Assignment 3. Solutions. Problems. February 22.


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1 Assignment. Solutions. Problems. February.. Find a vector of magnitude in the direction opposite to the direction of v = i j k. The vector we are looking for is v v. We have Therefore, v = =. v v = v = i + j + k.. Given P (, 4, 5) and P (4,, 7), find a) the direction of P P, b) the midpoint of the line segment P P. a) P P =, 6, P P = = 7. Therefore, the direction of P P is P P P P = 7, 6 7, 7. b) The midpoint of the line segment P P is ( + 4, 4, ) = (5,, 6).. Let u = i j, v = i + j, and w = i + j. Write u = u + u, where u is parallel to v and u is parallel to w. Since u is parallel to v and u is parallel to w, we have u = av and u = bw, for some numbers a and b. Thus, u = av + bw,
2 u = i j = a(i + j) + b(i + j) = (a + b)i + (a + b)j. Equating the corresponding coefficients, we get a system of two equations with unknowns a and b. { a + b =, a + b =. From here we get a = and b = 7. So, u = 6i 9j and u = 7i + 7j. 4. Find the coordinates of the point Q that divides the segment from P (x, y, z ) to P (x, y, z ) into two lengths whose ratio is p/q = r. We are given P Q / QP = r. So, P Q = r QP. Adding QP to both sides we get P P = (r + ) QP. So, QP = r + P P. Since P Q is parallel to P P, we see that So, P Q = r + P P = r + x x, y y, z z. OQ = OP + P Q = x, y, z + r + x x, y y, z z = r + x + r r + x, r + y + r r + y, Thus, the coordinates of the point Q are ( x + rx r +, y + ry r +, z ) + rz. r + r + z + r r + z. Problems. March.. v = i + j, u = i + j + k. Find a) v u, v, u, b) the cosine of the angle between v and u, c) the scalar component of u in the direction of v, d) the vector projection proj v u. a) v u = +, v =, u =.
3 b) cos θ = v u v u =. c) v u =. v d) proj v u = v u v v = ( i + j).. Find the measures of the angles between the diagonals of the rectangle whose vertices are A(, ), B(, ), C(, 4), D(4, ). The diagonals are AC =, 4 and BD = 4,. AC BD =. Therefore, the diagonals meet at 9.. u = j + k, v = i + j Write u as the sum of a vector parallel to v and a vector orthogonal to v. u = proj v u + (u proj v u) So, proj v u = v u v v = (i + j) = i + j. ( u proj v u = j + k i + ) j = i + j + k. u = ( i + ) j + ( i + ) j + k, where the first vector is parallel to v and the second vector is orthogonal to v. 4. Suppose that AB is the diameter of a circle with center O and that C is a point on one of the two arcs joining A and B. Show that CA and CB are orthogonal. We have CA = v + ( u), CB = v + u. Then CA CB = ( v u) ( v+u) = v v v u+u v u u = v u =, since both u and v are unit vectors. CA CB = means that CA and CB are orthogonal. Problems. March.
4 . u = i j + k, v = i + j + k. Find the length and direction of u v and v u. u v = = i j + k. Length of this vector: u v = =. u v Its direction: u v = i j + k. Since v u = u v, it has the same length and the opposite direction: i + j k.. P (,, ), Q(,, ), R(,, ). a) Find the area of the triangle determined by the points P, Q and R. b) Find a unit vector perpendicular to the plane P QR. P Q =,,, P R =,,. P Q P R = P Q P R = 6. The area of the triangle: P Q P R =. = 4i + 4j k. A unit vector perpendicular to the plane P QR: (or its opposite). P Q P R = P Q P R i + j k.. u = i + j k, v = i k, w = i + 4j k. Verify that (u v) w = (v w) u = (w u) v and find the volume of the parallelepiped determined by u, v, and w. (u v) w = 8. u v = = i + j + k.
5 (v w) u = 8. (w u) v = 8. v w = w u = 4 4 The volume of the parallelepiped is 8. = 4i 4j 4k. = 6i + j k. 4. Let u = i+j k, v = i+j+k, w = i+k, r = (π/)i πj+(π/)k. Which vectors are a) perpendicular? b) parallel? a) u v =, u w =, u r = π, v w =, v r =, w r =. Vectors with zero dot product are mutually orthogonal. b) u and r are parallel since r = π/u. All other vectors cannot be parallel, since they are mutually orthogonal. Problems. March 5.. Find parametric equations for the line through P (,, ) and Q(,, ). P Q =,,. Therefore the line is given by the equations x = t, y = t, z = + t, t R.. Find parametric equations for the line through the point (,, ) parallel to the line x = + t, y = t, z = t. The line given above is parallel to the vector,,. Therefore, the line we are looking for has equations x = + s, y = s, z = + s, s R.
6 . Find parametric equations for the line through (, 4, 5) perpendicular to the plane x + 7y 5z =. The line is parallel to the plane s normal vector, 7, 5. So, the line is given by the equations x = + t, y = 4 + 7t, z = 5 5t, t R. 4. Find parametric equations for the line segment joining the points (,, ) and (,, ). where t. r(t) = t,, + ( t),, = t, t, t, 5. Find the distance from the point (,, ) to the line x = t, y = +t, z = t. Denote P (,, ) and S(,, ). Clearly, S belongs to the line. Then SP =,,. Consider the vector v =,,, parallel to the line. The distance from P to the line equals d = SP v. v We have So, SP v = = i 6j + 4k. SP v = 56 = 4. v =. d = 4. Problems. March 8.
