MTH 104: Vector Equations and Loci I
July 27, 2021
ii
0.1
Vector Equations
In keeping with the usual practice in number algebra, we denote the unknown vector by
~x. In solving for ~x, we must find all values of ~x for which the equation is true.
Example 1. Solve for ~x in the equation
~a × ~x = ~a × ~b, ~a 6= ~0.
Solution:
Method I: Since
~a × ~x = ~a × ~b
we have
~a × (~x − ~b) = ~0.
Hence ~a is parallel to (~x − ~b), that is,
~x − ~b = λ~a,
where λ is a parameter. Thus, the solution is not unique, the general solution being
~x = ~b + λ~a.
Solution:
Method II: Let p~ = ~a × ~x. Hence p~ is perpendicular to both ~a, ~x. Also since p~ = ~a × ~b, p~
is also perpendicular to both ~a, ~b. It follows that ~x, ~a, ~b are coplanar. Therefore we may
express ~x in the form
~x = λ~a + µ~b.
Multiplying vectorially by ~a we have
~a × ~x = ~a × (λ~a + µ~b)
= µ(~a × ~b).
Since ~a × ~x = ~a × ~b, it follows that µ = 1. Thus the general solution is
~x = ~b + λ~a.
Example 2. Find the vector ~x which satisfies the equations
~a × ~x = ~a × ~b, ~a · ~x = k.
Hence solve for ~x where ~a = (−1, 1, −2), ~b = (2, −1, 3), k = 3.
0.1. VECTOR EQUATIONS
iii
Solution: From Example I, the solution of
~a × ~x = ~a × ~b
is
~x = ~b + λ~a.
Multiply scalarly by ~a, we have
~a · ~x = ~a · ~b + λ~a2 .
Hence
k = ~a · ~b + λ~a2 .
The general solution is therefore
~x = ~b +
k − ~a · ~b
~a, (~a 6= ~0).
a2
Evaluating
3 − (−9) − 1, 1, −2
6
= (0, 1, −1).
~x = (2, −1, 3) +
Example 3. Solve the equation
p~x + (~x · ~c)~a = ~b, (p 6= 0).
Solution: Forming the scalar product with ~c, we have
p(~x · ~c) + (~x · ~c)(~a · ~c) = (~b · ~c),
that is,
(~x · ~c) p + ~a · ~c = ~b · ~c.
Therefore
~x · ~c =
~b · ~c
(p + ~a · ~c 6= 0).
p + ~a · ~c
Eliminating ~x · ~c from the given equation we have
p~x +
Hence
~x =
~b · ~c
~a = ~b.
p + ~a · ~c
~b
~b · ~c
−
~a,
p p(p + ~a · ~c)
(1)
iv
provided p + ~a · ~c 6= 0. Hence if p + ~a · ~c 6= 0, we have a unique solution.
If p + ~a · ~c = 0, we see from the given equation that if a solution ~x exists it is a linear sum
of ~a and ~b. Therefore we write
~x = λ~a + µ~b.
Also if p + ~a · ~c = 0, it follows from (1) that ~b · ~c = 0.
Substituting for ~x in the given equation, we obtain
p λ~a + µ~b + λ~a · ~c + µ~b · ~c ~a = ~b.
But
p = −~a · ~c and ~b · ~c = 0.
Therefore
−(~a · ~c)(λ~a + µ~b) + λ(~a · ~c)~a = ~b,
giving
−µ(~a · ~c)~b = ~b.
Therefore
1
~a · ~c
provided ~a · ~c 6= 0. Thus the given equation is satisfied by
µ=−
~x = λ~a −
~b
~a · ~c
for any value of λ. Thus the solution is not unique when p = −~a · ~c and ~b · ~c = 0.
Furthermore if ~a · ~c = 0, p = 0 and therefore ~b = ~0. The given equation then reduces
to
(~x · ~c)~a = ~0.
Therefore
~x · ~c = 0 (~a 6= ~0),
that is, ~x is any vector perpendicular to ~c.
EXERCISES
1. Given that ~a and ~b are perpendicular vectors, show that the solution of the vector
equation
~x × ~a = ~b
is
~x =
1
(~a × ~b) + λ~a,
a2
0.1. VECTOR EQUATIONS
v
where λ is a parameter.
(Hint: Since ~a, ~b, ~a × ~b are mutually perpendicular vectors, we can write
~x = λ~a + µ~b + ν(~a × ~b),
where λ, µ, ν are numbers. Choose unit vectors ~i and ~j such that ~a = a~i and ~b = b~j.)
0.1.1
Vector equations containing unknown vectors
Example 4. Solve for ~x in the equation
~a × ~x = ~a × ~b, ~a 6= ~0.
Solution: Multiplying
~a × ~x = ~a × ~b
vectorially by ~a, we have
(~a × ~x) × ~a = (~a × ~b) × ~a.
Expanding
~a2 ~x − (~a · ~x)~a = ~a2~b − (~a · ~b)~a.
Therefore
~a · ~x − ~a · ~b
~x = ~b +
~a.
~a2
The form of this suggests that ~x is expressible as the sum of vectors in the directions of ~b
and ~a, that is
~x = ~b + λ~a,
where λ is a parameter.
Example 5. Find the vector ~x that satisfies the equations
~a × ~x = ~a × ~b, ~a · ~x = k.
Solution: Proceeding as in the previous example, we obtain
~a · ~x − ~a · ~b
~x = ~b +
~a.
~a2
But
~a · ~x = k.
Therefore
k − ~a · ~b
~x = ~b +
~a.
~a2
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Example 6. Solve the equation
~x × ~a = ~b (~a · ~b = 0)
Solution: The general solution of ~x × ~a = ~b, (~a · ~b = 0), may be obtained by multiplying
the equation vectorially by ~a
~a × (~x × ~a) = ~a × ~b
Expanding ,
(~a · ~a)~x − (~a · ~x)~a = ~a × ~b.
Thus this suggests writing
~x = λ~a + µ(~a × ~b).
Substituting in ~x × ~a = ~b, we get
µ(~a × ~b) × ~a = ~b (λ~a × ~a = 0).
