#Name: Irina Mukhametzhanova, Section 24
#This homework is OK to post

#12.3 Homework
#Exercise 1:
#<1,2,1> . <4,3,5> = 1*4 + 2*3 + 1*5 = 4 + 6 + 5 = 15

#Exercise 13:
#Vectors are orthogonal if their dot product equals to 0
#<1,1,1>*<1,-2,-2> = 1*1 + 1*(-2) + 1*(-2) = 1 - 2 - 2 = -3
#Because the dot product is negative, the angle between the two vectors is Obtuse

#Exercise 21:
#i + j -> <1,1,0>
#j + 2k -> <0,1,2>
with(LinearAlgebra);
DotProduct([1, 1, 0], [0, 1, 2]);
#Result: 1
#cos(angle) = dot product / product of vectors' magnitudes
Magnitude := v -> sqrt(v[1]*v[1] + v[2]*v[2] + v[3]*v[3]);
cosine := 1/(Magnitude([1, 1, 0])*Magnitude([0, 1, 2]));
#Result: sqrt(2)*sqrt(5)/10
#sqrt(2)*sqrt(5)/10 = sqrt(10)/10 = 1/sqrt(10)
#ANSWER: The cosine of the angle between the vectors is 1/sqrt(10)

#Exercise 29:
#a)
#<b,3,2>.<1,b,1> = b*1 + 3*b + 2*1 = 4b + 2 = 0
#4b = -2
#b = -0.5

#b)
#<4,-2,7>.<b^2,b,0> = 4*b^2 - 2*b + 7*0 = 4b^2 - 2b = 0
#b(4b-2) = 0
#b = 0, 0.5

#Exercise 31:
#Let vector 1 be v1 = <x,y,z>
#Then, <2,0,-3>.<x,y,z> = 2*x + 0*y - 3*z = 2x - 3z = 0
#ANSWER: Two vectors that fit this equation are <3,0,2> and <0,1,0>

#Exercise 57:
#Formula for projection of u onto v: DotProduct(u,v)*v/Magnitude(v)^2
#u = 5i + 7j - 4k = <5,7,-4>
#v = k = <0,0,1>
with:(LinearAlgebra):
DotProduct([5,7,-4],[0,0,1]);
#Result: -4
Magnitude := v -> sqrt(v[1]*v[1] + v[2]*v[2] + v[3]*v[3]);
answer := -4 * [0,0,1] / Magnitude([0,0,1])^2;
#ANSWER: [0,0,-4]

#Exercise 63:
#OP is a projection of u onto v
with:(LinearAlgebra):
DotProduct([3,5],[8,2]);
#Result: 34
Magnitude := v -> sqrt(v[1]*v[1] + v[2]*v[2]);
answer := 34 * [8,2] / Magnitude([8,2])^2;
#Result: [4,1]
OP = Magnitude([4,1]);
#ANSWER: OP has a length of sqrt(17)


#12.4 Homework
#Exercise 1:
#First, we can do this determinant by hand:
#1*3 - 4*2 = 3 - 8 = -5
#We can check this with Maple
with(linalg):
det([[1,2],[4,3]]);
#Result: -5

#Exercise 5:
#First, we can do this determinant by hand:
#1*(-3*1 - 0*0) - 2*(4*1 - 1*0) + 1*(4*0, - (-3)*1) = 1*(-3) - 2*(4) + 1*(3) = -3 - 8 + 3 = -8
#We can check this with Maple
with(linalg):
det([[1,2,1],[4,-3,0],[1,0,1]]);
#Result: -8

#Exercise 13:
#(i+j) x k = ixk + jxk
#Use the right hand rule to find the answer
#ixk + jxk = -j + i = i - j
#Answer: i - j

#Exercise 21:
#One property of cross product is u x (w+v) = uxw + wxv
#So, (u-2v) x (u+2v) can be written as:
#(u-2v) x u + (u-2v) x 2v = (u x u + (-2v) x u) + 2*((u-2v) x v) =
# = (0 - 2*(v x u)) + 2*(u x v + (-2v) x v) = -2*(v x u) + 2*(u x v + -2*(v x v)) = -2*(v x u) + 2*(u x v) = 2*(u x v) + 2*(u x v) =
# = 4*(u x v) = 4*[1,1,0] = 
# = [4,4,0]

