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## Welcome to Engineering Mathematics 2

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**Welcome to Engineering Mathematics 2**We will cover 2topics today 1. Section A Exam Questions 2. Section B Exam Questions**Exam Questions**1) The value of the determinant Is a) -96 b) -84 c) -72 d) 108**Exam Questions**2) The force F which has magnitude 6 and direction parallel to the vector (3, 1, -1) has components a) b) c) d) Therefore We need a magnitude of 6 hence,**Exam Questions**3) The angle between the vectors u = 2i – 3j + 6k and v = 2i – 2j – 3k is a) b) c) d) We use**Exam Questions**4) If the position vectors of the points A and B are (4, -1, 3) and (10, 8, -2) respectively, then the position vector of the point P on the line joining A to B such that AP = (1/3).AB is a) b) c) d)**Exam Questions**5) If u = (2, 3, 1) and v = (-1, 4, -3), then u x v has the components a) b) c) d)**Exam Questions**6) For any two vectors p and q, the expression (6p + 3q) x (q + 2p) can be simplified to a) 6p x q b) 12p x q c) -12p x q d) 0 Clearly, (6p + 3q) = 3.(q + 2p) hence, these vectors are parallel to each other. We know that the cross product of two parallel vectors is equal to zero.**Exam Questions**7) The Cartesian equation of the plane which is parallel to the vectors i + k and i + j and passes through the point (3, 1, 1) is a) b) c) d) To find the Cartesian equation we need to find the normal, hence Thus, Using the point (3, 1, 1)**Exam Questions**8) If the matrices A, B and C are defined by Then the element c22 is a) 0 b) 4 c) 6 d) 7**Exam Questions**9) If the matrices A, B and C are defined by Where x and y are constants, then C is an antisymmetric matrix only when a) x = 1 for all values of y b) x = 3, y = -2 c) x = -1, y = -2 d) x = -1 for all values of y If x = -1, y = -2 then**Exam Questions**10) If the matrix A and its inverse A-1 are defined to be Where x and y are constants, then a) b) c) d)**Exam Questions**11) Given that v1 = (1, 2) and v2 = (3, -5) are eigenvectors of the matrix A, the corresponding eigenvalues are a) 1/4 and 1/7 b) -1/4 and 1/7 c) 4 and 7 d) -4 and 7**Exam Questions**12) If the Jacobi iterative method is applied to the system of equations Starting from the initial guess x = y = z = 1, the result after one step is a) b) c) d) The rearranged equations are**Exam Questions**A radio transmitter T is situated at a point with coordinates (2, 2, 0), in appropriate units, with respect to Cartesian coordinates Oxyz where z is vertically upwards. Its signals are detected by receivers A and B each situated on the top of a building. If A has coordinates (4, 3, 1) and B has coordinates (−3, 1, 2) find (a) the position vector of the point Q on TA such that TQ = 0.4TA (b) the equation of the plane, labelled π, containing A, B and T (c) the perpendicular distance of O from π**Exam Questions**Let the position vectors of the points T, A & B be t, a & b respectively. Thus, (a) the position vector of the point Q on TA is**Exam Questions**(b) the equation of the plane, labelled π, containing A, B and T First we need to find the normal to the plane π Thus, the equation of the plane is given by 3x - 9y + 3z = C Using point T Hence, the equation is given by or**Exam Questions**(c) the perpendicular distance of O from π q is the position vector of the point Q about the origin and n is the vector normal to the plane π In an earlier lecture we proved that Where d is the perpendicular distance between a point and the plane. r is the vector connecting the point O and an arbitrary point on the plane. In our case we can use the vectors a, b, or t.**Exam Questions**Three planes have vector equations r · (−1, 3, −1) = 7, r · (0, −1, 2) = −3 and r · (2, −1, 0) = 1. Write these equations in Cartesian form and use the ROW REDUCTION method to find the position vector of the point Q where the three planes intersect. The points A, B and C have position vectors (7, 5, 1), (3, 5, 1) and (3, 5, 5) respectively. Find the volume of the tetrahedron with vertices at Q, A, B and C (a tetrahedron has 1/6th of the volume of a parallelepiped.**Exam Questions**The vector equations r · (−1, 3, −1) = 7, r · (0, −1, 2) = −3 and r · (2, −1, 0) = 1. Can be written in Cartesian form as Thus, in matrix form The augmented matrix is**Exam Questions**R3 = R3 + 2R1 R3 = R3 + 5R2 Hence; x = 2, y = 3 & z = 0 thus Q has the position vector A, B & C have position vectors (7, 5, 1); (3, 5, 1) & (3, 5, 5) respectfully. Let Volume =**Exam Questions**A coastguard helicopter A observes a sailing boat as it approaches the finishing line of a race. The finishing line is the line joining two points B and C situated on either side of a harbour. If the position vectors of A, B and C are (−2, 1, 3), (20, 5, 0) and (25, 10, 0) respectively with respect to Cartesian coordinates Oxyz, where z is vertically upwards, find a) the Cartesian equation of the plane containing A,B and C b) the vector equation of the line passing through B and C The boat is observed to have a position vector of (10, 3, 0) and to be travelling parallel to the vector (3, 1, 0). If it is assumed that the boat does not change direction find c) the vector equation of the path of the boat d) the position vector of the point where the boat crosses the finishing line**Exam Questions**Let a, b & c be the position vectors of A, B & C respectively, then a) To find the Cartesian equation of the plane we find its normal, let Now we can determine that 15x - 15y + 90z = Const. Using point A, hence, The equation of the plane is given by x – y + 6z = 15**Exam Questions**b) The line that passes through B and C is parallel to the vector Hence the line that passes through B and C is given by c) By the same logic, the path of the boat is given by**Exam Questions**d) The lines cross when In terms of the x-component In terms of the y-component Which can easily be solved i.e. λ = 2/5, μ = 4 Hence, that coordinates of the intersection are**Conclusion**Seminar!!! We have covered 2topics today 1. Section A Exam Questions 2. Section B Exam Questions