Expanding
~
~
µ (~a · ~a)b − (~a · b)~a = ~b.
Therefore
µ~a2~b = ~b, (~a · ~b = 0).
Hence
µ=
1
.
~a2
Thus the general solution of ~x × ~a = ~b, (~a · ~b = 0) is
~x = λ~a +
~a × ~b
,
~a2
where λ is a parameter.
When ~a = (3, −5, 4), ~b = (3, 1, −1), we have
~x = λ(3, −5, 4) +
(1, 15, 18)
,
50
that is,
1
(150λ + 1, −250λ + 15, 200λ + 18)
50
Again if ~a · ~x = −17, we have a unique solution. In this case
~x =
(3, −5, 4) ·
1
150λ + 1, −250λ + 15, 200λ + 18 = −17.
50
0.1. VECTOR EQUATIONS
vii
Evaluating the scalar product,
1
450λ + 3 + 1250λ − 75 + 800λ + 72 = −17.
50
Solving
λ=−
17
.
50
Therefore
1
− 50, 100, −50 = − 1, 2, −1 .
50
Example 7. Solve the equation
~x =
~x × ~a = ~b, ~x · ~c = p, (~a · ~b = 0.)
Solution: Forming a vector triple product with ~c, we have
(~x × ~a) × ~c = ~b × ~c
Expanding
(~x · ~c)~a − (~c · ~a)~x = ~b × ~c
. Substituting ~x · ~c = p, this becomes
p~a − (~a · ~c)~x = ~b × ~c.
Therefore
p~a + (~c × ~b)
~a · ~c
provided ~a · ~c 6= 0. Thus the solution is unique.
If however ~a · ~c = 0, then the general solution ~x × ~a = ~b, (~a · ~b = 0) takes the form
~x =
~x = λ~a +
~a × ~b
.
~a2
Forming the scalar product with ~c, we have
~x · ~c = λ~a · ~c +
(~a × ~b) · ~c
.
~a2
When ~a · ~c = 0, we have the condition
p=
(~a × ~b) · ~c
~a2
(2)
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for (2) to satisfy the equation ~x · ~c = p. Thus
~a × ~b
.
~a2
will satisfy the second equation ~x · ~c = p for any value of λ and in this case the solution is
not unique.
~x = λ~a +
In particular, the solution of
~x × ~a = ~b, ~x · ~a = −17
where ~a = (3, −5, 4), ~b = (3, 1, −1) is given by
p~a + (~a × ~b)
~a · ~a
−17(3, −5, 4) + (1, 15, 18)
=
50
(−50, 100, −50)
=
= (−1, 2, −1).
50
Example 8. Solve the simultaneous equations
~x =
p~x + q~y = ~a, ~x × ~y = ~b, (~a · ~b = 0).
Solution: Forming the vector product with ~x, the first equation becomes
(p~x + q~y ) × ~x = ~a × ~x
Expanding
q(~y × ~x) = ~a × ~x.
Using the second equation, we have
q~b = ~x × ~a.
Since ~a · ~b = 0, the general solution of this equation is given by
~a × q~b
~a2
~a × ~b
= λ~a + q 2 ,
~a
~x = λ~a +
where λ is a variable scalar.
Substituting in p~x + q~y = ~a, we obtain
~y =
Thus the solution is not unique.
1 − pλ
~a × ~b
~a − p 2 .
q
~a
0.1. VECTOR EQUATIONS
0.1.2
ix
Vector equations containing unknown scalars
Example 9. Find the scalars x, y, z that satisfy the equation
~
~a = x~b + y~c + z d,
where ~a = (−3, 2, 6), ~b = (−1, 1, 2), ~c = (3, 1, −1), d~ = (2, 1, 1).
Solution:
~
~a = x~b + y~c + z d.
Form a scalar triple product by multiplying scalarly by ~c×d~ each term of the given equation.
Thus we obtain
~ = x~b · (~c × d)
~ + y~c · (~c × d)
~ + z d~ · (~c × d).
~
~a · (~c × d)
~ = d~ · (~c × d)
~ = 0, we have
Since ~c · (~c × d)
x=
~
~a · (~c × d)
,
~b · (~c × d)
~
~ 6= 0.
provided ~b · (~c × d)
Similarly,
y=
~a · (d~ × ~b)
~a · (~b × ~c)
, z=
.
~c · (d~ × ~b)
d~ · (~b × ~c)
Thus the general solution of the equation
~a = x~b + y~c + z d~
is
x=
~
~ ~b]
[~a, ~c, d]
[~a, d,
[~a, ~b, ~c]
, y=
, z=
,
~
~
~
[~b, ~c, d]
[~b, ~c, d]
[~b, ~c, d]
~ =
provided [~b, ~c, d]
6 0.
Evaluating
−1 1 2
~b · (~c × d)
~ = 3 1 −1 = −(1 + 1) − (3 + 2) + 2(3 − 2) = −5.
2 1 1
−3 2 6
~
~a · (~c × d) = 3 1 −1 = −3(1 + 1) − 2(3 + 2) + 6(3 − 2) = −10.
2 1 1
x
−3 2 6
~a · (d~ × ~b) = 2 1 1 = −3(2 − 1) − 2(4 + 1) + 6(2 + 1) = 5.
−1 1 2
−3 2 6
~a · (~b × ~c) = −1 1 2 = −3(−1 − 2) − 2(1 − 6) + 6(−1 − 3) = −5.
3 1 −1
Hence
x = 2, y = −1, z = 1.
Example 10. Find the scalars x, y that satisfy the equation
~a × ~b = x~b + y~c,
where ~a = (3, 1, 1), ~b = (2, 0, −2), ~c = (1, −2, 0).
Solution:
Multiplying scalarly by ~c × ~a, we have
(~a × ~b) · (~c × ~a) = x~b · (~c × ~a) + y~c · (~c × ~a)
= x~b · (~c × ~a).