7 . Find an equation of the plane through (,, ) parallel to the plane x + y + z = 7. The desired plane has normal vector,,. Therefore, its equation is (x ) + (y + ) + (z ) =, x + y + z 5 =.. Find the plane determined by the intersecting lines. L : x = t, y = t, z = t; t R, L : x = + s, y = 4 + s, z = + s; s R. Let us find the point of intersection of these two lines. Equating the corresponding coordinates, we get the following three equations. t = + s, t = 4 + s, t = + s. Adding st and rd equations we get s =, therefore t =. We also see that these two values satisfy nd equation. Putting t = into the equations of line L, we get the point of intersection (,, ). In order to obtain a vector n normal to the plane, we compute the cross product of the vectors,, and,, that are parallel to the given lines. n = = i j + 4k. In fact, as a normal vector we can take another vector parallel to the latter one: i + j k. Therefore, the plane has the equation (x ) + (y ) (z + ) =, x + y z 7 =.. Find the distance from the point (,, ) to the plane x + y + z = 4. d = = 8.
8 4. Find the angle between the planes 5x+y z = and x y+z =. The angle between the planes is the acute angle between their normals n = 5,, and n =,,. cos θ = n n n n =. So, the planes are orthogonal (the angle is 9 ). Problems. March.. r(t) = (sec t)i + (tan t)j + 4 tk is the position of a particle in space at time t. Find the particle s velocity and acceleration vectors. Then find the particle s speed and direction of motion at t = π/6. Write the particle s velocity at that time as the product of its speed and direction. Velocity: v(t) = (sec t tan t)i + (sec t)j + 4 k Acceleration: a(t) = (sec t tan t + sec t)i + ( sec t tan t)j Velocity at t = π/6: v(π/6) = i + 4 j + 4 k 4 Speed at t = π/6: v(π/6) = =. Direction at t = π/6: v(π/6) v(π/6) = i + j + k Velocity at t = π/6 as the product of velocity and direction:. Evaluate the integral π/ [(sec t tan t)i + (tan t)j + ( sin t cos t)k] dt. ( i + j + ) k π/ [(sec t tan t)i + (tan t)j + ( sin t cos t)k] dt = [ (sec t)i (ln cos t)j + (sin t)k ] π/ = ( )i (ln ln )j + ( 4 )k = = i + ln j + 4 k
9 . Solve the initial value problem. Integrating we get Using we see that C =. So, d r = (i + j + k), dt r() = i + j + k, dr =. dt t= d r = (i + j + k), dt dr dt = (ti + tj + tk) + C. dr =, dt t= dr = (ti + tj + tk). dt Integrating one more time, we get r = (t i + t j + t k) + C. Using we see that Thus, r() = i + j + k, C = i + j + k. r = ( t )(i + j + k). 4. Find parametric equations for the line that is tangent to the curve r(t) = (cos t)i + (sin t)j + (sin t)k at t = π/. r(π/) = j So, the tangent line is v(t) = ( sin t)i + (cos t)j + ( cos t)k v(π/) = i k. x = s, y =, z = s, s R.
10 5. Show that the vectorvalued function ( r(t) = (i+j+k)+cos t i ) ( j +sin t i + j + ) k describes the motion of a particle moving in a circle of radius centered at the point (,, ) and lying in the plane x + y z =. Parametric equations of the given curve are x(t) = + cos t + sin t y(t) = cos t + sin t z(t) = + sin t Substituting these into the equation of the plane, we get ( + cos t + ) ( sin t + cos t + ) ( sin t + ) sin t =. This is true for all t. Thus the given curve lies in the given plane. Now compute the distance from the point (x(t), y(t), z(t)) to the point (,, ). d = (x(t) ) + (y(t) ) + (z(t) ) = = ( cos t + sin t) + ( cos t + ( ) sin t) + sin t cos t + cos t sin t + 6 sin t + cos t cos t sin t + 6 sin t + sin t = cos t + sin t =. So, all the points on the curve are at distance from the given point. Thus, the curve is a circle of radius. Problems. March.. Find the length of the curve r(t) = t i + t j + t k, t. L = v(t) = i + t j + t k v(t) = 4 + 4t + t 4 = + t. v(t) dt = ( + t )dt = (t + t ) = + = 7.
11 . Reparametrize the curve with respect to arc length measured from the point where t = in the direction of increasing t. r(t) = e t cos t i + j + e t sin t k. v(t) = (e t cos t e t sin t)i + (e t sin t + e t cos t) k. v(t) = (e t cos t e t sin t) + (e t sin t + e t cos t) = 4e 4t cos t 8e 4t cos t sin t + 4e 4t sin t + 4e 4t sin t + 8e 4t cos t sin t + 4e 4t cos t = 8e 4t = e t. So, s = t e τ dτ = e τ t e t = s +, t = ln( s + ), = ( e t ). t = ln( s + ). Substituting this into r(t) and using that we get e ln( s +) = s +, r(t(s)) = ( s + ) cos ln( s + ) i + j + ( s + ) sin ln( s + ) k.. Find T and κ for the curve r(t) = t i + (sin t t cos t) j + (cos t + t sin t) k, t >. v(t) = t i + (t sin t) j + (t cos t) k v(t) = 4t + t sin t + t cos t = 4t + t = t 5. T = v(t) v(t) = i + sin t j + cos t k dt dt dt dt = cos t j sin t k 5 5 cos = t + sin t 5 5 κ = dt dt / v(t) = 5t. = 5
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