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Welcome to Engineering Mathematics 2

Welcome to Engineering Mathematics 2. We will cover 4 topics today 1 . Lines 2 . Distance between a Point and a Line 3. Planes 4. Questions. Lecture Recap. The second way is known as the cross (or vector) product. .

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Welcome to Engineering Mathematics 2

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  1. Welcome to Engineering Mathematics 2 We will cover 4 topics today 1. Lines 2. Distance between a Point and a Line 3. Planes 4. Questions

  2. Lecture Recap The second way is known as the cross (or vector) product. Last week we learnt that the multiplication of 2 vectors is defined in 2 different ways. The first way is known as the dot (or scalar) product. The scalar quantity a.(b x c) is called a Scalar Triple Product.

  3. Lines We can define A line can be specified in two ways. Either using two points or a point and a direction. z Let t be an arbitrary scalar then, y P Po Which can be rewritten as v Or more simply if ro is the position vector of Po x

  4. Lines A line can be specified in two ways. Either using two points or a point and a direction. Hence we can derive the parametric equations for a line z y P Po v x

  5. Lines Find the parametric equations for the line through P(-3, 2, -3) and Q(1, -1, 4) using the vector PQ. Question Find the parametric equations for the line through (-2, 0, 4) parallel to v = 2.i +4.j – 2.k Thus Find the parametric equations using the vector QP.

  6. Lines Hence we can deduce that, We can generalise the previous result in the following way. Suppose we have two position vectors (a and b). We can write the equation of a straight line as Or, in other words This gives us the Cartesian form of a straight line in three dimensions.

  7. Distance between a Point and a Line There are two ways we can calculate the distance between point and a line. One uses a dot product and the other uses a cross product. We will use the cross product. The shortest distance between P and the line passing through AR is shown in red. Clearly the line PR is perpendicular to AR. Recall that A Thus, the perpendicular distance is given by R θ v P

  8. Distance between a Point and a Line Question Find the distance from the point S(1, 1, 5) to the line From the parametric equations we can see that the line passes through the point P(1, 3, 0) and runs parallel to v = i – j + 2.k Hence

  9. Distance between a Point and a Line Question Compute the distance between P(2, 0, 0) and the line passing through the point A(0, -1, 0) and parallel to the vector v(1, 2, 0).

  10. Distance between a Point and a Line Now we know the direction of the common perpendicular we can calculate the distance between the lines. If A and B are a point on either line then, The shortest distance between two non-intersecting lines occurs along their common perpendicular. We can calculate this distance. If u and v are direction vectors along the two lines then the common perpendicular is given by B With a corresponding unit length of v θ u A

  11. Distance between a Point and a Line Question The points A, B, C & D are at (2, 2, 2); (4, -1, 2); (1, -1, -1) and (5, 2, 0) respectively. What is the shortest distance between the lines AB and CD? Pick a point on either line. We will choose A and D.

  12. Planes Suppose that a plane passes through a point P0(x0,y0,z0) and is normal to the non-zero vector n. We can deduce that A plane can be specified in three ways. I.e. using one point and a normal direction, one point and two vectors or three points. Let A normal direction defines the ‘tilt’ or orientation of a plane. n Then P0(x0, y0, z0) r Or (in Cartesian form) P(x, y, z)

  13. Planes Question Find an equation for the plane through A(0, 0, 1); B(2, 0, 0) and C(0, 3, 0). Question Find an equation for the plane through P0(-3, 0, 7) and perpendicular to n = 5.i + 2.j - k First we find the normal to the plane The component equation is Now we can use one the points and the normal to find the equation of the plane. Let’s choose point A. Simplifying we obtain

  14. Planes The solution to the previous question gives the following general result. If we have a point and two vectors then we can find the Cartesian equation of the plane using In the solution to the previous question we derived our two in plane vectors using points. Hence, if we are given three points we can generalise the solution to Where A and B are positional vectors. Where v and u are the two in plane vectors. Notice that if Then This will be useful later

  15. Planes The angle between two planes is given by the angle between their normals. Question Find the angle between the planes 3.x - 6.y - 2.z = 15 & 2.x + y – 2.z = 5 We know that Thus Hence

  16. Planes If P is a point on a plane with a normal ‘n’, then the shortest distance from an arbitrary point ‘S’ to the plane is found using the dot product of PS and n (look at the calculation for the minimum distance between two lines). I.e. S θ P Plane

  17. Planes Question Find the distance from S(1, 1, 3) to the plane 3x + 2y + 6z = 6 The normal to the plane is given by We need to find a point in the plane, i.e. P on the previous slide. By examining the equation in bold above, we can immediately choose or or Hence,

  18. Questions Sometimes we want to know where a line and a plane intersect. This is particularly useful in computer graphics. Find the point where the line Intersects the plane 3x + 2y +6z = 6 Hence the point of intersection is

  19. Questions Find a vector parallel to the line of intersection of the planes The line of intersection is normal to both planes’ normal vectors. Hence, Thus,

  20. Questions Find the parametric equations for the line of intersection of the planes On the previous slide we determined that the vector parallel to the line of intersection is given by This only gives us the direction of the vector. To find the parametric equations we also need to find a point along the line of intersection. To do this choose z = 0 and simultaneously solve the equations above. Thus x = 3 & y = -1 and the common point is (3, -1, 0).

  21. Exam Question A building has a triangular roof. The top of the roof is at the point A(5, 7, 10) and the other two corners of the roof have coordinates B(0, 4, 6) and C(10, 4, 6). Find the area of the roof. Find a unit normal vector of the roof. Find the Cartesian equation for the plane of the roof. A telephone wire runs in a straight line from the point P = (12, 11, 14) to the top of the roof. Find the length of the wire. Find the angle between the wire and the roof.

  22. Exam Question The area of a triangle is given by The plane passes through A, B and C. Hence using B we can determine D. The equation of the plane is given by or

  23. Exam Question Let P = (12, 11, 14) Thus the length of wire is given by The angle between the wire and the roof is found by finding the angle between the wire and normal to the roof. Thus the angle between wire and the roof is 95.1o – 90o = 5.1o or 0.089 rad

  24. Conclusion COURSEWORK!!!! We have covered 4 topics today 1. Lines 2. Distance between a Point and a Line 3. Planes 4. Questions

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