Therefore, provided = ~a · (~b × ~c) 6= 0,
x=
(~a × ~b) · (~c × ~a)
~a · (~b × ~c)
Similarly, multiplying scalarly by ~a × ~b, we obtain
y=
(~a × ~b) · (~a × ~b)
.
~a · (~b × ~c)
Evaluating,
3 1
1
~a · (~b × ~c) = 2 0 −2 = 3(0 − 4) − (0 + 2) + (−4 − 0) = −18.
1 −2 0
(~a × ~b) · (~c × ~a) = (−2, 8, −2) · (−2, −1, 7) = −18.
(~a × ~b) · (~a × ~b) = (−2, 8, −2) · (−2, 8, −2) = 72.
Hence
x = 1,
y = −4.
0.1. VECTOR EQUATIONS
xi
EXERCISES
~ where ~a = 2~i+3~j −~k, ~b = 5~i+~j +2~k, ~c = ~i−2~j −4~k,
1. Find p, q such that ~a ×~b = p~c +q d,
~
~
~
~
d = i − j − k.
2. Find ~x which satisfies the equations
~x × ~a = ~b, ~x · ~a = 5,
where ~a = (2, 2, −3), ~b = (1, 17, 12).
3. A vector ~x satisfies the equations
~x × ~b = ~c × ~b, and ~x · ~a = 0.
Prove that ~x = ~c − (~a · ~c/~a · ~b)~b provided that ~a · ~b 6= 0.
4. If ~a, ~b are given vectors, ~b being perpendicular to ~a, and if k is a given scalar, show
that the solution of the equations
~a · ~x = k and ~a × ~x = ~b
for an unknown vector ~x is unique, and find it.
5. Given that ~x satisfies the equation
~x = p~a + ~x × ~a,
where p is a non-zero scalar, find ~x.
6. Find the vector ~x and scalar λ which satisfy the equations
~a × ~x = ~b + λ~a, ~a · ~x = 2,
where ~a = ~i + 2~j − ~k, ~b = 2~i − ~j + ~k
7. Find the vectors ~x and ~y that satisfy the simultaneous equations
~
~x + (~a × ~y ) = b,
~y + (~a × ~x) = c~,
where ~a, ~b, and ~c are given vectors.
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0.2
The Straight Line
Consider a variable point which moves under certain specified conditions. The Cartesian
form of the locus of the point is the equation connecting its coordinates, namely x, y if the
locus is two-dimensional or x, y, z if the locus is three-dimensional. The vector form of
the locus of the point is the equation connecting its position vector relative to some origin
with other given vector
If the point P (x, y, z) is on the locus and ~r is its position vector relative to the origin O,
then
~r = x~i + y~j + z~k.
Furthermore, if x, y, z are expressed in terms of a parameter u, that is
x = f (u), y = φ(u), z = ϕ(u),
we may write
~r = f (u)~i + φ(u)~j + ϕ(u)~k.
0.2.1
Direction-vector of a line
The direction of a line may be specified by referring to any other line parallel to it. In
particular, if we assign a direction and sense in describing a line, then these may be specified
by a vector.
In two dimensions the direction of a line is given by its gradient, which is defined as the
tangent of the angle which the line makes with the positive x-axis, the angle being measured
in the usual way from the positive x-axis
0.2.2
Vector equation of a straight line through a given point and in a
given direction
Let A(~r1 ) be a fixed point and I~ 6= 0 be a fixed vector. We require the equation of a
straight line L which passes through A and which is parallel to I~ (Fig. 1).
−→
~ we have
Let P (~r) be any variable point on the line L. Since AP is parallel to I,
−→
~
AP = tI,
where t is a number dependent on the position of P. Therefore
~
~r − ~r1 = tI,
0.2. THE STRAIGHT LINE
xiii
Figure 1:
xiv
that is,
~
~r = ~r1 + tI.
(3)
This is a parametric equation of the required line L since for any point P on L there is a
unique value of t, and for any value of t there is a unique point P on L.
If
~r = (x, y, z), ~r1 = (x1 , y1 , z1 ), I~ = (a, b, c),
equation (3) is
(x, y, z) = (x1 , y1 , z1 ) + t(a, b, c),
from which we obtain
x = x1 + ta, y = y1 + tb, z = z1 + tc.
(4)
This is a parametric form of the equations of a straight line through the point (x, y, z) and
having direction ratios (a, b, c). From (4), we deduce that
x − x1
y − y1
z − z1
=
=
(a 6= 0, b 6= 0, c 6= 0),
a
b
c
(5)
which is the standard Cartesian form of the equations of a straight line through the point
(x1 , y1 , z1 ) and having the direction ratios (a, b, c).
Example 11. Find the equations of a line through the point (1, 2, −3) and parallel to
3~i − 2~j + 4~k. Verify that the point (−2, 4, −7) is on the line.
Solution: The vector equation of the straight line is
~r = (x, y, z) = (1, 2, −3) + t(3, −2, 4).
Hence the standard Cartesian form of the equations of the straight line is
x−1
y−2
z+3
=
=
.
3
−2
4
Substituting x = −2 in these equations, we find y = 4, z = −7. Therefore the point
(−2, 4, −7) is on the line.
Example 12. Find the equations of the straight line joining the points
A(2, 1, −1) and B(0, 6, 3).
Solution: The direction-vector of the line is given by
−−→
AB = (−2, 5, 4).
0.2. THE STRAIGHT LINE
xv
The vector equation of the line is
~r = (x, y, z) = (2, 1, −1) + t(−2, 5, 4).
Hence the standard Cartesian form of the equations of the line is
x−2
y−1
z+1
=
=
.
−2
5
4
(6)
The vector equation of the line can also be written as
~r = (x, y, z) = (0, 6, 3) + t(−2, 5, 4),
and the Cartesian form is therefore
y−6
z−3
x
=
=
.
−2
5
4
(7)
Thus the equations of a given line are not unique.
Note that (6) may be put into the form of equation (7) by subtracting 1 from each of the
three expressions of (6).
Example 13. Find the perpendicular distance of the point A(5, 4, 1) from the line
x−6
y + 15
z − 14
=
=
5
1
8
and find the co-ordinates of the foot N of the perpendicular.
