In this section we study lines in R3 and show how to derive parametic equations to describe . We derive a formula by which to calculate the distance from a point to a line and the distance between two lines. We also show how to project a vector orthogonally onto a line and how to reflect a vector about a line.
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Theoretical Remarks
1.4.The parametric equation of a line inR3 is of the form
:
x = at + x1 y = bt + y1
z = ct + z1 for all t∈ R,
Figure 1.17: A line in R3
where (x1, y1, z1) is a point on the line and v = (a, b, c) is the vector that is parallel to the line . See Figure 1.17. Here t is a parameter that can take on any real value. That is, for every point (x, y, z) on the line , there exists a unique value of t, such that
(x, y, z) = (at + x1, bt + y1, ct + x1).
Problem
1.4.1.Find a parametric equation of the line inR3, where (−1, 1, 3) and (2, 3, 7) are two points on .
a) Establish which of the following three points, if any, are on this line :
(−4, −1, −1); (−1, 2, 3); (1 2, 2, 5)
b) Is the vector w = (−6, −4, −8) parallel to the line ? Explain.
Solution
1.4.1.We are given two points that are on the line , namely P1 : (−1, 1, 3) and P2 : (2, 3, 7).
Then the vector−−−→
P1P2 is parallel to and has the following coordinates:
−−−→P1P2= (3, 2, 4).
Hence the vector v which is parallel to is v =−−−→P1P2= (3, 2, 4).
Let P : (x, y, z) be an arbitrary point on . Then
−−→P1P = tv or (x + 1, y− 1, z − 3) = t(3, 2, 4) for all t ∈ R.
Comparing the x-, y−, and z-components of the above vector equation, we obtain x + 1 = 3t, y− 1 = 2t, z − 3 = 4t,
respectively. The parametric equation for is therefore
:
x = 3t− 1 y = 2t + 1
z = 4t + 3 for all t∈ R.
a) To find out whether the point (−4, −1, −1) is on the line, we use the obtained parametic equation for and find t. That is, t must satisfy the relations
−4 = 3t − 1, −1 = 2t + 1, −1 = 4t + 3.
This leads to a unique solution for t, namely t =−1. Hence the point (−4, −1, −1) is on .
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For the point (−1, 2, 3) we have
−1 = 3t − 1, 2 = 2t + 1, 3 = 4t + 3,
which cannot be satisfied for any value of t. Hence (−1, 2, 3) is not a point on .
The point (1/2, 2, 5) satisfies the parametric equation for t = 1/2, so that (1/2, 2, 5) is a point on .
b) The vector w = (−6, −4, −8) is indeed parallel to the line , since w =−2v,
where v is parallel to .
Problem
1.4.2.Find a parametric equation of the line in R3 which passes through the point (1,−1, 2) and which is orthogonal to the lines 1 and 2, given in parametric form by
1:
The point (1/2, 2, 5) satisfies the parametric equation for t = 1/2, so that (1/2, 2, 5) is a point on .
b) The vector w = (−6, −4, −8) is indeed parallel to the line , since w =−2v,
where v is parallel to .
Problem
1.4.2.Find a parametric equation of the line in R3 which passes through the point (1,−1, 2) and which is orthogonal to the lines 1 and 2, given in parametric form by
1:
The point (1/2, 2, 5) satisfies the parametric equation for t = 1/2, so that (1/2, 2, 5) is a point on .
b) The vector w = (−6, −4, −8) is indeed parallel to the line , since w =−2v,
where v is parallel to .
Problem
1.4.2.Find a parametric equation of the line in R3 which passes through the point (1,−1, 2) and which is orthogonal to the lines 1 and 2, given in parametric form by
1:
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Solution
1.4.2.Vector v1 = (2, 1, 1) is a vector that is parallel to 1 and v2 = (−3, 2, 4) is a vector that is parallel to 2. A vector that is orthogonal to both 1 and 2 is therefore
v = v1× v2
and this vector v is thus parallel to the line that we are seeking. We calculate v:
v = v1× v2=
A parametric equation for the line is therefore
:
Consider a line inR3 with the following parametric equation
: is a vector parallel to .
a) Assume that the point P0 : (x0, y0, z0) is not on the line . Find a formula for the distance from the point P0 to .
b) Find the distance from the point (−2, 1, 3) to the line , given by the parametric equation
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Solution
1.4.3.a) We find a formula for the distance from the point P0(x0, y0, z0) to the line , where
:
x = at + x1 y = bt + y1
z = ct + z1 for all t∈ R.
Here P1 : (x1, y1, z1) is a point on the line . Let s = −−−→
P0P2 denote the distance from P0 to , where P2 is a point on which is not known. We consider the right-angled triangle P1P2P0. See Figure 1.18.