#Exercise 25:
#Using the right hand rule, we see that v x w is equal to -u

#Exercise 27:
#The cross product must be orthogonal to both vectors.
#Because v is just on the x-axis, the cross product will have 0 as its x-component
#u = v x w = [0,n2,n3], where n2 and n3 are real integers
#u must be orthogonal to w:
#u . w = 0
#[0,n2,n3] . [0,1,-1] = n2 - n3 = 0
#So, n2 = n3
#Now we can use the magnitude to find out the components. The magnitude is also the area of a parallellogram formed by the two vectors.
#The area of a parallelogram is the product of its base and height:
#3 * sqrt(2)
#Now we go back to the equation for the cross product vector:
sqrt(n2^2 + n3^2) = sqrt(n2^2 + n2^2) = sqrt(2*n2^2) = sqrt(2)*sqrt(n^2) = sqrt(2)*n2 = 3*sqrt(2)
#So, n2 = n3 = 3
#The cross product of v and w is [0,3,3]

#Exercise 39:
#We can find the volume of the paralellipiped by finding the area of one of its sides (or a cross product of 2 of the vectors) and multiplying it by its thickness (component of 3rd vector perpendicular to that side)
with(LinearAlgebra):
CrossProduct([1,0,0],[0,2,0]);
#Result: [0,0,2]
Magnitude := v -> sqrt(v[1]*v[1] + v[2]*v[2] + v[3]*v[3]);
Magnitude([0,0,2]);
#Result: 2
#The component of 2 perpendicular to u and v is its z-component, or 2
Area := 2 * 2;
#Result: The volume of the paralellepiped is 4 cubic units

#Exercise 41:
with(LinearAlgebra):
CrossProduct([1,0,3],[2,1,1]);
#Result: [-3,5,1]
Magnitude := v -> sqrt(v[1]*v[1] + v[2]*v[2] + v[3]*v[3]);
Magnitude([-3,5,1]);
#Result: sqrt(35)

#Exercise 43:
#A triangle can be a paralellogram divided in half along its diagonal. So, we can find the area of the triangle by taking the magnitude of the cross product of 2 vectors and dividing it by 2
O := [0,0,0];
P := [3,3,0];
Q := [0,3,3];
V1 := P-O;
V2 := Q-O;
#Results: V1 = [3,3,0] and V2 = [0,3,3]
#The cross product of the two vectors is:
with(LinearAlgebra):
CrossProduct([3,3,0],[0,3,3]);
#Result: [9, -9, 9]
#Area of parallelogram = magnitude = sqrt(9^2 + (-9)^2 + 9^2) = 9*sqrt(3)
#Area of triangle = 0.5 * area of parallelogram = (9/2)*sqrt(3)

#Exercise 45:
#A triangle can be a paralellogram divided in half along its diagonal. So, we can find the area of the triangle by taking the magnitude of the cross product of 2 vectors and dividing it by 2
v1 := [1,2] - [-2,2];
v2 := [3,4] - [1,2];
#Result: v1 = [3,0], v2 = [2,2]
#The CrossProduct function needs 3D vectors, so, we can add 0 as the z-value to both v1 and v2
with(LinearAlgebra):
CrossProduct([3,0,0],[2,2,0]);
#Result: [0,0,6]
Magnitude := v -> sqrt(v[1]*v[1] + v[2]*v[2] + v[3]*v[3]);
Magnitude([0,0,6]);
#Result: 6
Area := 0.5 * 6
#ANSWER: 3