Solution: From the equation, the point (6, −15, 14) is on the line (Fig. 2).
The direction-vector I~ of the line is given by
I~ = (5, 1, 8).
−−→
The projection of AB on I~ is BN and this is given by
−−→ I~
BN = AB ·
~
|I|
= (1, −19, 13) ·
(5, 1, 8) √
√
= 90.
90
From Pythagoras’ theorem,
AN 2 = AB 2 − BN 2
= (1 + 361 + 169) = 441.
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Figure 2:
Therefore
AN = 21,
that is, the length of the perpendicular is 21 units.
The equation of a straight line is given by
~
~r = ~r1 + tI.
−−→
−−→
So, taking ON = ~r, OB = ~r1 , we have
−−→
ON = (6, −15, 14) + t(5, 1, 8)
= (6 + 5t, −15 + t, 14 + 8t),
where t is a parameter to be determined.
Now
−−→ −−→ −→
AN = ON − OA
= (1 + 5t, −19 + t, 13 + 8t).
−−→
~ we have
Since AN is perpendicular to I,
−−→ ~
AN · I = 0.
Therefore
5(1 + 5t) + (−19 + t) + 8(13 + 8t) = 0,
0.2. THE STRAIGHT LINE
xvii
giving
t = −1.
Hence
−−→
ON = (1, −16, 6),
that is, N is the point (1, −16, 6).
0.2.3
Vector equation of a straight line through two given points
Suppose that A(r~1 ), B(r~2 ) are the two given points. The direction-vector of the line joining
−−→
A, B is given by AB where
−−→
AB = r~2 − r~1 .
The equation of line passing through A(r~1 ) and having direction-vector is I~
~
~r = ~r1 + tI.
−−→
Therefore the equation of the line passing through A(r~1 ) and having direction-vector AB
is
~r = ~r1 + t(~r2 − ~r1 ),
that is,
~r = (1 − t)~r1 + t~r2 .
(8)
If
~r = (x, y, z), ~r1 = (x1 , y1 , z1 ), ~r2 = (x2 , y2 , z2 ),
we have the equivalent parametric form
x = x1 + (x2 − x1 )t, y = y1 + (y2 − y1 )t, z = z1 + (z2 − z1 )t,
or the equivalent Cartesian form
y − y1
z − z1
x − x1
=
=
.
x2 − x1
y2 − y1
z2 − z1
Note: A necessary and sufficient condition for three points to be collinear is that the algebraic sum of the coefficients of their position vectors vanishes.
Example 14. Find a vector equation of the line passing through the points P (1, −2, 1) and
Q(3, 1, 1). What are the corresponding parametric equations and the equations in symmetric
(Cartesian) form?
xviii
Solution:
~r = ~r1 + t(~r2 − ~r1 ),
= (1, −2, 1) + t[(3, 1, 1) − (1, −2, 1)]
= (1, −2, 1) + t(2, 3, 0)
= (1 + 2t)~i + (−2 + 3t)~j + ~k
By equating the corresponding components of the vector equation, we obtain the parametric
equations
x = 1 + 2t, y = −2 + 3t, z = 1.
By eliminating the parameter t, we obtain the equation in symmetric form
x−1
y+2
=
, z = 1.
2
3
The line lies on the plane z = 1.
0.2.4
The angle between two straight lines
Definition: The angle between two straight lines is defined as the angle between their
direction-vectors.
Let I~1 , I~2 be the direction-vectors of two lines. Then the angle θ between them is given
by
I~1 · I~2
cos θ =
.
|I~1 ||I~2 |
If the lines are parallel we have
I~1 = λI~2 ,
where λ is a constant, and if the lines are perpendicular we have
I~1 · I~2 = 0.
Example 15. Find the point of intersection of the two lines defined by the vector equations
~r1 = ~i − ~j + 2~k + t(~i + ~j + ~k)
~r2 = 3~i − 3~j + ~k + s(−~i + 3~j + 2~k)
and determine the angle between them.
0.2. THE STRAIGHT LINE
xix
Solution: We need to find a value of the parameter t and a value of the parameter s such
that
~r1 = ~r2 .
If no such values exist, then the lines do not intersect.
Equating the corresponding components of the equation ~r1 = ~r2 , we find
1 + t = 3 − s, −1 + t = −3 + 3s, 2 + t = 1 + 2s.
Solving for t and s, we get t = 1 and s = 1.
Then the lines intersect at the point (2, 0, 3).
From the vector equations of the lines,
I~1 = ~i + ~j + ~k and I~1 = −~i + 3~j + 2~k
cos θ =
I~1 · I~2
4
4
=√ √ =√ .
~
~
3 14
42
|I1 ||I2 |
Hence the angle between the lines is
√
θ = cos−1 (4/ 42) = 51.9◦
Example 16. Find the angle between the two lines:
~r1 = ~i − 2~j + 3~k + t(2~i − 3~j + 6~k)
(i)
~r2 = 2~i−7~j+10~k+s(~i+2~j+2~k).
(ii)
Solution: The angle between the lines depend only upon their directions
2
3 6
Line (i) has direction cosines
7, −7, 7.
1 2 2
Line (ii) has direction cosines
3, 3, 3.
The angle θ between the lines is given by
cos θ =
2 1 3 2 6 2 +
7 3
7 3
7 3
8
θ = arc cos
.
21
+
−
Note: The angle between two lines with direction cosines l1 , m1 , n1 and l2 , m2 , n2 is given
by
cos θ = l1 l2 + m1 m2 + n1 n2 .
xx
Example 17. Two lines have equations
x−4
y−3
z−7
=
=
2
1
2
and
x+1
y
z+1
= =
.
3
2
6
Show that the lines intersect.
Solution: Introducing a parameter t into the equation of the first line, we get
x−4
y−3
z−7
=
=
=t
2
1
2
so that
x = 2t + 4, y = t + 3, z = 2t + 7.
Similarly, the equations of the second line become
x = 3s − 1, y = 2s, z = 6s − 1.
