Download Solutions for Homework 2

M243. Fall 2011. Homework 2. Solutions.
H2.1 Given a cube ABCDA1 B1 C1 D1 , with sides AA1 , BB1 , CC1 and DD1 being parallel (can
think of them as “vertical”).
(i) Find the angle between diagonal AC1 of a cube and diagonal AB1 of its face
(ii) Repeat part (i) replacing AB1 by A1 B
(iii) Let M denote the center of the square ABCD and let N be a point of the segment
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BB1 such that NBN
B1 = 2 . Find the angle between lines M C1 and AN .
Solution
We introduce the coordinate system such that A(0, 0, 0), D(1, 0, 0), B(0, 1, 0), and A1 (0, 0, 1).
In each part we denote the measure of the unknown angle between the corresponding vectors by γ. As we know, if γ ∈ [0, π/2], then the angle between the corresponding lines is
γ. If γ ∈ (π/2, π], it is π − γ.
−−→
−−→
(i) Since C1 (1, 1, 1) and B1 (0, 1, 1), then AC1 = h1, 1, 1i, and AB1 = h0, 1, 1i. Hence
√ √
−−→ −−→ −−→ −−→
cos γ = AC1 · AB1 /|AC1 ||AB1 | = 2/( 3 2) ≈ .8164965809, and γ ≈ .6154797087 radians.
−−→
−−→ −−→ −−→ −−→
(ii) Since A1 (0, 0, 1), then A1 B = h0, 1, −1i. Therefore cos γ = AC1 · A1 B/|AC1 ||A1 B| = 0,
and γ = π/2 radians.
(iii) Coordinates of points M and N are easy to find from a drawing, but general formuli from H1.1 (ii) can be used if it creates difficulties. One gets M (1/2, 1/2, 0) and
−−−→
−−→
N (0, 1, 3/5). Therefore M C1 = h1/2, 1/2, 1i, and AN = h0, 1, 3/5i. Hence cos γ =
√ √
−−−→ −−→ −−−→ −−→
M C1 · AN /|M C1 ||AN | = 1.1/( 1.5 1.36) ≈ .7701540461, and γ ≈ .6917137050 radians.
H2.2 Let A1 , A2 , A3 , A4 be four points in R3 . Suppose lines A1 A2 and A3 A4 are perpendicular,
and lines A1 A3 and A2 A4 are perpendicular. Prove that then lines A1 A4 and A2 A3 are
also perpendicular.
Solution
−−−→
To simplify the notations, denote vectors A1 Ai by e~i , i = 2, 3, 4. Then the given perpendicularity conditions can be rewritten as
−−−→ −−−→
−−−→ −−−→
A1 A2 ⊥ A3 A4 ⇔ A1 A2 · A3 A4 = 0 ⇔ e~2 · (e~4 − e~3 ) = 0 ⇔ e~2 · e~4 = e~2 · e~3
(1)
Similarly (or just interchanging indices 2 and 3), we have
−−−→ −−−→
A1 A3 ⊥ A2 A4 ⇔ e~3 · e~4 = e~3 · e~2
We want to show that
−−−→ −−−→
A1 A4 ⊥ A2 A3 ⇔ e~4 · e~3 = e~4 · e~2
Obviously (1) and (2) imply (3) and we are done.
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(2)
(3)
H2.3
(i) If ~c = |~a|~b + |~b|~a, where ~a, ~b, ~c are all nonzero vectors, show that ~c bisects the angle
between ~a and ~b.
(ii) Let OA, OB, OC be three rays in space, and let rays OD, OE, OF bisect the angles AOB, BOC, COA, respectively. Prove that the three angles formed by the rays
OD, OE, OF are either all acute, or all right, or all obtuse.
Solution
(i) Since | |~a|~b | = |~a||~b| = |~b||~a| = | |~b|~a |, then the parallelogram with sides determined by
vectors |~a|~b and |~b|~a is a rhombus and ~c corresponds to its diagonal. But a diagonal of a
rhombus bisects its angle: the obtained two triangles are congruent by SSS. Clearly the
same argument gives a more general statement: the sum of two vectors of equal length
bisects the angle between them. The proof is finished.
Suppose one wants to prove that the sum ~z = ~x + ~y of two vectors ~x and ~y of equal lengths
forms equal angles with each of them without using geometry. It also can be easily done.
Let α denote the measures of the angle between ~x and ~z, and β denote the measures of
the angle between ~y and ~z. Since |~x| = |~y |, then
cos α =
~x · ~z
~x · ~x + ~x · ~y
|~x|2 + ~x · ~y
|~y |2 + ~x · ~y
~y · ~x + ~y · ~y
~y · ~z
=
=
=
=
=
= cos β
|~x||~z|
|~x||~z|
|~x||~z|
|~y ||~z|
|~y ||~z|
|~y ||~z|
By definition of the angle between two vectors, both α and β are in [0, π]. Therefore the
equality of their cosines, implies the equality of the angles, and the proof is finished.
(ii) The statement of the problem can be rephrased as the following: show that the cosines
of the angles determined by the three rays are either all positive, or all zero, or all negative.
It is also clear that the sign of the cosine of the angle between two vectors is determined
by the sign of their dot product:
~x · ~y
cos γ =
,
|~x||~y |
and the denominator is always positive for nonzero vectors.
This suggest the following solution. Consider three unit vectors: e~A =
−→
−→
1 −
1 −
−
−
→ OB, and e~C = −
−
→ OC.
|OB|
1 −→
−→ OA,
|OA|
e~B =
|OC|
Since eA , eB , eC are all of equal length ( =1 ), vectors e~A + e~B , e~B + e~C and e~C + e~A are
parallel to rays OD, OE, OF and bisect the angles AOB, BOC, COA, respectively, as was
explained in part (i). Therefore the sign of cos(∠DOE) is determined by the sign of
(e~A + e~B ) · (e~B + e~C ) = e~A · e~B + (e~B )2 + e~B · e~C + e~C · e~A =
1 + e~A · e~B + e~B · e~C + e~C · e~A
Since the final expression of the dot product above is symmetric with respect to all permutations of letters A, B, C, the signs of cosines of two other angles will be determined by the
sign of exactly the same expression! (In case you have doubts, just repeat the computation
for two other angles). This ends the proof.
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H2.4 Suppose points A1 , A2 , ...An lie on a unit circle centered at point O and divide it into n
congruent arcs.
(i) Find
n
X
−−−→
|A1 Ai |2
i=1
(ii) Find the sum of squares of lengths of all segments Ai Aj , 1 ≤ i < j ≤ n.
(Hint: Use Problem H1.4)
Solution
−−→
−−−→
(i) Let O be the center of the circle and P
~ei = OAi . Then A1 Ai = ~ei − ~e1 and by the
statement of Problem H1.4, we know that ni=1 ~ei = ~0. Therefore
n
n
n
X
X
X
−−−→
|A1 Ai |2 =
(~ei − ~e1 )2 =
(~ei 2 − 2~ei · ~e1 + ~e1 2 ) =
i=1
i=1
i=1
n
X
n
X
i=1
i=1
(2 − 2~ei · ~e1 ) = 2n − 2~e1 ·
~ei = 2n − 2~e1 · ~0 = 2n − 0 = 2n
Therefore the answer is 2n.
(ii) It is clear from the symmetry of the regular n-gon, that replacing the vertex A1 in the
P
−−−→
the sum nj=1 |A1 Aj |2 by any other vertex Ai , leads to the same answer 2n. Therefore
P
2
1≤j≤n |Ai Aj | = 2n. Hence we have


