 0
ā · b̄ × c̄ = 4
1
3
5 Therefore, the vectors are not coplanar.
Bonus problem: Find parametric equations of a line such that cosine of the angle between the
2
1
line and the x-axis is , cosines of angles between the line and the y-axis and z-axes both are .
3
3
Solution: The line has to pass through the origin, the point of intersection of all axes. Assume
that the angles between the line and the corresponding axis are θx , θy , and θz , and that the
direction vector of the line is v̄ = ha, b, ci. Then
cos θx =
1
a
v̄ · ī
= ,
=√
3
|v̄||ī|
a2 + b 2 + c 2
cos θy = √
b
2
= ,
3
a2 + b 2 + c 2
cos θz = √
c
2
=
3
a2 + b 2 + c 2
Hence, a = 1, b = 2, c = 2. The parametric equations of the line are
x = t,
y = 2t,
z = 2t
Bonus problem:
Find a symmetric equation of a line such that cosines of angles between
1
the line and each of coordinate axes all are √ .
3
Solution: The line has to pass through the origin, the point of intersection of all axes. Assume
that the angle between the line and each of the axes is θ. Then the direction vector of the line
is v̄ = ha, a, ai. Indeed,
v̄ · ī
a
a
1
cos θ =
=√
= √ =√
|v̄||ī|
a 3
3
3a2
We can assume that a = 1. Then the parametric equations of the line are
x = t, y = t, z = t. Its symmetric equation is
x=y=z
2