Question-   MATH1141 - HIGHER MATHEMATICS 1A 2020 T1 ASSIGNMENT

Solution-

Answer -MATH1141 - HIGHER MATHEMATICS 1A 2020 T1 ASSIGNMENT 1.

In this question you will find the intersection of two planes using two different methods. You are given two planes in parametric form, Π1 :   x1 x2 x3   =   0 −2 −1   + λ1   1 −1 1   + λ2   2 0 −1   Π2 :   x1 x2 x3   =   1 −2 −1   + µ1   2 −1 1   + µ2   3 0 −1   , where x1, x2, x3, λ1, λ2, µ1, µ2 ∈ R. Let L be the line of intersection of Π1 and Π2. (a) Find vectors n1 and n2 that are normals to Π1 and Π2 respectively and explain how you can tell without any extra calculations that Π1 and Π2 must intersect in a line. As the plane Π1 is parallel to the vectors   1 −1 1   and   2 0 −1  , we will take their cross product to find n1, so n1 =   1 −1 1   ×   2 0 −1   =   (−1) · (−1) − 1 · 0 1 · 2 − 1 · (−1) 1 · 0 − (−1) · 2   =   1 3 2   . Similarly, we will calculate the normal to Π2 as follows: n2 =   2 −1 1   ×   3 0 −1   =   (−1) · (−1) − 1 · 0 1 · 3 − 2 · (−1) 2 · 0 − (−1) · 3   =   1 5 3   . Since n1 and n2 shares exactly one common coordinate, x1, they cannot be non-zero multiples of each other and hence are not parallel. Therefore, the planes Π1 and Π2 are also not parallel and must intersect in a line. 1 MATH1141 - HIGHER MATHEMATICS 1A 2020 T1 ASSIGNMENT 2 (b) Find the Cartesian Equations for Π1 and Π2. Using the results from part (a), the point-normal equation of Π1 is   1 3 2   ·   x1 x2 x3   −   0 −2 −1   ! = 0, where   0 −2 −1   is a given coordinate vector of a point on Π1. By simplifying, we have   1 3 2   ·   x1 x2 + 2 x3 + 1   = 0, and evaluating the dot product gives x1 + 3(x2 + 2) + 2(x3 + 1) = 0. Finally, by expanding and collecting like terms, the Cartesian Equation of Π1 is x1 + 3x2 + 2x3 + 8 = 0. Similarly, for the plane Π2,   1 5 3   ·   x1 x2 x3   −   1 −2 −1   ! = 0,   1 5 3   ·   x1 − 1 x2 + 2 x3 + 1   = 0, (x1 − 1) + 5(x2 + 2) + 3(x3 + 1) = 0 and therefore, the Cartesian Equation for Π2 is given by x1 + 5x2 + 3x3 + 12 = 0. (c) For your first method, assign one of x1, x2 or x3 to be the parameter ω and hence write down a parametric vector form of the line of intersection L. To find the parametric vector form of L, we will first solve the Cartesian Equations of Π1 and Π2 simultaneously. From part (b), these equations are x1 + 3x2 + 2x3 + 8 = 0 (1), and x1 + 5x2 + 3x3 + 12 = 0 (2). Subtracting equation (1) from equation (2), we have 0 + 2x2 + x3 + 4 = 0, x3 = −2x2 − 4 (3). MATH1141 - HIGHER MATHEMATICS 1A 2020 T1 ASSIGNMENT 3 Substituting x3 from equation (3) into equation (1) gives x1 + 3x2 + 2(−2x2 − 4) + 8 = 0, x1 − x2 = 0, x1 = x2. (4) Now, by letting x2 = ω and using equations (3) and (4), we have the parametric equations x1 = ω, x2 = ω and x3 = −2ω − 4. Therefore, a possible parametric vector form of L is L =   w w −2w − 4   , L =   0 0 −4   + ω   1 1 −2   . (d) For your second method, substitute expressions for x1, x2 or x3 from the parametric forms of Π2 into your Cartesian equation for Π1 and hence find a parametric vector form of the line of intersection L. The parametric forms of Π2 can be written as three separate equations, namely x1 = 1 + 2µ1 + 3µ2, x2 = −2 − µ1 and x3 = −1 + µ1 − µ2. Substituting these expressions into the Cartesian equation of Π1 from part (b), we have (1 + 2µ1 + 3µ2) + 3(−2 − µ1) + 2(−1 + µ1 − µ2) + 8 = 0, 1 + 2µ1 + 3µ2 − 6 − 3µ1 − 2 + 2µ1 − 2µ2 + 8 = 0, 1 + µ1 + µ2 = 0, µ2 = −µ1 − 1. To find a possible