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`1. What is a point? What is a vector? How are they related?`
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`2. Find the dot products for the following pairs of vectors:`
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`(a) (1,1,1) . (1,1,-1)`
`(b) (2,5,3) . (2,5,3)`
`(c) (1,0,0) . (10,1,5)`
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`a = 1x1 + 1x1 + 1x-1 = 1`
`b = 38`
`c = 10`
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`3. Find the length of the vector (2,5,3).`
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`root (2x2 + 5x5 + 3x3) = root(38) = 6.16`
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`4. Let i = (1,0,0), j = (0,1,0) and k = (0,0,1). By considering the`
`meaning of cross product (*), write down the answers to:`
`i*i, j*j, k*k, i*j, j*i, i*k, k*i, j*k, k*j.`
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` `
`i*i = 0 0 0`
`j*j = 0 0 0`
`k*k = 0 0 0`
`i*j = 0 0 1`
`j*i = 0 0 -1`
`i*k = 0 -1 0`
`k*i = 0 1 0`
`j*k = 1 0 0`
`k*j = -1 0 0`
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`i*j Method1 (1,0,0)*(0,1,0)`
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`x = y1z2 - y2z1 = 0x0 - 1x0 = 0`
`y = z1x2 - z2x1 = 0x1 - 0x1 = 0`
`z = x1y2 - x2y1 = 1x1 - 0x0 = 1       = (0,0,1)`
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`i*j Method2 (1,0,0)*(0,1,0)`
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`reminder that the determinant of 2x2 matrix:`
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`         | a b |`
`         | c d | = ad - cb`
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`consider unit vectors`
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`        ux=(1,0,0) uy=(0,1,0) uz=(0,0,1)`
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`then the determinant of the following 3x3 matrix`
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`         | ux uy uz |    | ux uy uz | `
`         | ix iy iz | =  | 1  0  0  | `
`         | jx jy jz |    | 0  1  0  | `
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`that can be simplified to 2x2 determinants`
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`        det|0 0|ux - det|1 0|uy + det|1 0|uz`
`           |1 0|        |0 0|        |0 1|`
` `
`        det|0 0| = (0x0 - 1x0) = 0`
`           |1 0|`
`        `
`      - det|1 0| = (1x0 - 1x0) = 0`
`           |0 0|`
` `
`        det|1 0| = (1x1 - 0x0) = 1`
`           |0 1|`
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`giving the cross product`
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`        0ux - 0uy + 1yz = (0,0,1)`
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`Note that this is the same as calculated previously!`
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`5. Using the result in 4. and the fact that (x,y,z) = x i + y j + z k`
`find an expression for p*q where p = (x1,y1,z1) and q = (x2,y2,z2).`
`Hence find the cross products for each of (a), (b), (c) in 2.`
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` `
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`(a) -2 2 0`
`(b) 0 0 0`
`(c) 0 -5 1`
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`6. The equation of a plane is ax+by+cz = d. (Under what circumstances`
`will d=0?). `
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`When the plane goes through the origin`
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`Find a,b,c,d for each of the cases:`
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`(a) The plane is the XY plane.`
`(b) The plane is the XZ plane.`
`(c) The plane is the YZ plane.`
`(d) The plane is parallel to the XY plane and of distance D from the origin.`
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`Reminder: In the plane equation the a,b,c is the plane norm (the vector perpendicular`
`to the plane) that gives the precise position of the plane. The d is the translation `
`distance of the plane from the origin. The x,y,z is a point in 3D space. So, values `
`of x,y,z that satisfy the equation are on the plane.`
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`Visualise the plane as the XY axis (using a piece of paper helps). A vector that is `
`perpendicular to this plane does not travel in the X or Y direction but will travel `
`in the Z direction. Such a vector is thus (x=0,y=0,z=1).`
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`(a) a=0  b=0  c=1  d=0   (a+b+c=1 in a norm)`
`(b) a=0  b=1  c=0  d=0`
`(c) a=1  b=0  c=0  d=0`
`(d) a=0  b=0  c=1  d=D`
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` `
`Describe the geometric meaning of the planes:`
`(a) x=y                Reformulate as x-y=0`
`(b) x+y = 0`
`(c) x+y+z = 0`
`(d) x+y+z = 1.`
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` `
`Imagine values of x y and z that will satisfy this (a) plane equation: `
`(x=+ve, y=+ve, z=any).`
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`This corresponds to (a=+ve,b=+ve,c=0) as the plane norm with d=0. From`
`this we can deduce the position of the plane.`
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`(a) a=1  b=-1  c=0   d=0`
`(b) a=1  b=1   c=0   d=0`
`(c) a=1  b=1   c=1   d=0`
`(d) a=1  b=1   c=1   d=1`
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`7. Find the equation of the plane that passes through the following`
`three points: (1,0,0), (0,1,0), (0,0,1).`
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`Simply visualise the three points. The plane that goes through them`
`has a norm that seems to point straight away from (or at, according`
`to in which direction you're looking at) the origin. The direction of `
`this norm vector is clearly something like (1,1,1) meaning a plane`
