Questions and Answers #3 Vectors and Spaces

Questions and Answers Sheet 3 Vectors and Spaces Question #1 Identify the surface with the given vector equation. 𝑟(𝑠, 𝑡) =< 𝑠, 𝑡, 𝑡 2 − 𝑠 2 > Answer: The idea here is to first write down the parametric equations of the given surface. Then we try to identify some rule or regularity between them to find out what the given surface is. The parametric equations are: x=s y=t 𝑧 = 𝑡 2 − 𝑠2 The first thing one should notice is that the equality 𝑧 = 𝑦 2 − 𝑥 2 holds for any point on the surface. But we know for the face that the given equation corresponds to a quadratic surface which is hyperbolic paraboloid. Question #2 The three components of velocity in a flow field are given by 𝑢 = 𝑥 2 + 𝑦2 + 𝑧2 𝑣 = 𝑥𝑦 + 𝑦𝑧 + 𝑧 2 𝑤 = −3𝑥𝑧 − 𝑧 2 /2 + 4 (a) Determine the volumetric dilatation rate and interpret the results. (b) Determine an expression for the rotation vector. Is this an irrotational flow field? Answer: The rate of volumetric deformation is known as volumetric dilation rate is defined as the rate of increase of volume of a fluid element per unit volume. In an incompressible fluid, the volumetric dilation rate is zero because the fluid element volume cannot change without change in fluid density. Now we can use general equation to determine volumetric dilatation rate, where we have; 𝑢 = 𝑥 2 + 𝑦2 + 𝑧2 𝑣 = 𝑥𝑦 + 𝑦𝑧 + 𝑧 2 𝑤 = −3𝑥𝑧 + ⃗ = △⋅ 𝑉 ⃗ = △⋅ 𝑉 𝑑𝑢 𝑑𝑥 𝑑 𝑑𝑥 + 𝑧2 2 𝑑𝑣 𝑑𝑦 (𝑥 2 +4 + 𝑑𝑤 𝑑𝑧 2 + 𝑦 + 𝑧2) + 𝑑 𝑑𝑦 (𝑥𝑦 + 𝑦𝑧 + 𝑧 2 ) + 𝑑 𝑑𝑧 (−3𝑥𝑧 + 𝑧2 2 + 4) ⃗ = 2𝑥 + 𝑥 + 𝑧 − 3𝑥 − 𝑧 △⋅ 𝑉 ⃗ =0 △⋅ 𝑉 Volumetric dilatation rate is equal zero, than flow is incompressible. And now we will determine and expression for the rotation vector. The rotation, of the element about the one axis is defined as the average of the angular velocities and of the two mutually perpendicular lines. ⃗⃗⃗ = 1 ((𝑑𝑤 − 𝑑𝑣) 𝑖̂ + (𝑑𝑢 − 𝑑𝑤 ) 𝑗̂ + (𝑑𝑣 − 𝑑𝑢) 𝑖̂) 𝑊 2 𝑑𝑥 𝑑𝑧 𝑑𝑧 𝑑𝑥 𝑑𝑥 𝑑𝑦 1 ⃗⃗⃗ ≠ (−2𝑧𝑖̂ + 5𝑧𝑗̂ − 𝑦𝑖̂) 𝑊 2 Since vector W is not zero everywhere the flow field is not irrotational. Question #3 Find the coordinate vector of w relative to the basis 𝑆 = {𝑢1 , 𝑢2 } for 𝑅𝟚 𝑢1 = (1, −1), 𝑢2 = (1,1); 𝑤 = (1,0) Answer: Consider the basis 𝑆 = {𝑢1 , 𝑢2 } for 𝑅𝟚 where 𝑢1 = (1, −1), 𝑢2 = (1,1); 𝑤 = (1,0) In order to find the coordinate vector of w relative to the basis S we must find scalars a and b such that 𝑎𝑣1 + 𝑏𝑣2 = 𝑤 Since 𝑎𝑣1 + 𝑏𝑣2 = 𝑎(1, −1) + 𝑏(1,1) = (𝑎, −𝑎) + (𝑏, 𝑏) = (𝑎 + 𝑏, −𝑎 + 𝑏) and w=(1,0) we must find scalars a and b such that (a+b,-a+b)=(1,0) which leads us to the linear system a+b=1 -a+b=0 1 a+b=1 and -a+b=0 gives us that a+b-a+b=1+0 which gives us that 2b=1 and therefore 