parametric vector form for L, we will substitute µ2 into the parametric form of Π2, which gives L =   1 −2 −1   + µ1   2 −1 1   + (−µ1 − 1)   3 0 −1   =   1 + 2µ1 − 3µ1 − 3 −2 − µ1 −1 + µ1 + µ1 + 1   =   −2 − µ1 −2 − µ1 2µ1   , and thus L =   −2 −2 0   + µ1   −1 −1 2   . MATH1141 - HIGHER MATHEMATICS 1A 2020 T1 ASSIGNMENT 4 (e) If your parametric forms in parts (c) and (d) are different, check that they represent the same line. If your parametric forms in part (c) and (d) are the same, explain how they could have been different while still describing the same line. From parts (c) and (d), we have obtained two different parametric equations for the line L. However, observe that the lines are parallel with µ1 = −ω. Also, as   0 0 −4   and   −2 −2 0   are arbitrarily chosen vector coordinates of two points that satisfy L, both equations must therefore represent the same line. (f) Find m = n1 × n2 and show that m is parallel to the line in parts (c) and (d). Recall that n1 =   1 3 2   and n2 =   1 5 3   from part (a). Hence, m =   1 3 2   ×   1 5 3   =   9 − 10 2 − 3 5 − 3   . =   −1 −1 2   . Since m is a non-zero multiple of ω   1 1 −2  , m is therefore parallel to the line L. (g) Give a geometric explanation of the result in part (f). The normal n1 is perpendicular to all lines in the plane Π1. By taking the converse of this statement, every line that is perpendicular to n1 must be parallel to Π1. Similarly, every line that is perpendicular to n2 must be parallel to Π2. As m is the cross product of n1 and n2, it is perpendicular to both n1 and n2 and hence parallel to Π1 and Π2. Since L is line of intersection between Π1 and Π2, m must therefore be parallel to L. 2. Show that the equation: e −8x + 6 cos(15x) = 0 Has a unique solution for x ∈ [0, π 15 ] Define f(x) = e −8x + 6 cos(15x) in the interval [0, π 15 ]. Since e −8x and 6 cos(15x) are both continuous for x ∈ [0, π 15 ], this property holds for f(x). MATH1141 - HIGHER MATHEMATICS 1A 2020 T1 ASSIGNMENT 5 Next, observe that f(0) = e −8·0 + 6 cos(15 · 0) = 7 > 0, and f( π 15 ) = e −8· π 15 + 6 cos(15 · π 15 ) = −5.8 < 0. By applying the Intermediate Value Theorem, there is at least one real number c ∈ [0, π 15 ] such that f(c) = 0. Now, by differentiating f(x), we have f 0 (x) = −8e −8x − 90 sin(x) = −(8e −8x + 90 sin(x)). Since e −8x > 0 for all x ∈ R and sin(x) ≥ 0 for x ∈ [0, π], we can deduce that f 0 (x) < 0 for x ∈ [0, π 15 ] and hence f(x) is strictly decreasing in the interval for which it is defined. Therefore, f(x) = e −8x + 6 cos(15x) = 0 has one unique solution for x ∈ [0, π 15 ]. 3. Use the Mean Value Theorem to estimate the error when approximating (8.035)2/3 by 82/3 . The precise absolute error in this approximation is given by | Error | = | 8.035 2 3 − 8 2 3 | . Now, let f(x) = x 2 3 , and so f 0 (x) = 2 3 x − 1 3 . By applying the Mean Value Theorem on f(x) in the interval [8, 8.035], we have 8.035 2 3 − 8 2 3 8.035 − 8 = 2 3 x − 1 3 , where 8 < c < 8.035. Simplifying and rearranging this equation gives 8.035 2 3 − 8 2 3 = 0.035 · 2 3 c − 1 3 = 7 200 · 2 3 · 1 c 1 3 . Hence, we can find the precise absolute error by | Error | = | 8.035 2 3 − 8 2 3 | = 7 200 · 2 3 · 1 c 1 3 since c > 8 > 0, which also implies that | Error | < 7 200 · 2 3 · 1 8 1 3 = 7 200 · 2 3 · 1 2 = 7 600 . Therefore, the magnitude of the error for the approximation of 8.035 2 3 by 8 2 3 is less than 7 600 .

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