#12.5 Homework
#Exercise 1:
#If you have a normal vector [n1,n2,n3] perpendicular to the plane, and a point (a,b,c) ON that plane, then the equation of the plane is:
#n1*(x-a) + n2*(y-b) + n3*(z-c) = 0
#Our normal vector is [1,3,2] and out point is (4,-1,1), so, we plug them into the equation:
#1*(x-4) + 3*(y+1) + 2*(z-1) = 0
#We expand the equation to get to the scalar form:
#x-4 + 3*y+3 + 2*z-2 = 0
#Put the constants onto the right side of the equation:'
#x + 3*y + 2*z = 4 - 3 + 2
#The equation of the plane is x + 3*y + 2*z = 3

#Exercise 5
#If you have a normal vector [n1,n2,n3] perpendicular to the plane, and a point (a,b,c) ON that plane, then the equation of the plane is:
#n1*(x-a) + n2*(y-b) + n3*(z-c) = 0
#Our normal vector is i (or [1,0,0]) and out point is (3,1,-9), so, we plug them into the equation:
#1*(x-3) + 0*(y-1) + 0*(z+9) = 0
#We expand the equation to get to the scalar form:
#x-3 = 0
#Put the constants onto the right side of the equation:'
#x = 3
#The equation of the plane is x = 3

#Exercise 9:
#Because the plane has to go through the origin, we would need to use (0,0,0) as a point ON the plane.
#The plane's normal vector can point in any direction, so, we can create any vector
#One example of a plane going through the origin is:
#x + y + z = 0

#Exercise 11:
#a) If a plane is parallel to the yz-plane, then its normal vector would be parallel to the x-axis. Vector [0,0,1] is not parallel to the x-axis, but is to the z-axis. So, this is FALSE
#b) Following the reasoning in answer to a), this is TRUE, because it is parallel to the x-axis
#c) The plane parallel to the yz-plane shouldn't be affected by its y and z values. So, its equation cannot contain y or z, and this statement is FALSE
#d) Following the reasoning in answer to c), this statement is TRUE

#Exercise 13:
#We can get the normal vector from the coefficients of x, y, and z.
#The normal vector of the plane 9*x - 4*y - 11*z = 2 is [9,-4,-11]

#Exercise 15:
#If you have a normal vector [n1,n2,n3] perpendicular to the plane, and a point (a,b,c) ON that plane, then the equation of the plane is:
#n1*(x-a) + n2*(y-b) + n3*(z-c) = 0
#So, the normal vector of the plane 3*(x-4) - 8*(y-1) + 11*z = 0 is [3,-8,11]

#Exercise 17:
#To find the equation, we would also need to find the norma vector.
#To find it, we can take the cross product of any 2 vectors from the 3 points, since it will perpendicular to them both
P := [2,-1,4];
Q := [1,1,1];
R := [3,1,-2];
V1 := Q-P;
V2 := R-P;
#Results: V1 = [-1,2,-3] and V2 = [1,2,-6]
with(LinearAlgebra):
CrossProduct(V1,V2);
#Result: [-6,-9,-4]
#We now use that normal vector and any of the 3 points (Q) to write the equation of the plane
#-6*(x-1) - 9*(y-1) - 4*(z-1) = 0
#-6*x+6 - 9*y+9 - 4*z+4 = 0
#-6*x - 9*y - 4*z = -19
#We can multiply both sides by -1 to get rid of the negatives
#So, the equation of the plane is 6*x + 9*y + 4*z = 19

#Exercise 19:
#To find the equation, we would also need to find the norma vector.
#To find it, we can take the cross product of any 2 vectors from the 3 points, since it will perpendicular to them both
P := [1,0,0];
Q := [0,1,1];
R := [2,0,1];
V1 := Q-P;
V2 := R-P;
#Results: V1 = [-1,1,1] and V2 = [1,0,1]
with(LinearAlgebra):
CrossProduct(V1,V2);
#Result: [1,2,-1]
#We now use that normal vector and any of the 3 points (Q) to write the equation of the plane
#1*(x-0) + 2*(y-1) - 1*(z-1) = 0
#x + 2*y-2 - z+1 = 0
#x + 2*y - z = 2 - 1 = 1
#So, the equation of the plane is x + 2*y - z = 1