If the two lines intersect it will be at a point P (x, y, z) whose co-ordinates satisfy both set
of equations.
Equating x co-ordinates
2t + 4 = 3s − 1
Equating y co-ordinates
t + 3 = 2s
Hence
s=1
and
t = −1.
Using these values for s and t, the co-ordinates become
First line
2t + 7 = 5
Second line
6s − 1 = 5.
So the lines intersect at a point where s = 1
giving
x = 2, y = 2, z = 5.
Example 18. (a) Find the parametric and symmetric equations of the line that passes
through the points A(2, 4, −3) and B(3, −1, 1)
(b) At what point does this line intersect the xy-plane?
0.2. THE STRAIGHT LINE
xxi
Solution: (a) We are not explicitly given a vector parallel to the line, but observe that
−−→
the vector ~v with representation AB is parallel to the line and
~v = 3 − 2, −1 − 4, 1 − (−3) = 1, −5, 4 .
Thus direction numbers are a = 1, b = −5, and c = 4. Taking the point (2, 4, −3), we have
the following parametric equations:
x = 2 + t, y = 4 − 5t, x = −3 + 4t
and symmetric equations are
x−2
y−4
z+3
=
=
.
1
−5
4
(b) The line intersects the xy-plane when z = 0, so we put z = 0 in the symmetric equation
to obtain
x−2
y−4
3
=
=
1
−5
4
This gives x =
11
4
1
and y = 41 , so the line intersects the xy-plane at the point ( 11
4 , 4 , 0).
EXERCISES
1. Find (a) the vector equation, (b) the Cartesian equations of a straight line passing
through the point A(2, −1, 3) and parallel to the line through the points B(3, 2, −1),
C(−1, 1, 2). Verify that the point (−2, −2, 6) is on the required line.
2. Obtain the co-ordinates of the point of intersection of the lines
x−2
y+5
z
=
= ,
5
−9
7
x+1=y−6=
z+3
.
2
3. Find the equation of a straight line joining the points A(2, −3, 4), B(1, 6, −1).
Obtain the point of intersection of this line with the xOy plane.
4. Find the equations of a line through the point A(6, 2, −4) and having a direction
ratios A(2, −1, 3). If this line meets the xOy plane in the point B, find (i) the length
of AB, (ii) the angle AB makes with the yOz plane.
5. Find the co-ordinates of the two points on the line
y−1
z−2
x+3
=
=
2
3
−1
which are
√
33 units of length from the point (2, 3, 4).
xxii
6. Find the angle between each of the following pairs of lines
(a)
r~1 = 3~i + 2~j − 4~k + λ(~i + 2~j + 2~k)
r~2 = 5~j − 2~k + µ(3~i + 2~j + 6~k)
(b) A line with direction ratios 4 : 4 : 2
A line joining (3, 1, 4) to (7, 2, 12).
(c)
x+4
y−1
z+3
=
=
3
5
4
y−4
z−5
x+1
=
=
1
1
2
7. Two lines have equations
r~1 = 2~i + 9~j + 13~k + λ(~i + 2~j + 3~k)
r~2 = a~i + 7~j − 2~k + µ(−~i + 2~j − 3~k)
If they intersect, find the value of a and the position vector of the point of intersection.
8. Verify that the point (6, −5, −1) lies on the line
~r = 2(1 + t)~i − (1 + 2t)~j − (3 − t)~k.
If Q is the foot of the perpendicular from the point P (4, 7, −9) to the line, find
(i) the co-ordinates of Q,
(ii) the length of P Q,
(iii) the equation of the perpendicular.
0.2.5
The shortest distance between two skew lines
Definition 1. Non-intersecting, non-parallel straight lines in space are known as skew
lines.
A property of skew lines is that they have a common perpendicular, the length of which is
the least distance between the lines.
The shortest distance between two skew lines is the length of the common perpendicular
P1 P2 of the lines. Let L1 , L2 be two lines, I~1 , I~2 their direction-vectos and A1 (r~1 ), A2 (r~2 )
be two points on L1 , L2 respectively as shown in Figure 4. The common perpendicular has
0.2. THE STRAIGHT LINE
xxiii
Figure 3:
Figure 4:
xxiv
direction-vector I1 × I2 . The shortest distance is given by the projection of A1 A2 on the
unit vector in the direction of P1 P2 , that is, by
A1 A2 ·
I1 × I2
I1 × I2
= (r~2 − r~1 ) ·
.
|I1 × I2 |
|I1 × I2 |
Denoting the numerical value of this distance by p we have
p=
|(r~1 − r~2 ) · (I1 × I2 )|
.
|I1 × I2 |
Example 19. Show that the lines L1 and L2 with parametric equations
y = −2 + 3t,
x = 1 + t,
x = 2s,
y = 3 + s,
z =4−t
z = −3 + 4s
are skew lines.
Solution: The lines are not parallel because the corresponding vectors 1, 3, −1 and
2, 1, 4 are not parallel. (Their components are not proportional.) If L1 and L2 had a
point of intersection, there would be values of t and s such that
1 + t = 2s,
−2 + 3t = 3 + s,
4 − t = −3 + 4s
8
But if we solve the first two equations, we get t = 11
5 and s = 5 , and these values don’t
satisfy the third equation. Therefore there are no values of t and s that satisfy the three
equations, so L1 and L2 do not intersect. Thus L1 and L2 are skew lines.
Example 20. Find
(i) the shortest distance,
(ii) the co-ordinates of the feet of the common perpendicular,
(iii) the equation of the common perpendicular, for the skew lines
L1 ≡
y−5
z−4
x+7
=
=
−8
3
1
and
L2 ≡
x+4
y
z − 19
= =
.
4
3
−2
Solution: If I~1 , I~2 are the direction-vectors of L1 , L2 respectively, we have
I~1 = (−8, 3, 1), I~2 = (4, 3, −2).
