Solutions to old Exam 1 problems

Hi students!
I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections of the book will actually be covered. Enjoy!!
Best, Bill Meeks
PS. There are probably errors in some of the solutions presented here and for a few problems you need to complete them or simplify the answers; some questions are left to you the student. Also you might need to add more detailed explanations or justifications on the actual similar problems on your exam. I will keep updating these solutions with better corrected/improved versions.

Problem 1(a) - Fall 2008
Find parametric equations for the line L which contains A(1, 2, 3) and B(4, 6, 5).
Solution:
To get the parametric equations of L you need a point through which the line passes and a vector parallel to the line.
−→

Take the point to be A and the vector to be the AB.
The vector equation of L is
−→

−→

r(t) = OA + t AB = 1, 2, 3 + t 3, 4, 2 = 1 + 3t, 2 + 4t, 3 + 2t ,

where O is the origin.
The parametric equations are: x = 1 + 3t y = 2 + 4t, z = 3 + 2t

t ∈ R.

Problem 1(b) - Fall 2008
Find parametric equations for the line L of intersection of the planes x − 2y + z = 10 and 2x + y − z = 0.
Solution:
The vector part v of the line L of intersection is orthogonal to the normal vectors 1, −2, 1 and 2, 1, −1 . Hence v can be taken to be: i j k v = 1, −2, 1 × 2, 1, −1 = 1 −2 1 = 1i + 3j + 5k.
2 1 −1
Choose P ∈ L so the z-coordinate of P is zero. Setting z = 0, we obtain: x − 2y = 10
2x + y = 0.
Solving, we find that x = 2 and y = −4. Hence,
P = 2, −4, 0 lies on the line L.
The parametric equations are: x =2+t y = −4 + 3t z = 0 + 5t = 5t.

Problem 2(a) - Fall 2008
Find an equation of the plane which contains the points
P(−1, 0, 1), Q(1, −2, 1) and R(2, 0, −1).
Solution:
Method 1
−→
−→
Consider the vectors PQ = 2, −2, 0 and PR = 3, 0, −2 which lie parallel to the plane.
Then consider the normal vector: i j k −→
−→
n = PQ × PR = 2 −2 0
3 0 −2

= 4i + 4j + 6k.

So the equation of the plane is given by:
4, 4, 6 · x + 1, y , z − 1 = 4(x + 1) + 4y + 6(z − 1) = 0.

Problem 2(a) - Fall 2008
Find an equation of the plane which contains the points
P(−1, 0, 1), Q(1, −2, 1) and R(2, 0, −1).
Solution:
Method 2
The plane consists of all the points S(x, y , z) ∈ R3 , such that
−→ −→

−→

PS, PQ and PR are in the same plane (coplanar).
But this happens if and only if their box product is zero.
So the equation of the plane is: x +1 y z −1
2
−2
0
3
0
−2

= 4(x + 1) + 4y + 6(z − 1) = 0.

Problem 2(b) - Fall 2008
Find the distance D from the point (1, 6, −1) to the plane
2x + y − 2z = 19.
Solution:
1 +by1 +cz1
Recall the distance formula D = |ax√a2 +b2 +c+d| from a point
2
P = (x1 , y1 , z1 ) to a plane ax + by + cz + d = 0.
In order to apply the formula, rewrite the equation of the plane in standard form: 2x + y − 2z − 19 = 0.
So, the distance from (1, 2, −1) to the plane is:
|(2 · 1) + (1 · 6) + (−2 · −1) − 19|
| − 9|
D=
= √ = 3.
9
22 + 12 + (−2)2

Problem 2(c) - Fall 2008
Find the point Q in the plane 2x + y − 2z = 19 which is closest to the point (1, 6, −1). (Hint: You can use part b) of this problem to help find Q or first find the equation of the line L passing through
Q and the point (1, 6, −1) and then solve for Q.)
Solution:
The line L in the Hint passes through (1, 6, −1) and is parallel to n = 2, 1, −2 .
So, L has parametric equations: x = 1 + 2t y = 6 + t, t ∈ R. z = −1 − 2t
L intersects the plane 2x + y − 2z = 19 if and only if
2(1 + 2t) + (6 + t) − 2(−1 − 2t) = 19 ⇐⇒ 9t = 9 ⇐⇒ t = 1.
Substituting t = 1 in the parametric equations of L gives the point Q = (3, 7, −3).

Problem 3(a) - Fall 2008
Find the volume V of the parallelepiped such that the following four points A = (3, 4, 0), B = (3, 1, −2), C = (4, 5, −3),
D = (1, 0, −1) are vertices and the vertices B, C , D are all adjacent to the vertex A.
Solution:
The parallelepiped is determined by its edges
−→

−→

−→

AB = 0, −3, −2 , AC = 1, 1, −3 , AD = −2, −4, −1 .
Its volume can be computed as the absolute value of the box
−→

−→

−→

product AB · (AC × AD), i.e.,

V=

0 −3 −2
1
1 −3
−2 −4 −1

= |3(−1 − 6) − 2(−4 + 2)| = |−17| = 17.