X
X
X
X
1
1

|Ai Aj |2 =
|Ai Aj |2  =
|Ai Aj |2 =
2
2
1≤i<j≤n
1≤i≤n
1≤i6=j≤n
1≤j≤n
1 X
1
(2n) = n(2n) = n2
2
2
1≤i≤n
The constant 1/2 in front of the second sum above is due to the fact that every addend
from the first sum appears in the second sum exactly two times.
H2.5 Problem # 38 from Section 13.4
Solution
We know that |~u × ~v | = |~u||~v | sin γ = (3)(5) sin γ = 15 sin γ, where γ is the measure of the
angle between ~u and ~v . Since by definition γ ∈ [0, π], the maximum value of the |~u × ~v | is
15 and the minimum is 0. Since ~u rotates in the xy-plane and ~v = 5~j, the direction of ~u ×~v
is always parallel to z-axis (unless it is ~0). Whether it is positive or negative direction of
the z-axis, depends on the quadrant in the xy-plane where the endpoint of ~u lies. It is
positive if it is in the 4-th or the 1-st quadrant, and it is negative in the 2-nd and the 3-rd
quadrant.
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H2.6 Problem # 42 from Section 13.4. (If you wish, you can use Maple to solve this problem).
Solution
This is a good exercise for Maple.
Our goal is to prove the property of the triple cross product of vectors
a x (b x c) = (a dot c) b - (a dot b) c (*)
It is clear that this can be done by using components of vectors a,b,c.
We start with loading the package linalg which contains commands related to vectors and
matrices. To supress the display of the long list of commands containing in the package
we end the line with ” : ”
> with(linalg):
Next we write the vector by using components
> a:= vector( [a1,a2,a3] ); b := vector( [b1,b2,b3] ); c:= vector( [c1,c2,c3] );
a := [a1, a2, a3]
b := [b1, b2, b3]
c := [c1, c2, c3]
The next line computes the left hand side of (*), ie a x (b x c):
> LHS := simplify( crossprod(a, crossprod(b,c)));
LHS := [a2 b1 c2 - a2 b2 c1 - a3 b3 c1 + a3 b1 c3,
a3 b2 c3 - a3 b3 c2 - a1 b1 c2 + a1 b2 c1,
a1 b3 c1 - a1 b1 c3 - a2 b2 c3 + a2 b3 c2]
The following three lines compute the right hand side of (*):
> Y := simplify( scalarmul(b,dotprod(a,c,‘orthogonal‘)) );
Y := [(a1 c1 + a2 c2 + a3 c3) b1, (a1 c1 + a2 c2 + a3 c3) b2,
(a1 c1 + a2 c2 + a3 c3) b3]
> Z := simplify (scalarmul(c,(-1)*dotprod(a,b,‘orthogonal‘)) );
Z := [-(a1 b1 + a2 b2 + a3 b3) c1, -(a1 b1 + a2 b2 + a3 b3) c2,
-(a1 b1 + a2 b2 + a3 b3) c3]
> RHS := simplify( matadd(Y,Z));
RHS := [a2 b1 c2 - a2 b2 c1 - a3 b3 c1 + a3 b1 c3,
a3 b2 c3 - a3 b3 c2 - a1 b1 c2 + a1 b2 c1,
a1 b3 c1 - a1 b1 c3 - a2 b2 c3 + a2 b3 c2]
Now we compare the expressions on both sides of (*) by subtracting them:
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> simplify( matadd(LHS, scalarmul(RHS,-1)));
[0, 0, 0]
Since the result is zero vector, LHS = RHS, and the statement is proven.
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