0.2. THE STRAIGHT LINE
xxv
The points (−7, 5, 4), (−4, 0, 19) are on L1 , L2 and the position vectors of any points
~ 1, L
~ 2 respectively are
P1 (r~1 ), P2 (r~2 ) on L
~r1 = (−7, 5, 4) + s(−8, 3, 1) = (−7 − 8s, 5 + 3s, 4 + s)
~r2 = (−4, 0, 19) + t(4, 3, −2) = (−4 + 4t, 3t, 19 − 2t),
where s, t are parameters.
Therefore
−−−→
P1 P2 = (3 + 8s + 4t, −5 − 3s + 3t, 15 − s − 2t)
We require P1 P2 to be perpendicular to both L1 , L2 . This is so when
−−−→ ~
−−−→
P1 P2 · I1 = 0 and P1 P2 · I~2 = 0.
Evaluating the scalar products we obtain
−24 − 74s − 25t = 0 − 33 + 25s + 29t = 0.
The solution of these equations is
s = −1,
t = 2.
By substituting these values, we obtain the co-ordinates of the feet of the perpendicular,
and they are
P1 (1, 2, 3), P2 (4, 6, 15).
The shortest distance is given by
p
−−−→
|P1 P2 | = 32 + 42 + 122 = 13
The common perpendicular passes through the point (1, 2, 3) and has direction-vector
(3, 4, 12). Hence the equations of the common perpendicular are
x−1
y−2
z−3
=
=
.
3
4
12
Note: The shortest distance can also be determined by using the the formula
p=
|(r~1 − r~2 ) · (I1 × I2 )|
,
|I1 × I2 |
where
Hence
Therefore
r~1 = (−7, 5, 4),
I~1 = (−8, 3, 1),
r~1 − r~2 = (−3, 5, −15), I~1 × I~2
p=
r~2 = (−4, 0, 9),
I~2 = (4, 3, −2).
= (−9, −12, −36).
|(−3, 5, −15) · (−9, −12, −36)|
507
=
= 13.
|(−9, −12, −36)|
39
xxvi
EXERCISES
1. Find the length and the equations of the shortest distance between the lines
x−6
y+4
z−2
=
=
,
2
5
1
x+1
y−9
z−5
=
=
.
−4
5
7
2. Which of the following four lines are parallel? Are any of them identical?
L1 : x = 1 + 6t, y = 1 − 3t, z = 12t + 5,
L2 : x = 1 + 2t, y = t, z = 1 + 4t,
L3 : 2x − 2 = 4 − 4y = z + 1,
L4 : ~r = 3, 1, 5 + t 4, 2, 8 .
3. Find the distance between the skew lines with parametric equations
x = 1 + t, y = 1 + 6t, z = 2t, and x = 1 + 2s, y = 5 + 15s, z = −2 + 6s.
4. Let L1 be the line through the origin and the point (2, 0, −1). Let L2 be the line
through the points (1, −1, 1) and (4, 1, 3). Find the distance between L1 and L2 .
5. Determine whether the lines L1 and L2 are parallel, skew, or intersecting. If they
intersect, find the point of intersection.
(i)
L1 : x = 3 + 2t, y = 4 − t, z = 1 + 3t
L2 : x = 1 + 4s, y = 3 − 2s, z = 4 + 5s
(ii)
L1 : x = 5 − 12t, y = 3 + 9t, z = 1 − 3t
L2 : x = 3 + 8s, y = −6s, z = 7 + 2s
(iii)
y−3
z−1
x−2
=
=
1
−2
−3
x−3
y+4
z−2
L2 =
=
=
1
3
−7
L1 =
0.3. THE PLANE
xxvii
(iv)
x
y−1
z−2
=
=
1
−1
3
x−2
y−3
z
L2 =
=
=
2
−2
7
L1 =
6. Two skew lines have vector equations
r~1 = 3~j + s(2~i − 5~j + 3~k) and r~2 = 2~i + ~k + t(~j + ~k).
Show that 4~i + ~j − ~k is a vector which is perpendicular to both lines and hence find
the shortest distance between the lines.
7. Find the shortest distance between two lines L1 and L2 if the vector equations of L1
and L2 are
r~1 = 7~i + ~j − 2~k + s(3~i − 5~j + 2~k) and r~2 = ~i + 9~j + 5~k + t(7~i + ~j − 8~k).
0.3
0.3.1
The Plane
Vector equation of a plane through a given point and parallel to
two given lines
Let A(r~1 ) be a given point, P (~r) any point in the plane and I~1 , I~2 the direction-vectors of
−→
the two given lines (Fig. 5). Since AP is coplanar with I~1 and I~2 , we may write
−→
AP = sI~1 + tI~2 ,
where s, t are the numbers depending on the position of P as it moves over the plane.
Therefore
~r − ~r1 = sI~1 + tI~2 ,
i.e.
~r = ~r1 + sI~1 + tI~2 .
This is the vector equation of the required plane in terms of the parameters s and t.
Example 21. Find the equation of a plane containing the line
x−1
y+4
z−1
=
=
3
4
1
and which is parallel to the line
x
y
z
= = .
4
3
12
xxviii
Figure 5:
Solution: The point (1, −4, 1) is in the plane. The plane is parallel to the directionvectors of the lines , that is, to the vectors (3, 4, 1) and (4, 3, 12). The vector equation of
the plane is therefore
~r = (x, y, z) = (1, −4, 1) + s(3, 4, 1) + t(4, 3, 12).
Hence
x = 1 + 3s + 4t, y = −4 + 4s + 3t, z = 1 + s + 12t.
On eliminating s and t, we obtain
45x − 32y − 7z = 166.
0.3.2
Vector equation of a plane containing three given non-collinear
points
Let A(r~1 ), B(r~2 ), C(r~3 ) be the given points on the required plane and P (~r) be any point
on the plane (Fig. 6).
−→ −−→ −→
Since the vectors AP , AB, AC are coplanar, we may write
−→
−−→
−→
AP = sAB + tAC,
where s, t are numbers depending on the position of P, as P moves across the plane.
Therefore
~r − r~1 = s(r~2 − r~1 ) + t(r~3 − r~1 ).