`equation of x+y+z=1`
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`8. In general given the points (x1,y1,z1), (x2,y2,z2), (x3,y3,z3) find`
`the equation of the plane which passes through these points, by using`
`the plane equation, ax+by+cz = d.`
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`Let p1,p2,p3 be our three points in the plane. We can construct two vectors`
`that go between the points that definately exist on the plane. To compute the`
`plane equation we simply want to find the norm of these two vectors (which will`
`give us the norm of the plane as well!). `
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`So, the two vectors, (using the points from the previous example)`
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`v1 = p2 - p1 = (-1,1,0)               (point 1 to point 2)`
`v2 = p3 - p1 = (-1,0,1)        (point 1 to point 3)`
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`both lie in the plane. If we take their cross product we find our normal vector:`
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`n = v1 x v2 = (1,1,1)`
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`and our displacement from the origin:`
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`d = n.p1 = (1,1,1).(1,0,0) = 1`
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`so the equation of the plane is x+y+z=1`
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`9. Let p0, p1, p2 be three distinct points in 3D. Then consider the`
`cross product n = (p2-p0)*(p1-p0). What does this vector mean`
`geometrically? `
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`It is the norm to the plane on which the three points lie.`
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`Let p=(x,y,z) be any point on the plane formed by the`
`three points. What is the geometric relationship between the vector`
`(p-p0) and the vector n? `
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`They are orthogonal.`
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`Hence show that the equation of the plane`
`must be the dot product: n.(p-p0) = 0. `
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`The equation for the plane consists of all vectors r satisfying (r-a).n=0 where`
`a is a known point on the plane and n is the plane norm.`
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`By explicitly evaluating the`
`expressions, show that this is the same plane equation as you have`
`found in question 8.`
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`10. Using the argument in 9., show that n must be normal to the plane,`
`and that in question 8 the normal to the plane is (a,b,c). What are`
`the normals to the planes given in question 6 (note that there are at`
`least two answers in each, case, give both)?`
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`11. Evaluate the following product of a row vector and a matrix:`
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`   (1, 2, 3, 4) | 1 2  3 4 |`
`               | 2 1  0 3 |`
`               | 2 0  1 3 |`
`               | 1 2 -1 1 |`
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`1x1 + 2x2 + 3x2 + 4x1  = 15`
`1x2 + 2x1 + 3x0 + 4x2  = 12`
`1x3 + 2x0 + 3x1 + 4x-1 = 2`
`1x4 + 2x3 + 3x3 + 4x1  = 23`
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`= (15, 12, 2, 23)`
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`12. Evaluate the following product of two matrices:`
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` | 1 2  3 4 | * | 2 1 3  4 |`
` | 2 1  0 3 |   | 1 2 2  0 |`
` | 2 0  1 3 |   | 1 1 1  1 |`
` | 1 2 -1 1 |   | 2 3 1 -1 |`
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`eg: `
`at pos 1,1 we have 1x2 + 2x1 + 3x1  + 4x2  = 15`
`at pos 4,4 we have 1x4 + 2x0 + -1x1 + 1x-1 = 2`
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`so:`
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` | 15 20 14 3 |`
` | 11 13 11 5 |`
` | 11 12 10 6 |`
` |  5  7  7 2 |`
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`13. What is the transpose of the matrix in question 11?`
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` | 1  2  3  4 |         | 1  2  2  1 |`
` | 2  1  0  3 |         | 2  1  0  2 |`
` | 2  0  1  3 | becomes | 3  0  1 -1 |`
` | 1  2 -1  1 |         | 4  3  3  1 |`
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`obtained by switching its rows and columns`
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`14. What is the inverse of a matrix? How can you define it?`
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`The inverse is written A^-1  (A to the power -1). The definition is:`
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`A . A^-1 = A^-1 . A = I`
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`where I is the identity matrix:`
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` | 1 0 0 0 |`
` | 0 1 0 0 |`
` | 0 0 1 0 |`
` | 0 0 0 1 |`
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`If A has an inverse then A is called a nonsingular matrix. An inverse only`
`exists if and only if: det(A) != 0. We compute the inverse of a matrix using `
`the Gauss-Jordan elimination. But we're not going to do that here:-)`
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`15. Suppose that A is a square matrix, and its inverse and transpose`
`are equal.  (Such a matrix is called orthogonal). Show that, for any`
`angle q, the following matrix is orthogonal:`
` `
`  cos(t) sin(t)`
` -sin(t) cos(t)`
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`Suppose further that x and b are vectors of the appropriate dimension`
`and that xA=b.  How would you find x? In the example, suppose that b =`
`(1,2) find x.`
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