𝑏 = -a+b=0 gives us that a=b and therefore 𝑎 = 1 2 1 1 Hence (𝑤)𝑆 = ( , ) 2 2 Question #4 Find the curvature of 𝑟(𝑡) =< 9𝑡, 𝑡 2 , 𝑡 3 > at the point (9, 1, 1). Answer: Given: 𝑟(𝑡) =< 9𝑡, 𝑡 2 , 𝑡 3 > at the point (9,1,1) The objective is to find the curvature The curvature us |𝑟 ′×𝑡 ′′′ | |𝑟 ′|3 It is given that, 𝑟(𝑡) =< 9𝑡, 𝑡 2 , 𝑡 3 > On differentiating, 𝑟 ′ (𝑡) =< 9,2𝑡, 3𝑡 2 > On further differentiating, 𝑟 ′′ (𝑡) =< 0,2,6𝑡 > 𝑟 ′ × 𝑟 ′′ =< 6𝑡 2 , −54𝑡, 18 > At t=1 |𝑟 ′ × 𝑟 ′′ | = √62 + 542 + 182 =57.24 |r'|=9.7 |𝑟 ′ ×𝑟 ′′ | |𝑟 ′|3 = 57.24 9.73 =0.063 Question #5 How do I Find two linearly independent vectors perpendicular to the vector? It is very complicated. 8 v =  −7   −5 Answer: How do I Find two linearly independent vectors perpendicular to the vector? It is very complicated. 8 v =  −7   −5 2 Question #6 Find a vector equation and parametric equations for the line. The line through the point (1,0,8) and perpendicular to the plane x+3y+z=3 Answer: The line passing through the point (1,0,8) and perpendicular to the plane x+3y+z=3 Let 𝑟0 = 𝑖 + 0𝑗 + 8𝑘 A line perpendicular to the plane x+3y+z=3 has the same direction as a normal vector to the plane, which is n=<1,3,1> So, take v=i+3j+k By using the vector eqaution for the line 𝑟 = 𝑟0 + 𝑡𝑣 𝑟 = (𝑖 + 0𝑗 + 8𝑘) + 𝑡(𝑖 + 3𝑗 + 𝑘) = (1 + 𝑡)𝑖 + (0 + 3𝑡)𝑗 + (8 + 𝑡)𝑘 = (1 + 𝑡)𝑖(3𝑡)𝑗 + (8 + 𝑡)𝑘 Thus, the vector equation of the line is 𝑟(𝑡) = (1 + 𝑡) + 𝑖(3𝑡)𝑗 + (8 + 𝑡)𝑘 Obtain the parametric equation of line as follows. Let 𝑃0 = (𝑥0 , 𝑦0 , 𝑧0 ) = (1,0,8) and 𝑣 =< 𝑎, 𝑏, 𝑐 ≥< 1,3,1 > By the parametric equation for the line through the point (𝑥0 , 𝑦0 , 𝑧0 )and parallel to the direction vector are 𝑥 = 𝑥0 + 𝑎𝑡, 𝑦 = 𝑦0 + 𝑏𝑡, 𝑧 = 𝑧0 + 𝑐𝑡 That implies, 𝑥 = 1 + 𝑡, 𝑦 = 0 + 3𝑡, 𝑧 = 8 + 𝑡 𝑥 = 1 + 𝑡, y = 3𝑡, 𝑧 = 8 + 𝑡 Thus, the parametric equation of the line is 𝑥(𝑡) = 1 + 𝑡, 𝑦(𝑡) = 3𝑡, 𝑧(𝑡) = 8 + 𝑡 Question #7 Find a basis for the set of vectors in 𝑅3 in the plane 𝑥 + 2𝑦 + 𝑧 = 0 Answer: Solving for x=-2y-z , we can plug that into x to get rid on the uneeded variable x=-2y-z y=y z=z Because x=-2y-z we can plug that into x to get rid of the uneeded variable.  x   −2 y − z   y =  y  =      z   z   −2  −1   y  1  + z  0   0   1  We have found spanning set with no unneeded variables, which by the spanning set theorem, is the basis.  −2   −1  1  ,  0 }      0   1  Question #8 Find the derivative of the vector function. 𝑟(𝑡) = 𝑎 + 𝑡𝑏 + 𝑡 2 𝑐 Answer: 𝑟(𝑡) = 𝑎 + 𝑡𝑏 + 𝑡 2 𝑐 a,b,c are any constant vectors, like i, j, k, but not necessarily pointing in the same directions or having the same magnitudes. Take the derivative of whatever is multiplying each vector, as usual. 