0.3. THE PLANE
xxix
Figure 6:
The required equation of the plane is
~r = r~1 + s(r~2 − r~1 ) + t(r~3 − r~1 )
or
~r = (1 − s − t)r~1 + sr~2 + tr~3 .
Note: A necessary condition for four points to be coplanar is that the algebraic sum of the
coefficients of their position vectors vanishes.
Example 22. A plane passes through the points A(1, 2, 3), B(−1, 2, 0), and C(2, −1, −1).
Find its equation.
Solution: The vector equation of the plane is
~r = (x, y, z) = (1 − s − t)(1, 2, 3) + s(−1, 2, 0) + t(2, −1, −1),
where s and t are parameters.
Hence
x = 1 − s − t − s + 2t = 1 − 2s + t,
y = 2 − 2s − 2t + 2s − t = 2 − 3t,
z = 3 − 3s − 3t − t = 3 − 3s − 4t.
On eliminating s and t, we obtain
9x + 11y − 6z = 13.
xxx
Figure 7:
0.3.3
Normal-vector of a plane
Consider a plane π and let A be a point in the plane. Let ~n be a vector such that ~n is
−→
perpendicular to the vector AP for all points P in the plane. We say that the vector ~n is
perpendicular to the plane π, and we shall refer to the vector ~n as the normal vector of
the plane.
0.3.4
Vector equation of a plane containing a given point and normal to
a given line
Let ~n be the normal-vector of a plane, A a fixed point and P any point on the plane.
−→
−−→
Taking O aS the origin, let OA = r~1 , OP = ~r (Fig. 7).
−→
Since AP is perpendicular to the normal-vector of the plane we have
−→
AP · ~n = 0.
0.3. THE PLANE
xxxi
But
−→
AP = ~r − r~1 .
Therefore
(~r − r~1 ) · ~n = 0.
(9)
This is the vector equation of a plane containing the point whose position vector is ~r and
having ~n as the normal-vector. In terms of Cartesian co-ordinates, if
~r = (x, y, z), r~1 = (x1 , y1 , z1 ), ~n = (a, b, c),
we have
a(x − x1 ) + b(y − y1 ) + c(z − z1 ) = 0
(10)
which is the equation of a plane containing the point (x1 , y1 , z1 ) and with direction ratios
(a, b, c).
If the normal-vector is the unit vector n̂, then the projection of r~1 on n̂, namely r~1 · n̂, is
numerically equal to the perpendicular distance of the plane from the origin. Since this
distance is constant, we may write (9) as
~r · n̂ = p,
(11)
where p is a constant, numerically equal to the perpendicular distance of the plane from
the origin. More generally, we may write
~r · ~n = k,
(12)
lx + my + nz = p,
(13)
where k is a constant.
The Cartesian equivalent of (11) is
where l, m, n are the direction cosines of n̂, that is, of the plane. Equation (13) is usually
known as the perpendicular form of the equation of the plane.
The Cartesian equivalent of (12) is
ax + by + cz = k,
(14)
where a, b, c are the direction ratios of the plane.
Note: Of the various forms of the equations of a plane, the form ~n ·~r = k, i.e. ax+by+cz =
k, is the most useful.
Example 23. Find the equation of a plane containing the point (1, −1, 2) and perpendicular to the vector 3~i − 2~j + ~k.
xxxii
Solution:
Method I: The point r~1 = (1, −1, 2) lies on the plane. The normal-vector of the plane is
~n = (3, −2, 1).
The equation of the plane is given by
~n · (~r − r~1 ) = 0,
that is,
~n · ~r = ~n · r~1 .
Therefore
(3, −2, 1) · (x, y, z) = (3, −2, 1) · (1, −1, 2).
Hence the required equation is
3x − 2y + z = 7.
Method II: The normal-vector of the plane is ~n = (3, −2, 1). The equation of a plane is
given by
~n · ~r = k.
Since the point (1, −1, 2) satisfies this equation, we have
~n · (1, −1, 2) = k.
Therefore (3, −2, 1) · (x, y, z) = (3, −2, 1) · (1, −1, 2).
Hence the required equation is
3x − 2y + z = 7.
Example 24. Find the co-ordinates of the foot of the perpendicular from the point (3, −1, 2)
to the plane 5x − 6y − 30z = 23.
Solution: The direction-vector of the perpendicular is given by the normal-vector of the
plane, that is, by (5, −6, −30). Since the point (3, −1, 2) is on this perpendicular, the
vector equation of the perpendicular is
~r = (3, −1, 2) + t(5, −6, −30),
i.e.
~r = (3 + 5t, −1 − 6t, 2 − 30t).
The equation of the plane may be written as
(5, −6, −30) · ~r = 23
from which we have
5(3 + 5t) − 6(−1 − 6t) − 30(2 − 30t) = 23.
0.3. THE PLANE
On solving we obtain
Hence the point
xxxiii
t=
2
31 .
10
12
60
3 + , −1 − , 2 −
31
31
31
!
=
!
103
43 2
.
,− ,
31
31 31
43 2
Therefore 103
,
−
,
31
31 31 is the foot of the perpendicular.
Example 25. A plane passes through the points A(1, 2, 3), B(−1, 2, 0), and C(2, −1, −1).
Find its equation.
−−→
−→
Solution: The normal-vector is perpendicular to both AB and AC. Therefore the vector
−−→
−→
product of AB and AC may be taken as the normal-vector ~n. Thus we have
~n = (−2, 0, −3) × (1, −3, −4) = (−9, −11, 6).
The equation of the plane is then given by
~n · ~r = k, i.e.
− 9x − 11y + 6z = k.
Since A(1, 2, 3) is a point on the plane, we have
−9 − 22 + 18 = k,
k = −13. Thus the equation of the plane
i.e.
is
9x + 11y − 6z = 13.
Example 26. Find an equation of the plane that passes through the points P (1, 3, 2),
P (3, −1, 6), and R(5, 2, 0).
−−→
−→
Solution: The vectors ~a and ~b corresponding to P Q and P R are
~a = 2, −4, 4
~b = 4, −1, −2 .
