answers

Vector Integral and Differential Calculus
(ACM 20150) – Assignment 2
Issue Date: Monday 22nd September 2014
Due Date: 17:00 Monday 29th September 2014
Instructions for MSC tutors: MSC tutors may help students with the assignment (e.g.
revision of appropriate sections in the lecture notes, or working through problems similar
to those on this assignment sheet), but may not solve the exact problems contained
herein, as this assignment counts towards the students’ final grade.
1. Compute the point of intersection of the lines
rL (t) = (2, 1, 3) + t(1, 0, 2),
rM (u) = (−1, 1, 0) + u(2, 0, 1).
Form the matrix
(r0 − s0 ) · e
[(2, 1, 3) − (−1, 1, 0)] · (1, 0, 2)
9
=
=
,
(r0 − s0 ) · f
[(2, 1, 3) − (−1, 1, 0)] · (2, 0, 1)
9
and the matrix
−|e|2 e · f
−e · f |f |2
=
−5 4
−4 5
,
and the inverse
−|e|2 e · f
−e · f |f |2
−1
1
=−
9
5 −4
4 −5
.
Hence,
t0
u0
−1 −|e|2 e · f
(r0 − s0 ) · e
,
−e · f |f |2
(r0 − s0 ) · f
1 5 −4
9
−
,
9
9 4 −5
1 5·9−4·9
−
,
9 4·9−5·9
−1
.
1
=
=
=
=
1
Vector Integral and Differential Calculus
The geometry of lines and planes
Substitute:
rL (t0 ) = (2, 1, 3) − (1, 0, 2) = (1, 1, 1).
Cross-check:
rM (u0 ) = (−1, 1, 0) + (2, 0, 1) = (1, 1, 1).
Hence,
P = (1, 1, 1)
is the point of intersection of the lines L and M .
2. Let
L : r(t) = r0 + te,
M : r(u) = s0 + uf ,
(1a)
(1b)
be the equations of two distinct lines in R3 . Here, all of the symbols have their
usual meaning. In particular, t and u are parameters that vary independently
over the whole of R.
State conditions that are necessary and sufficient for the lines L and M to
have a point of intersection. Hence, consider Equation (1) equipped with the
following definite values for r0 , e, s0 , and f :
For L: r0 = (1, 0, 1), e =
(−1, 2, 0)
For M : s0 = (2, 2, 0), f =
(0, 0, 1).
Do L and M intersect in this instance? Provide a sketch to illustrate your
answer.
Conditions:
e × f 6= 0,
(r0 − s0 ) · (e × f ) = 0.
We have
x
ˆ
ˆ
ˆ
y
z
e × f = −1 2 0 0 0 1 = (2, 1, 0),
so the lines are definitely non-parallel. Also,
r0 − s0 = (1, 0, 1) − (2, 2, 0) = (−1, −2, 1).
Hence,
(r0 − s0 ) · (e × f ) = (−1, −2, 1) · (2, 1, 0) = −2 − 2 = −4 6= 0,
so the lines are skew (non-intersecting).
These calculations should be accompanied by a sketch of the lines L and M in
three-dimensional space, with a set of Cartesian axes.
2
Vector Integral and Differential Calculus
The geometry of lines and planes
3. Find the point at which the line r = (1, 1, 0) + t(1, 0, 0) intersects the plane
r · (1, 0, −1) = 5.
Substitute the equation of the line into the equation of the plane and solve for t:
[(1, 1, 0) + t(1, 0, 0)] · (1, 0, −1) = 5,
Hence,
(1 + t, 1, 0) · (1, 0, −1) = 5,
1 + t = 5,
t = 4.
The point of intersection is this
r = (1, 1, 0) + 4(1, 0, 0) = (5, 1, 0).
4. Given two planes
Π1 = {r|r · (2, 1, −1) = 1},
Π2 = {r|r · (2, 3, 1) = 3},
compute their line of intersection. Are there any circumstances in which the
line of intersection between two arbitrary planes
Π1 = {r|r · n1 = k1 },
Π2 = {r|r · n2 = k2 },
does not exist? What is the mathematical condition on (n1 , n2 ) for the line
of intersection to exist?
Write out these conditions in Cartesian form:
Π1 :
Π2 :
2x + y − z = 1,
2x + 3y + z = 3.
We will solve for y and z in terms of x. Add the equations: 4x + 4y = 4,
y = 1 − x.
Substitute back into the equation for Π1 : 2x + (1 − x) − z = 1, or
2x + y − z
2x + y − 1
2x + (1 − x) − 1
x
=
=
=
=
1,
z,
z,
z.