𝑟 ′ (𝑡) = (1)′ 𝑎 + (𝑡)′ 𝑏 + (𝑡 2 )′ 𝑐 = (0)𝑎 + (1)𝑏 + (2𝑡)𝑐 = 𝑏 + 2𝑡𝑐 Result: 𝑟 ′ (𝑡) = 𝐾 + 2𝑡𝑐 Question #9 Hi i know this is a really really simple question but it has me confused. I want to calculate the cross product of two vectors 𝑎×𝑟 The vectors are given by 𝑎 = 𝑎𝑧, 𝑟 = 𝑥𝑥 + 𝑦𝑦 + 𝑧𝑧 The vector 𝑟 is the radius vector in cartesian coordinares. I want to calculate the cross product in cylindrical coordinates, so I need to write 𝑟 in this coordinate system. The cross product in cartesian coordinates is 𝑎 × 𝑟 = −𝑎𝑦𝑥 + 𝑎𝑥𝑦, however how can we do this in cylindrical coordinates? Answer: The radius vector 𝑟 in cylindrical coordinates is 𝑟 = 𝑝𝑝 + 𝑧𝑧 . Calculating the cross-product is then just a matter of vector algebra: 𝑎 × 𝑟 = 𝑎𝑧 × (𝑝𝑝 + 𝑧𝑧) = 𝑎(𝑝(𝑧 × 𝑝) + 𝑧(𝑧 × 𝑧)) = 𝑎𝑝(𝑧 × 𝑝) = 𝑎𝑝𝜙⃗, where in the last line we've used the orthonormality of the triad {𝑝, 𝜙⃗, 𝑧} Question #10 Let 𝑈 = 𝑆𝑝{(3,3,1)} How can I find the Orthogonal complement ? I'm not sure how to calculate it. In the book I'm learning from it's saying that I need to write the vectors of U in Ax=0 where the lines of A are the vectors of U. But since U has only one vector I'm not sure how could this help me to find the orthogonal complement Answer: The plane of equation 𝑃 ≡ 3𝑥 + 3𝑦 + 1𝑧 = 0 It has dimension equal to 2 and every vector 𝑢 = (𝑥, 𝑦, 𝑧) ∈ 𝑃 is such that < 𝑢, 𝑣 ≥ 0 where 𝑣 = (3,3,1) Question #11 I'm having trouble with these types of questions. I have the following vector u=(4,7,-9) and it wants me to find 2 vectors that are perpendicular to this one. I know that (4,7, −9) ⋅ (𝑥, 𝑦, 𝑧) = 0 Answer: Take vector 𝑛 = (𝑥, 𝑦, 𝑧) . As you said by yourself, dot product should vanish. So (4,7, −9) ⋅ (𝑥, 𝑦, 𝑧) = 4𝑥 + 7𝑦 − 9𝑧 = 0 As you might see, all points that lie on the plane 4𝑥 + 7𝑦 − 9𝑧 = 0 satisfy the condition of perpendicularity. If you want two linear independent vectors, just pick two different points. So, pick any two different triples 𝑛1 = (𝑥1 , 𝑦1 , (4𝑥1 + 7𝑦1 )/9) and 𝑛2 = (𝑥2 , 𝑦2 , (4𝑥2 + 7𝑦2 )/9) where (𝑥1 , 𝑦1 ) ≠ (𝑥2 , 𝑦2 ) , and you're done. Question #12 How to calculate the intersection of two planes ? These are the planes and the result is gonna be a line in 𝑅𝟛 : 𝑥 + 2𝑦 + 𝑧 − 1 = 0 2𝑥 + 3𝑦 − 2𝑧 + 2 = 0 Answer: You need to solve the two equations 𝑥 + 2𝑦 + 𝑧 − 1 = 0 2𝑥 + 3𝑦 − 2𝑧 + 2 = 0 Notice that, these are two equations in three variables, so you have a free variable say z=t, then we have x+2y=1-t 2x+3y=2t-2 Solving the last system gives {𝑥 = −7 + 7𝑡, 𝑦 = 4 − 4𝑡} Then the parametrized equation of the line is given by (𝑥, 𝑦, 𝑧) = (−7 + 7𝑡, 4 − 4𝑡, 𝑡) = (−7,4,0) + (7, −4,1)𝑡 Question #13 Find the points on the surface 𝑦 2 = 9 + 𝑥𝑧 that are closest to the origin. Answer: Let (x,y,z) be a point on the surface of 𝑦 2 = 9 + 𝑥𝑧 . The distance of this point from the origin is given by: 𝑑 = √(𝑥 − 0)2 + (𝑦 − 0)2 + (𝑧 − 0)2 = √𝑥 2 + 𝑦 2 + 𝑧 2 To minimise the distance, it is simpler to minimise the function 𝑑 2 . 