Figure 1.18: The distance from a point P0 : (x0, y0, z0) to the line inR3
It follows that s = −−−→
P0P2 = −−−→
P1P0 sin θ. (1.4.1)
On the other hand we have, from the definition of the cross product, that
−−−→P1P0× v = −−−→P1P0 v sin θ. (1.4.2) Solving||−−−→P1P0|| sin θ from (1.4.2) and inserting it into (1.4.1), we obtain the following formula for the distance:
s = −−−→
P1P0× v
v .
b) The given line passes through the point P1: (1,−4, 2) and is parallel to the vector v = (1, 3, 5). Thus for the point P0: (−2, 1, 3), we have
−−−→P1P0 = (−3, 5, 1)
and
−−−→P1P0× v =
e1 e2 e3
−3 5 1
1 3 5
= 22e1+ 16e2− 14e3 = (22, 16,−14).
Calculating the lengths of the vectors−−−→
P1P0× v and v, we obtain
−−−→
P1P0× v =
(22)2+ (16)2+ (−14)2 = 6√ 26
v =
12+ 32+ 52 =√ 35,
so that the distance from the point P0 to the given line is
s = −−−→
P1P0× v
v = 6√
√26 35 .
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Problem
1.4.4.Find a formula for the distance between two lines inR3 and use your formula to find the distance between the following two lines:
1:
x = 2t + 1 y = t− 1
z = 3t + 1 for all t∈ R,
2 :
x = t y = 2t + 2
z = 1 for all t∈ R.
Solution
1.4.4.Assume that P1 : (x1, y1, z1) is a point on the line 1 and that P2 : (x2, y2, z2) is a point on another line 2. Let v1 denote a vector that is parallel to 1 and v2 a vector that is parallel to 2 (see Figure 1.19). Now v = v1× v2 is a vector that is orthogonal to both
Figure 1.19: Distance s between two lines inR3
v1 and v2, and therefore v is orthogonal to the lines 1 and 2. To find the distance s between 1 and 2, we project−−−→
P1P2 orthogonally onto the vector v. This leads to s =projv−−−→
P1P2 = |−−−→
P1P2· v|
v = |−−−→
P1P2· (v1× v2)|
v1× v2 .
For the given line 1 we have v1 = (2, 1, 3) with a point P1 : (1,−1, 1) ∈ 1 and for the
given line 2 we have v2= (1, 2, 0) with a point P2: (0, 2, 1)∈ 2. Thus
−−−→P1P2= (−1, 3, 0), v1× v2 =
e1 e2 e3
2 1 3
1 2 0
=−6e1+ 3e2+ 3e3= (−6, 3, 3),
so that the distance s between 1 and 2 is s = |(−1, 3, 0) · (−6, 3, 3)|
√36 + 9 + 9 = 5
√6.
Problem
1.4.5.Consider the following two vectors inR3: u = (−1, 3, 3), v = (2, −1, 4).
Consider now the line inR3, such that contains the point (2,−1, 4) and the zero-vector 0 = (0, 0, 0).
a) Find the orthogonal projection of the vector u onto the line , i.e. calculate proj u.
b) Find the distance between the point (−1, 3, 3) and the line .
c) Find the reflection of the vector u about the line .
Solution
1.4.5.a) We aim to obtain the vector w which is the orthogonal projection of the vector u onto the line , i.e. w =proj u. This can be achieved by projection u onto any position vector that is lying on this line , for example vector v. See Figure 1.20.
Thus
w = proj u = projvu = (u· ˆv)ˆv where ˆv = v
v. For u = (−1, 3, 3) and v = (2, −1, 4), we have
w =
u· v
v2
v = u · v v· v
v = (−1)(2) + (3)(−1) + (3)(4)
22+ (−1)2+ 42 (2,−1, 4) = 1
3(2,−1, 4).
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Figure 1.20: The orthogonal projection of u onto .
b) The distance d between the point (−1, 3, 3) and the line is −−→AB (see Figure 1.20).
By vector addition we then have
−−→AB = u− w = (−1, 3, 3) − (2 3,−1
3,4
3) = (−5 3,10
3 ,5 3).
Thus
−−→AB =
25 9 +100
9 +25 9 = 5
3
√6.
c) The reflection of the vector u about the line is given by the vector−−→
OC. See Figure 1.21. By vector addition we have
−−→OC +−→
CA +−−→
AB = u.
However, −→CA =−−→AB, so that
−−→OC = u− 2−−→AB,
PROBLEMS, THEORY AND SOLUTIONS IN
LINEAR ALGEBRA: PART 1 EUCLIDEAN SPACE40 CHAPTER 1. VECTORS, LINES AND PLANES INVeCtors, lines and Planes in r3R3
Figure 1.21: The reflection of u about .
where u = (−1, 3, 3) and−−→AB = (−5 3,10
3 ,5
3) (see part a) above). Thus the reflection about is
Figure 1.21: The reflection of u about .
where u = (−1, 3, 3) and−−→
AB = (−5 3,10
3 ,5
3) (see part a) above). Thus the reflection about is
Figure 1.21: The reflection of u about .
where u = (−1, 3, 3) and−−→
AB = (−5 3,10
3 ,5
3) (see part a) above). Thus the reflection about is
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