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Vector Integral and Differential Calculus
The geometry of lines and planes
Gather the equations together:
x = x,
y = 1 − x,
z = x.
Hence, the line of intersection is
r = (x, 1 − x, x) = (0, 1, 0) + x(1, −1, 1),
where x ranges over all real numbers. In other words, the line of intersection is
r(t) = (0, 1, 0) + t(1, −1, 1),
t ∈ R.
From a geometrical point of view, the planes cannot intersect if they are parallel,
that is, if n1 and n2 are parallel, or
n1 × n2 = 0.
This answer suffices. However, we consider also the algebraic point of view, in
which we are to solve the equations
Π1 :
Π2 :
nx x + ny y + nz z = k1 ,
mx x + my y + mz z = k2 ,
Suppose w.l.o.g. that nz 6= 0. We therefore focus on writing z in terms of x and
y. Multiplying Eq. (1) by mz and (2) by nz and subtracting gives
(nx mz − mx nz )x + (ny mz − my nz )y = k1 mz − k2 nz .
But this is
ˆ + y [(n1 × n2 ) · x]
ˆ = k2 nz − k1 mz .
−x [(n1 × n2 ) · y]
In general, the only way for this to have a solution y = y(x) or x = x(y) is if
n1 × n2 6= 0.
ˆ 6= 0. Then,
Suppose, again w.l.o.g. that (n1 × n2 ) · x
y=
ˆ x + k2 nz − k1 mz
[(n1 × n2 ) · y]
:= Ax + B.
ˆ
(n1 × n2 ) · x
Substitute back into Eq. (1).
nx x + ny (Ax + B) + nz z = k1 =⇒ nz z = (k1 − ny B) + x(−nx − ny A).
Since nz 6= 0 by assumption,
z=
(k1 − ny B) + x(−nx − ny A)
= Cx + D.
nz
Thus, the line of intersection is
r = (x, Ax + B, Cx + D) = (0, B, D) + x(1, A, C).
4
Vector Integral and Differential Calculus
The geometry of lines and planes
5. The vectors a = (2, −6, −3) and b = (4, 3, −1) lie entirely in a certain plane.
Determine a unit vector perpendicular to the plane.
Consider
x
ˆ yˆ zˆ a × b = 2 −6 −3 ,
4 3 −1 ˆ − 10yˆ + 30z,
ˆ
= 15x
ˆ − 2yˆ + 6z.
ˆ
∝ 3x
√
√
The length of the vector here is 32 + 22 + 62 = 49 = 7. Hence, the required
unit vector is
b = 17 (3x
ˆ − 2yˆ + 6z)
ˆ .
n
ˆ − 2yˆ + 6z)
ˆ is an equally good answer.
Note that −(1/7) (3x
6. If a and b are arbitrary vectors in R3 , show that
(a) |a + b| ≤ |a| + |b| (Hint: Use the result a · b = |a||b| cos θ);
(b) |a − b| ≥ |a| − |b| (Hint: Define the vector w := a − b and apply the
result in (1) to the vectors w and b);
(c) ||a| − |b|| ≤ |a − b|.
(a) Square up:
iff
iff
iff
iff
|a + b| ≤ |a| + |b|,
|a + b|2 ≤ (|a| + |b|)2 ,
(a + b)2 ≤ |a|2 + |b|2 + 2|a||b|,
|a|2 + |b|2 + 2a · b ≤ |a|2 + |b|2 + 2|a||b|,
a · b ≤ |a||b|.
But
a · b ≤ |a · b| ≤ |a||b|| cos θ| ≤ |a||b|,
and the implications therefore go both ways.
(b) Let w = a − b. By the first result,
|w + b|
|(a − b) + b|
|a|
|a| − |b|
≤
≤
≤
≤
|w| + |b|,
|a − b| + |b|,
|a − b| + |b|,
|a − b|.
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Vector Integral and Differential Calculus
The geometry of lines and planes
Square up:
iff
iff
iff
iff
||a| − |b|| ≤ |a − b|,
(|a| − |b|)2 ≤ (a − b)2 ,
|a|2 + |b|2 − 2|a||b| ≤ |a|2 + |b|2 − 2a · b,
−|a||b| ≤ −a · b,
|a||b| ≥ a · b
(note the change in the direction of the inequality). But
a · b ≤ |a · b| ≤ |a||b|| cos θ| ≤ |a||b|,
and the implications therefore go both ways.
6

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