𝑑2 = 𝑥 2 + 𝑦2 + 𝑧 2 ; ∵ 𝑦 2 = 9 + 𝑥𝑧 𝑑 2 = 𝑓(𝑥, 𝑧) = 𝑥 2 + 9 + 𝑥𝑧 + 𝑧 2 𝑓𝑥 = 2𝑥 + 𝑧; 𝑓𝑧 = 𝑥 + 2𝑧 Now when we solve 𝑓𝑥 = 0 and 𝑓𝑧 = 0 , we get only once critical point (0,0). 𝑓𝑥𝑥 = 2; 𝑓𝑥𝑧 = 1; 𝑓𝑧𝑧 = 2 2 𝐷 = 𝑓𝑥𝑥 𝑓𝑧𝑧 − 𝑓𝑥𝑧 𝐷(0,0) = 3 − 1 = 3 > 0; 𝑓𝑥𝑥 > 0 Hence the point (0,0) is a point of minima using the second derivatives test. After substituting we get 𝑦 2 = 9 ⇒ 𝑦 = ±3 . So the 2 points that are closest to the surface are (0,3,0) and (0, -3, 0). Result: (0,3,0) and (0, -3, 0) Question #14 Find the angle between the vectors v and w in the following: 𝑣 = (4,1)𝑇 , 𝑤 = (3,2)𝑇 . Answer: 𝑣 = (4,1)𝑇 and 𝑤 = (3,2)𝑇 . Let 0 ≤ 0 ≤ 𝜋 be the angle between v and w. We have that cos 0 = 𝑣𝑇 𝑤 . ||𝑣||||𝑤|| Since ||𝑣|| = √42 + 12 = √16 + 1 = √17, ||𝑤|| = √32 + 22 = √9 + 4 = √13 and 𝑣 𝑇 𝑤 = 4 ∗ 3 + 1 ∗ 2 = 12 + 2 = 14 we have that cos 0 = 14 √17√13 = 14√221 221 and then 0 = arccos ( 14√221 221 ). Result: The angle between v an w is arccos ( 14√221 221 ). Question #15 Find the values of x such that the vectors << 3, 2, x>> and\<< 2x, 4, x>> are orthogonal. Answer: Since 𝑎 ∗ 𝑏 = |𝑎||𝑏| cos 0, the dot product is zero when the vectors are perpendicular. Therefore, we need to find x, such that ≪ 3,2, 𝑥 ≫ ∗ ≪ 2𝑥, 4, 𝑥 ≫ = 0 Remember that: ≪ 𝑎1 , 𝑏1 , 𝑐1 ≫∗≪ 𝑎2 , 𝑏2 , 𝑐2 ≫= 𝑎1 𝑎2 + 𝑏1 𝑏2 + 𝑐1 𝑐2 This is because when it comes to the unit vectors i,j,k, their dot product with themselves is 1 and their dot product with the other two is 0. (3)(2𝑥) + (2)(4) + (𝑥)(𝑥) = 0 6𝑥 + 8 + 𝑥 2 = 0 Factorize (𝑥 + 2)(𝑥 + 4) = 0 Using the zero product property, we can write 𝑥 + 2 = 0, 𝑥 + 4 = 0 𝑥 = −2, 𝑥 = −4 Result: 𝑥 = −2, 𝑥 = −4 Question #16 Determine whether the following vectors are linearly independent in 𝑃3 : 𝑥 + 2, 𝑥 + 1, 𝑥 2 − 1 Answer: Consider the vector 𝑥 + 1, 𝑥 + 2, 𝑥 2 − 1 in 𝑃3 . Take 𝑐1 , 𝑐2 , 𝑐3 ∈ 𝑅 such that 𝑐1 (𝑥 + 1) + 𝑐2 (𝑥 + 2) + 𝑐3 (𝑥 2 − 1) = 0. Then we have that 𝑐1 𝑥 + 𝑐1 + 𝑐2 𝑥 + 2𝑐2 + 𝑐3 𝑥 2 − 𝑐3 = (𝑐1 + 2𝑐2 − 𝑐3 ) + (𝑐1 + 𝑐2 )𝑥 + 𝑐3 𝑥 2 = 0 + 0𝑥 + 0𝑥 2 . Which gives us that 𝑐1 + 2𝑐2 − 𝑐3 = 0, 𝑐1 + 𝑐2 and 𝑐3 = 0. Since 𝑐1 + 2𝑐2 − 𝑐3 = 0 and 𝑐3 = 0 we have that 𝑐1 + 2𝑐2 = 0. Since 𝑐1 + 2𝑐2 = 0 and 𝑐1 + 𝑐2 = 0 we have that (𝑐1 + 2𝑐2 ) − (𝑐1 + 𝑐2 ) = 0 − 0 and then 𝑐2 = 0. Since 𝑐1 + 𝑐2 = 0 and 𝑐2 = 0 we have that 𝑐1 = 0. Hence 𝑐1 = 𝑐2 = 𝑐3 = 0 and then x+1, x+2 and 𝑥 2 − 1 