Since both ~a and ~b lie in the plane, their cross product ~a × ~b is orthogonal to the plane and
can be taken as the normal-vector. Thus
~n = ~a × ~b = 12~i + 20~j + 14~k.
With the point P (1, 3, 2) and the normal vector ~n, an equation of the plane is
12(x − 1) + 20(y − 3) + 14(z − 2) = 0
or
6x + 10y + 7z = 50.
xxxiv
Example 27. Find the point at which the line with parametric equations x = 2 + 3t,
y = −4t, z = 5 + t intersects the plane 4x + 5y − 2z = 18.
Solution: We substitute the expressions for x, y, and z from the parametric equations
into the equation of the plane:
4(2 + 3t) + 5(−4t) − 2(5 + t) = 18.
This simplifies to −10t = 20, so t = −2. Therefore the point of intersection occurs when
the parameter value is t = −2. Then x = 2 + 3(−2) = −4, y = −4(−2) = 8, z = 5 − 2 = 3
and so the point of intersection is (−4, 8, 3).
Note: Two planes are parallel if their normal vectors are parallel. For instance, the
planes x + 2y − 3z = 4 and 2x + 4y − 6z = 3 are parallel because their normal vectors are
n~1 = 1, 2, −3 and n~2 = 2, 4, −6 and n~2 = 2n~1 . If two planes are not parallel, then
they intersect in a straight line and the angle between the two planes is defined as the acute
angle between their normal vectors.
Example 28. Find the distance between the parallel planes 10x + 2y − 2z = 5 and 5x +
y − z = 1.
Solution: First we note that the planes are parallel because their normal vectors 10, 2, −2
and 5, 1, −1 are parallel. To find the distance D between the planes, we choose any point
on one plane and calculate its distance to the other plane. In particular, if we put y = z = 0
in the equation of the first plane, we get 10x = 5 and so ( 21 , 0, 0) is a point in this plane.
By using the formula
ax1 + by1 + cz1 + d
√
D=
,
a2 + b2 + c2
the distance between ( 12 , 0, 0) and the plane 5x + y − z = 1 is
√
3
5( 21 ) + 1(0) − 1(0) − 1
3
2
p
.
D=
= √ =
6
3 3
52 + 12 + (−1)2
√
So the distance between the planes is
3
6 .
EXERCISES
1. Find the equation of a plane which contains the lines
x−1
y−2
z−3
=
=
,
−1
2
4
x+1
y−4
z−2
=
=
.
2
−2
1
0.3. THE PLANE
xxxv
2. A plane passes through the points A(1, 1, −2), B(2, 5, 3), and C(−2, −1, 3). Find
its equation.
3. Find the equation of a plane containing the point (−2, −5, 2) and the line
x−7
y
z−1
= =
.
1
3
−5
4. Find the equation of a plane which passes through the point (2, 1, −3) and is normal
to the vector ~i − 2~j − 4~k.
5. Find the equation of a plane containing the point (1, 2, 3) and which is normal to
the line
x+2
y+7
z+5
=
=
.
2
3
−2
Find the point of intersection of the line and the plane.
6. Find the length of the perpendicular from the point (2, 3, 4) to the plane 3x + 4y −
12z = 9.
7. (a) Find the point at which the given lines intersect:
~r = 1, 1, 0 + t 1, −1, 2
~r = 2, 0, 2 + s − 1, 1, 0 .
(b) Find an equation of the plane that contains these lines.
8. Find the distance between the given parallel planes.
(a) 2x − 3y + z = 4, 4x − 6y + 2z = 3,
(b) 6z = 4y − 2x, 9z = 1 − 3x + 6y.
9. Which of the following four planes are parallel? Are any of them identical?
P1 : 3x + 6y − 3z = 6,
P3 : 9y = 1 + 3x + 6z,
0.3.5
P2 : 4x − 12y + 8z = 5
P4 : z = x + 2y − 2.
The angle between two planes
Definition 2. The angle between two planes is defined as the angle between two lines, one
drawn in each plane, at right angles to the line of intersection of the two planes.
xxxvi
Figure 8:
In Figure 8 (i), P1 O and P2 O are lines in planes π1 and π2 respectively such that each is
perpendicular to AB. Then angle P1 OP2 is the angle between the planes. In Figure 8 (ii),
P1 O and P2 O are the intersections of the planes with the plane of the paper, the line of
intersection of AOB being perpendicular to the planes of the paper. The normal-vectors
n~1 , n~2 of the planes then lie in the plane of the paper.
Alternative Definition: The angle between two planes is defined as the angle between their
normal-vectors.
The angle θ between the planes n~1 · ~r = k1 and n~2 · ~r = k2 is given by
cos θ =
n~1 · n~2
.
|n~1 ||n~2 |
If the planes are parallel, we have
n~1 = λn~2 ,
where λ is a constant and if the planes are perpendicular, we have
n~1 · n~2 = 0.
0.3.6
The angle between a line and a plane
Definition 3. The angle between a line and a plane is defined as the angle between the
line and the projection of the line on the plane.
In the Figure 9, AB is a line, AN is its projection on the plane, θ is the required angle
and φ is the angle between the line and the perpendicular BN to the plane. Hence if
0.3. THE PLANE
xxxvii
Figure 9:
~ ~n are the directio-vector and normal-vector of the line and plane respectively, then we
I,
have
I~ · ~n
cos φ =
.
~ n|
|I||~
But
1
φ = π − θ.
2
Hence
sin θ =
I~ · ~n
.
~ n|
|I||~
When θ = 0, the line is parallel to the plane and we have
I~ · ~n = 0
(line parallel to plane).
When θ = 21 π, the line is perpendicular to the plane and this occurs when
I~ = λ~n
(line perpendicular to plane),
where λ is a parameter.
EXERCISES
1. Find the acute angle between the planes
(i) 2x − y + z + 8 = 0, 3x + 2y − z − 1 = 0.
(ii) 5x − y − 2z = 5, y = 0.
xxxviii
2. Find the acute angle between the line
x−1
y+7
z−5
=
=
3
4
12
and the plane
2x − 6y + 3z = 8.