#Exercise 25:
#If you have a normal vector [n1,n2,n3] perpendicular to the plane, and a point (a,b,c) ON that plane, then the equation of the plane is:
#n1*(x-a) + n2*(y-b) + n3*(z-c) = 0
#Our normal vector is i+k (or [1,0,1]) and out point is (-2,-3,5), so, we plug them into the equation:
#1*(x+2) + 0*(y+3) + 1*(z-5) = 0
#We expand the equation to get to the scalar form:
#x+2 + z-5 = 0
#Put the constants onto the right side of the equation:'
#x + z = -2 + 5 = 3
#The equation of the plane is x + z = 3

#Exercise 31:
plot3d(4 - x - y, x = -5 .. 5, y = -5 .. 5);

#Exercise 53:
#The equation of a plane is a*x + b*y + c*z = d
#To find where the plane crosses the xz-plane, or the y-intercept, we plug y = 0 into the equation:
#a*x + c*z = d
#So, we need to see for what values of a, c, and d the equation is equal to 3*x + 2*z = 5
#a can be a multiple of 3
#b can be any value
#c can be a multiple of 2
#d can be a multiple of 5
#So, the final equation of the plane that contains that line is:
#(3*u)*x + b*y + (2*u)*z = 5*u, where u =/= 0


#13.1 Homework
#Exercise 5:
#To find the parametric equations of the line, we can use this formula:
#r(t) = P + t*v
#Where P is a point on the line, and v is a vector parallel to the line
#We have a point P(3,-5,7) and the vector[3,0,1] that we can use:
#r(t) = (3,-5,7) + t*[3,0,1]
#r(t) = [3+3*t, -5, 7+t]
#The vector parametrization of the line is (3+3*t)i - 5*j + (7+t)*k

#Exercise 17:
#To find the radius of the circle, we can find the magnitude of the position vector of that circle:
#radius = sqrt(9^2 * (cos(t))^2 + 9^2 * (sin(t))^2) = sqrt(9^2((cos(t))^2 + (sin(t))^2) = sqrt(9^2) = 9
#There aren't any translations in the vector, so, the center of the circle is at (0,0,0)
#The plane containing the circle has the origin, and a normal vector of k (or [0,0,1]). So, its equation is:
#z = 0 -> the xy-plane
#So, this position vector represents a circle with a radius 9, placed at the origin, lying on the xy-plane


#13.2 Homework
#Exercise 3:
#To evaluate the limit of this vector, we evaluate the limit of each component:
#lim_t->0(e^(2*t)) = e^(2*0) = 2^0 = 1
#lim_t->0(ln(t+1)) = ln(1) = 0
#lim_t->0(4) = 4
#The limit of the equation is i + 4k

#Exercise 5:
#Using the equation provided in the exercise, we take the limit of all of the components of the vector:
#i-component: lim_h->0((r(t+h) - r(t))/h) = lim_h->0(((t+h)^(-1)*i + sin(t+h) + 4 - t^(-1) - sin(t) - 4)/h) = lim_h->0(((t+h)^(-1) - t^(-1))/h) = lim_h->0((1/(t+h) + 1/t)/h) = lim_h->0(((t-t-h)/t*(t+h))/h) = lim_h->0(-h/(t*(t+h))/h) = lim_h->0(-1/(t*(t+h))) = -1/(t^2)
#j-component: lim_h->0((sin(t+h) - sin(t))/h) 
#Use the rule: sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b)
#lim_h->0((sin(t)*cos(h) + cos(t)*sin(h) - sin(t))/h) = lim_h->0((sin(t)*(cos(h)-1) + cos(t)*sin(h))/h) 
#We can use the rules: lim_h->0((cos(h)-1)/h) = 0,
#and lim_h->0(sin(h)/h) = 1
#sin(t)*lim_h->0((cos)h)-1)/h) + cos(t)*lim_h->0(sin(h)/h) = cos(t)
#k-component: lim_h->0(4) = 0
#The limit of the equation is [-1/(t^2), cos(t), 0]


#Exercise 7:
#The derivative of this vector is the derivative of each of its components:
#d/dt t = 1
#d/dt t^2 = 2*t
#d/dt t^3 = 3*t^2
#So, the derivative of r(t) = [t,t^2,t^3] is [1,2*t,3*t^2]