are linearly independent. Result: x+1, x+2 and 𝑥 2 − 1 are linearly independent. Question #17 At what point on the paraboloid 𝑦 = 𝑥2 + 𝑧2 is the tangent plane parallel to the plane 𝑥 + 2𝑦 + 3𝑧 = 1? Answer: Given 𝑦 = 𝑥 2 + 𝑧 2 which is parallel to plane 𝑥 + 2𝑦 + 3𝑧 = 1 Let 𝐹(𝑥, 𝑦, 𝑧) = 𝑥 2 − 𝑦 + 𝑧 2 The gradient is the normal vector, or the direction vector of F. gradient of 𝐹 =< 𝑓𝑥 , 𝑓𝑦 , 𝑓𝑧 > =< 2𝑥, −1,2𝑧 > Since F is parallel to the plane, their direction vectors are multiples of each other. c is a constant. < 2𝑥, −1,2𝑧 ≥< 1,2,3 > 𝑐 -1=2c 1 𝑐=− 2 2x=c 1/2 1 𝑥=− =− 2 4 2z=3c 1 3 2𝑧 = 3 (− ) = − 2 2 3 𝑧=− 4 Plug x and z into original given equation to find y. 1 2 3 2 5 𝑦 = 𝑥 2 + 𝑧 2 = (− ) + (− ) = 4 4 8 1 5 3 Answer: (− , , − ) 4 8 4 1 5 3 4 8 4 Result: Answer: (− , , − ) Question #18 Find the dimension of each of the following vector spaces. (a) The vector space of all diagonal 𝑛 × 𝑛 matrices. (b) The vector space of all symmetric n \times n matrices. (c) The vector space of all upper triangular 𝑛 × 𝑛 matrices. Answer: Because a diagonal matrix will have only n entries in total so dimension of diagonal matrix will be equal to square of its no. of diagonal elements. (a) = 𝑛2 As for symmetric matrix 𝐴’ = 𝐴 so two elements will be same for this matrix & the no. of all such combinations will be any 2 out of n i.e. (nC2) (b) = 𝑛(𝑛+1) 2 Similarly for a tringular matrix no. of all possible combinations will be the same i.e. 𝑛(𝑛 + 1)/2 (c) = 𝑛(𝑛+1) 2 Question #19 Determine the velocity vector of the given path. 𝑐(𝑡) = (sin 3 𝑡)𝑖 + (cos 3 𝑡)𝑗 + 2𝑡 3/2 𝑘 Answer: We can rewrite the given parametrization: 𝑐(𝑡) = (sin(3𝑡) , cos(3𝑡) , 2𝑡 3/2 ) To obtain the velocity vector calculate the derivatives component-wise: 𝑐′(𝑡) = (( sin (3𝑡))′, ( cos (3𝑡)′, (2𝑡^{3/2})′) = (3 cos (3𝑡), −3 sin (3𝑡),3𝑡^{1/2}) Result: To obtain the velocity vector calculate the derivatives component-wise: ′ 𝑐 ′ (𝑡) = ((sin(3𝑡))′ , (cos(3𝑡))′ , (2𝑡 3/2 ) ) = (3 cos(3𝑡) , −3 sin(3𝑡) , 3𝑡 1/2 ) Question #20 Find a vector function that represents the curve of intersection of the two surfaces. The cylinder 𝑥 2 + 𝑦 2 = 4 and the surface z=xy Answer: Using the aforementioned values, we can find the parameterization for the intersection of the circular sylinder 𝑥 2 + 𝑦 2 = 4 in 𝑅𝟛 and the plane z=xy ny substituting 𝑥 = 𝑟 𝑠𝑖𝑛 and 𝑦 = 𝑟 𝑐𝑜𝑠 with 𝑟 = √4 = 2 to get: z=xy = (2 sin 𝑡)(2 cos 𝑡) = 4 sin 𝑡 𝑐𝑜𝑠 = 2 sin 2 𝑡 Thus we can write the vector function r(t) as follows: 𝑟(𝑡) =< 2 sin 𝑡 , 2 cos 𝑡 , 2 sin 2 𝑡 > There are many possible answers to this. Another way of parameterizing a circle is by letting 𝑥 = 𝑟 𝑐𝑜𝑠 and 𝑥 = 𝑟 sin