MA1S11 DISCRETE MATHEMATICS: SOLUTIONS TO …

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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).
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

4

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

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