#Exercise 15:
#To find the tangent vector, we take the derivative of that vector:
#r1'(t) = [(t)',(t^2)'] = [1,2*t]
#So, at t=1, the tangent vector is equal to:
#[1,2*1] = [1,2]
#The curve that this parametric equation represents a parabola on the origin.
#And here is the sketch (the magnitude of the vector going along the line is sqrt(5)):
plot([x^2, 2*x - 1], x = 0 .. 2);
#The tangent vector of r2, following the same logic, would be:
r2'(t) = [(t^3)',(t^6)'] = [3*t^2, 6*t^5]
#So, at t=1, the tangent vector is equal to:
#[3*(1)^2,6*(1)^5] = [3,6] 
#The curve that this parametric equation represents is also a parabola on the origin.
#And here is the sketch (the magnitude of the vector going along the line is 3*sqrt(5)):
plot([x^2, 2*x - 1], x = 0 .. 2);

#Exercise 31:
#To find the tangent line, we first need to find the tangent vector that it is parallel to. To find that, we take the derivative of r(t), and all of its components:
#r'(t) = [(1-t^2)',(5*t)',(2*t^3)'] = [-2*t,5,6*t^2]
#Find the vector at t=2:
#r'(2) = [-2*2,5,6*2^2] = [-4,5,24]
#We would also need a point on the line. For that, we plug in t=2 into the position equation:
#r(2) = [1-2^2,5*2,2*2^3] = [-3,10,16]
#Now we can find the parametrization of the tangent line using the formula P + t*v
#(-3,10,16) + t*[-4,5,24] = [-3-4*t,10+5*t,16+24*t]
#The parametrization of the tangent line is l(t) = [-3-4*t,10+5*t,16+24*t]

#Exercise 33:
#To find the tangent line, we first need to find the tangent vector that it is parallel to. To find that, we take the derivative of r(s), and all of its components:
#r'(s) = [(4*s^(-1))',(0)',(-(8/3)*s^(-3))'] = [-4*s^(-2),0,8*s^(-4)]
#Find the vector at s=2:
#r'(2) = [-4*2^(-2), 0, 8*2^(-4)] = [-1,0,1/2]
#We would also need a point on the line. For that, we plug in s=2 into the position equation:
#r(2) = [4*2^(-1),0,-(8/3)*2^(-3)] = [2,0,-1/3]
#Now we can find the parametrization of the tangent line using the formula P + t*v
#(2,0,-1/3) + t*[-1,0,1/2] = [2-t,0,-1/3 + (1/2)*t]
#The parametrization of the tangent line is l(t) = [2-t,0,-1/3 + (1/2)*t]

#Exercise 41:
#To find the intergal of a vector, we take the integral of all of its components:
#i-component: integral from -2 to 2 of u^3 = (u^4/4) from -2 to 2 = ((2)^4/4) - ((-2)^4/4) = 0
#j-component: integral from -2 to 2 of u^5 = (u^6/6) from -2 to 2 = ((2)^6/6) - ((-2)^6/6) = 0
#So, the integral of the vector u^3*i + u^5*j is 0, or [0,0]
#Maple confirms our results:
int(t^5 + t^3, t = -2 .. 2);
#Result: 0

#Exercise 49:
#To find the general solution of the differential equation (or r(t), in this case), we take the integral of r'(t):
int((t^2)*i + (5*t)*j + k);
#Result: r(t) = ((1/3)*t^3)*i + ((5/2)*t^2)*j + t*k + C
#We now would need to find the arbitrary constant C. Use what we already know:
#r(1) = ((1/3)*1^3)*i + ((5/2)*1^2)*j + k + C = (1/3)*i + (5/2)*j + k + C = j + 2*k
#C = -(1/3)*i + j - (5/2)*j + 2*k - k
#C = -(1/3)(i - (3/2)*j + k
#The general solution of the differential equation is:
#r(t) = ((1/3)*t^3 - (1/3))*i + ((5/2)*t^2 - (3/2))*j + (t+1)*k