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Review Test 1
Math 2243
Name
Id
Read each question carefully. Avoid simple mistakes. Put a box around the final answer to a question. (Use the back of the page if necessary). You must show your work in order to get credits or partial credits.
- Which of the points P (6, 2 , 0), Q(− 5 , 0 , 4), and R(− 1 , − 1 , 10) is closest to the plane y = x? Which point lies on the xy-plane?
- Determine the center and radius of the sphere represented by the following equation. x^2 + y^2 + z^2 = 4x − 2 y
- Given two vectors: ~a = 4i + 2j − k and ~b = −i − j + 2k, compute Projba and Compba, that is, the projection of ~a onto the normal direction that is perpendicular to ~b (such that ~a = P rojba + Compba)
- Evaluate (a + b) • (2a − 3 b) where a =< 3 , − 2 , − 4 > and b =< − 1 , 0 , 1 >. Also, find b × a.
- a) Locate the point (x 0 , y 0 , z 0 ) at which the line intersects the given plane:
x = 1 − 2 t, y = − 3 , z = 2 − t; x − y + 2z + 2 = 0
b) What is the direction angles of v = − 2 i − k.
- Find: a) A vector orthogonal to the plane containing the points: P (1, 0 , −1), Q(− 2 , 1 , 1), and R(4, − 2 , 0). b) equation of the plane; and, c) the area of the triangle 4 P QR.
- Find a unit vector that is orthogonal to both i − j and j + k
- (optional) Find the parametric equations for the line that passes through (1, 2 , 1) and is perpendicular to the line `: x = 1 + t, y = 2, z = 3 − t with the assumption that these two lines are in the same plane.
- Find an equation of the plane that passes through A = (1, − 1 , 2) and contains the line of intersection of the planes: x+y−z = 0 and y+2z −2 =
- Prove the identity a · (b × c) = −(a × c) · b by using the properties of cross product.
- Find r(t) if r′(t) = i + sin tj −
tk and r(1) = i − j.
- Locate the point of intersection between the two space curves: r 1 (t) = < 2 t, t − 1 , 1 + 2t^2 > and r 2 (s) =< 2 s + 2, 1 , (s + 1)^2 >. What is their angle of intersection?
- Find the length of the curve r(t) =
2 ti + etj + e−tk, 0 ≤ t ≤ ln 2.
- Reparametrize the curve with respect to arc length measured from the point where t = 1 in the direction of increasing t,
r(t) = cos(2t)i + sin(2t)j − 8 tk.
Use your reparametrized space curve equations to find the point on the curve where the arc length is 2 ft from the point t = 1.
- (Bonus) Compute a) the distance from the point P (1, − 1 , 1) to the line x 2 =^
y− 1 3 =^ z; b) the distance from (0,0,1) to the plane x + y + z = 1. c) Find the distance in R^3 between the x-axis(the line y = z = 0) and the line y = z = 6
x − 3 =
y − 2 2
= z + 1
d) How about the distance between the x-axis and the line y = z = 6
- Compute a) the unit tangent T, unit norm N (also called principal norm) and binormal B = T × N, b) curvature κ, c) acceleration a = aT + aN of r(t) =< t, t^2 , t^3 > at the point (− 1 , 1 , −1).
Solution to Review Test 1 (The main credits come from your idea, work, steps and the answer that are correct)
- R=(-1,-1,10); P=(6,2,0)
- Try complete the square to write the quadratic equation in an (equivalent) stand from (x − a)^2 + (y − b)^2 + (z − c)^2 = r^2
(~a × ~c) · ~b = det
a 1 a 2 a 3 c 1 c 2 c 3 b 1 b 2 b 3
From linear algebra or property of evaluating determinant of a matrix, switch two rows results in a negative sign for the determinant.
- Integrating we get r(t) =
r′(t)dt =
i +
sin tj −
tk = ti − cos tj − 2 3 t
3 / (^2) k + C. Plugging in t = 1 in r(t) above and use r(1) = i − j we can solve
the constant vector C.
- We need to find the point (a,b,c) that lies on both curves, or, satisfies both parametric equations. To do so we write down the equations:
2 t = 2s, t − 1 = 1, 1 + 2t^2 = (s + 1)^2.
Then solve it to obtain t=2, s=2. The angle means the angle between two tangents at the point (4,1,9), which can be computed using dot product (for the cosine), .... blah, blah, blah How to find the tangents at (4,1,9)? ←− use the standard formula T = dr/dt (here it is not divided by its magnitude, you can do that if you wish). The rest is straightforward.
L =
∫ (^) ln 2
0
‖r′(t)‖dt
∫ (^) ln 2
0
2)^2 + (et)^2 + (−e−t)^2 dt
∫ (^) ln 2
0
(2 + e^2 t^ + e−^2 t)dt =
∫ (^) ln 2
0
(et^ + e−t)^2 dt
∫ (^) ln 2
0
(et^ + e−t)dt = · · ·
- r(t) = r(t(s)) := r(s). Need to find t = t(s) which is an inverse function of s = s(t). But
s =s(t) =
∫ (^) t
0
(−2 sin 2t)^2 + (2 cos 2t)^2 + 64dt
∫ (^) t
0
8 + 64dt = 6
2 t ⇒ t = t(s) = s/ 6
Substituting t = s/ 6
2 into r(t) we obtain the parametric equation r = r(s), where s denotes the arc length variable. 15* c) Given two lines 1 and 2 , there exists a vector AB perpendicular to both lines 1 and 2. If we draw a third line 3 , passing through B and parallel to 1 , then AB is also perpendicular to 3. Therefore AB ⊥ the plane passing through 2 and 3. From the graph we know that AB is the distance between 1 and the plane passing through 2 , 3.
But it is easy to find the distance between a line and a plane. In our case, take a point P on the line 1 : x− 1 3 = y− 2 2 = z+1 1 ; take another point Q on the line 2 , which is the x-axis. Say, P = (3, 2 , −1), Q = (0, 0 , 0), PQ = (− 3 , − 2 , 1). The normal direction N of the plane determined by 2 and 3 is given by
N = L 2 × L 3 = L 2 × L 1 =
i j k 1 0 0 1 2 1
= −j + 2k,
where L 2 is the direction of 2 , L 2 = (1, 2 , 1) and L 1 is the direction of x-axis, L 1 = (1, 0 , 0). We have, Distance between 1 and the plane
= length of projNPQ = ‖
PQ · N
‖N‖^2
N‖
|PQ · N|
‖N‖
(−1)^2 + 2^2
which is also the distance between 1 and 2.
T =
dr ds
dr/dt |dr/dt|
N =
κ
dT /ds =
dT /dt |dT /dt|
κ = |
dT ds
|v|
dT dt
a = aT T + aN N
aT =
d^2 s dt^2
aN = κ(
ds dt
)^2 =
‖v‖^2 R^2
(R equal to the radius of the osculate circle, or called radius of curvature)
