Download MA1S11 DISCRETE MATHEMATICS

MA1S11 DISCRETE MATHEMATICS:
SOLUTIONS TO TUTORIAL 4
1. Find parametric equations for the line in space which passes through
the points P (2, 4, 1) and Q(1, 3, 5).
Solution: To write down parametric equations for a line we need to
know a point (x0 , y0 , z0 ) on the line and a vector ha, b, ci which is
→
parallel to the line. Let’s use the point P and the vector P Q. In
→
component form P Q is
→
→
→
P Q = OQ − OP
= h1, 3, 5i − h2, 4, 1i
= h−1, −1, 4i
The parametric equations are

x = x0 + t a 



=⇒
y = y0 + t b




z = z0 + t c

x=2−t 



y =4−t




z = 1 + 4t
where the parameter t is any real number.
2. Find an equation for the plane which passes through the points
P (1, 2, 3), Q(4, −1, −3) and R(0, 0, 1).
Solution: The general equation of a plane is ax + by + cz = d where
a, b, c, d are constants. The three points must satisfy this equation
1
2
MA1S11 DISCRETE MATHEMATICS: SOLUTIONS TO TUTORIAL 4
so we get a system of three linear equations.
a + 2b + 3c = d
4a − b − 3c = d
c=d
Rewriting this with the four unknowns a, b, c, d on the left we have
a + 2b + 3c − d = 0
4a − b − 3c − d = 0
c−d=0
This is a homogeneous sytem. We will use Gauss Jordan elimination
to solve the system. The augmented matrix is


1 2
3 −1 0




 4 −1 −3 −1 0 


0 0
1 −1 0
Add −4 times Row 1 to Row 2.

1
2
3
−1 0





 0 −9 −15 3 0 


0 0
1 −1 0
Add −3 times Row 3 to Row 1.

1
2
0
2
0





 0 −9 −15 3 0 


0 0
1 −1 0
MA1S11 DISCRETE MATHEMATICS: SOLUTIONS TO TUTORIAL 4
Add 15 times Row 3 to Row 2.

1 2 0 2 0


 0 −9 0 −12 0

0 0 1 −1 0
3





Multiply Row 2 by − 19 .

1 2 0
2
0





 0 1 0 43 0 


0 0 1 −1 0
Add −2 times Row 2 to Row 1.

− 23

1 0 0
0




 0 1 0 43 0 


0 0 1 −1 0
We have reached the reduced row echelon form. The general solution
to the system is
a = 23 d
b = − 43 d
c=d
where d is a free variable. It is convenient here to choose d = 3,
so a particular solution is (2, −4, 3, 3). The equation of the plane is
2x − 4y + 3z = 3.
3. Compute the distance between the point P (5, 7) and the line y =
4x − 3.
Solution: To begin we take any point Q(x1 , y1 ) on the line and look
→
→
at the vector QP . Since QP is most likely not orthogonal to the
line we need to project it along a normal vector n. Recall that the
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MA1S11 DISCRETE MATHEMATICS: SOLUTIONS TO TUTORIAL 4
general equation for a line is ax + by = c where a, b, c are constants
and a normal vector to this line is n = ha, bi. So in our case we have
n = h4, −1i. Our distance formula is
→
distance = kprojn QP k
→
|n · QP |
=
knk
Note that since Q lies on the line we have
→
n · OQ = h4, −1i · hx1 , y1 i = 4x1 − y1 = 3
We also have
→
n · OP = h4, −1i · h5, 7i = 20 − 7 = 13
knk =
p
42 + (−1)2 =
√
17
Combining these results gives
→
→
→
n · QP = n · (OP − OQ)
→
→
= n · OP − n · OQ
= 13 − 3
= 10
10
=⇒ distance = √
17
4. Compute the distance between the parallel planes 2x + 3y − z = 8
and 2x + 3y − z = 2.
Solution: We use basically the same method as in Q.3. Choose any
point in the first plane, let’s say P (4, 0, 0). If Q(x1 , y1 , z1 ) is any
point in the second plane and n is orthogonal to the second plane
MA1S11 DISCRETE MATHEMATICS: SOLUTIONS TO TUTORIAL 4
then
5
→
|n · QP |
distance = kprojn QP k =
knk
→
Recall that the general equation for a plane is ax + by + cz = d
where a, b, c, d are constants and a normal vector to this plane is
n = ha, b, ci. In our case we have n = h2, 3, −1i. Note that since Q
lies in the second plane we have
→
n · OQ = h2, 3, −1i · hx1 , y1 , z1 i = 2x1 + 3y1 − z1 = 2
We also have
→
n · OP = h2, 3, −1i · h4, 0, 0i = 2(4) + 3(0) − 1(0) = 8
knk =
p
√
22 + 32 + (−1)2 = 14
Combining these results gives
→
→
→
n · QP = n · (OP − OQ)
→
→
= n · OP − n · OQ
= 8−2
= 6
6
=⇒ distance = √
14