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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). 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