 # Equations of lines planes and spheres

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Equations of lines planes and spheres
```VECTOR ALGEBRA
Find the volume V of the parallelepiped with sides a = i + 2j + 3k, b = 4i + 5j + 6k and
c = 7i + 8j + 10k.
We have already found that a × b = −3i + 6j − 3k, in subsection 7.6.2. Hence the volume
of the parallelepiped is given by
V = |a · (b × c)| = |(a × b) · c|
= |(−3i + 6j − 3k) · (7i + 8j + 10k)|
= |(−3)(7) + (6)(8) + (−3)(10)| = 3. Another useful formula involving both the scalar and vector products is Lagrange’s identity (see exercise 7.9), i.e.
(a × b) · (c × d) ≡ (a · c)(b · d) − (a · d)(b · c).
(7.36)
7.6.4 Vector triple product
By the vector triple product of three vectors a, b, c we mean the vector a × (b × c).
Clearly, a × (b × c) is perpendicular to a and lies in the plane of b and c and so
can be expressed in terms of them (see (7.37) below). We note, from (7.25), that
the vector triple product is not associative, i.e. a × (b × c) = (a × b) × c.
Two useful formulae involving the vector triple product are
a × (b × c) = (a · c)b − (a · b)c,
(7.37)
(a × b) × c = (a · c)b − (b · c)a,
(7.38)
which may be derived by writing each vector in component form (see exercise 7.8).
It can also be shown that for any three vectors a, b, c,
a × (b × c) + b × (c × a) + c × (a × b) = 0.
7.7 Equations of lines, planes and spheres
Now that we have described the basic algebra of vectors, we can apply the results
to a variety of problems, the ﬁrst of which is to ﬁnd the equation of a line in
vector form.
7.7.1 Equation of a line
Consider the line passing through the ﬁxed point A with position vector a and
having a direction b (see ﬁgure 7.12). It is clear that the position vector r of a
general point R on the line can be written as
r = a + λb,
226
(7.39)
7.7 EQUATIONS OF LINES, PLANES AND SPHERES
R
b
r
A
a
O
Figure 7.12 The equation of a line. The vector b is in the direction AR and
λb is the vector from A to R.
since R can be reached by starting from O, going along the translation vector
a to the point A on the line and then adding some multiple λb of the vector b.
Diﬀerent values of λ give diﬀerent points R on the line.
Taking the components of (7.39), we see that the equation of the line can also
be written in the form
y − ay
z − az
x − ax
=
=
= constant.
(7.40)
bx
by
bz
Taking the vector product of (7.39) with b and remembering that b × b = 0 gives
an alternative equation for the line
(r − a) × b = 0.
We may also ﬁnd the equation of the line that passes through two ﬁxed points
A and C with position vectors a and c. Since AC is given by c − a, the position
vector of a general point on the line is
r = a + λ(c − a).
7.7.2 Equation of a plane
The equation of a plane through a point A with position vector a and perpendicular to a unit position vector n̂ (see ﬁgure 7.13) is
(r − a) · n̂ = 0.
(7.41)
This follows since the vector joining A to a general point R with position vector
r is r − a; r will lie in the plane if this vector is perpendicular to the normal to
the plane. Rewriting (7.41) as r · n̂ = a · n̂, we see that the equation of the plane
may also be expressed in the form r · n̂ = d, or in component form as
lx + my + nz = d,
227
(7.42)
VECTOR ALGEBRA
n̂
A
a
R
d
r
O
Figure 7.13 The equation of the plane is (r − a) · n̂ = 0.
where the unit normal to the plane is n̂ = li + mj + nk and d = a · n̂ is the
perpendicular distance of the plane from the origin.
The equation of a plane containing points a, b and c is
r = a + λ(b − a) + µ(c − a).
This is apparent because starting from the point a in the plane, all other points
may be reached by moving a distance along each of two (non-parallel) directions
in the plane. Two such directions are given by b − a and c − a. It can be shown
that the equation of this plane may also be written in the more symmetrical form
r = αa + βb + γc,
where α + β + γ = 1.
Find the direction of the line of intersection of the two planes x + 3y − z = 5 and
2x − 2y + 4z = 3.
The two planes have normal vectors n1 = i + 3j − k and n2 = 2i − 2j + 4k. It is clear
that these are not parallel vectors and so the planes must intersect along some line. The
direction p of this line must be parallel to both planes and hence perpendicular to both
normals. Therefore
p = n1 × n2
= [(3)(4) − (−2)(−1)] i + [(−1)(2) − (1)(4)] j + [(1)(−2) − (3)(2)] k
= 10i − 6j − 8k. 7.7.3 Equation of a sphere
Clearly, the deﬁning property of a sphere is that all points on it are equidistant
from a ﬁxed point in space and that the common distance is equal to the radius
228
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