Problem 3(b) - Fall 2008
Find the center and radius of the sphere x 2 − 4x + y 2 + 4y + z 2 = 8.
Solution:
Completing the square we get x 2 −4x +y 2 +4y +z 2 = (x 2 −4x +4)−4+(y 2 +4y +4)−4+(z 2 )
= (x − 2)2 − 4 + (y + 2)2 − 4 + z 2 = 8
⇐⇒
(x − 2)2 + (y + 2)2 + z 2 = 16.
This gives:
Center = (2, −2, 0)

Radius = 4

Problem 4(a) - Fall 2008
The position vector of a particle moving in space equals
1
r(t) = t 2 i − t 2 j + 2 t 2 k at any time t ≥ 0.
a) Find an equation of the tangent line to the curve at the point
(4, −4, 2).
Solution:
The parametrized curve passes through the point (4, −4, 2) if and only if
1
t 2 = 4, −t 2 = −4, t 2 = 2 ⇐⇒ t 2 = 4 ⇐⇒ t = ±2.
2
Since we have that t ≥ 0, we are left with the choice t0 = 2.
The velocity vector field to the curve is given by r (t) = 2t, −2t, t hence r (2) = 4, −4, 2 .
The equation of the tangent line in question is: x = 4 + 4t y = −4 − 4t, t ≥ 0. z = 2 + 2t

Problem 4(b) - Fall 2008
The position vector of a particle moving in space equals
1
r(t) = t 2 i − t 2 j + 2 t 2 k at any time t ≥ 0.
(b) Find the length L of the arc traveled from time t = 1 to time t = 4.
Solution:
The velocity field is: v(t) = r (t) = 2t, −2t, t .
Since t ≥ 0, the speed is:
√
r (t) = 9t 2 =⇒ r (t) = 3t.
Therefore, the length is:
4

L=

3t dt =
1

3 2 t 2

4

=
1

3
3
3
45
· 16 − · 1 = · 15 = .
2
2
2
2

Problem 4(c) - Fall 2008
Suppose a particle moving in space has velocity v(t) = sin t, 2 cos 2t, 3e t and initial position r(0) = 1, 2, 0 . Find the position vector function r(t).
Solution:
One can recover the position, by integrating the velocity: t r(t) =

v(τ )dτ + r(0)
0

Carrying out this integral yields: r(t) =

− cos τ

t

t

, sin 2τ
0

0

, 3e t

t

+ 1, 2, 0
0

= 2 − cos t, 2 + sin 2t, 3e t − 3 .

Problem 5(a) - Fall 2008
Consider the points A(2, 1, 0), B(3, 0, 2) and C (0, 2, 1). Find the area of the triangle ABC . (Hint: If you know how to find the area of a parallelogram spanned by 2 vectors, then you should be able to solve this problem.)
Solution:
Recall that:
Area =
−→

1 −→ −→
AB × AC .
2
−→

Since AB = 1, −1, 2 and AC = −2, 1, 1 ,
1
Area =
2

i j k
1 −1 2
−2 1 1

=

1
1√
| −3, −5, −1 | =
35.
2
2

Problem 5(b) - Fall 2008
Three of the four vertices of a parallelogram are P(0, −1, 1),
Q(0, 1, 0) and R(2, 1, 1). Two of the sides are PQ and PR. Find the coordinates of the fourth vertex.
Solution:
Denote the fourth vertex by S. Then
−→

−→

−→

OS = OQ + PR = 0, 1, 0 + 2, 2, 0 = 2, 3, 0 , where O is the origin. That is,
S = (2, 3, 0).

Problem 6(a) - Spring 2008
Find an equation of the plane through the points A = (1, 2, 3),
B = (0, 1, 3), and C = (2, 1, 4).
Solution:
Since a plane is determined by its normal vector n and a point on it, say the point A, it suffices to find n. Note that:
−→

−→

n = AB × AC =

i j k
−1 −1 0
1 −1 1

= −1, 1, 2 .

So the equation of the plane is:
−(x − 1) + (y − 2) + 2(z − 3) = 0.

Problem 6(b) - Spring 2008
Find the area of the triangle ∆ with vertices at the points
A = (1, 2, 3), B = (0, 1, 3), and C = (2, 1, 4).
(Hint: the area of this triangle is related to the area of a certain parallelogram) Solution:
Consider the points A = (1, 2, 3), B = (0, 1, 3) and C = (2, 1, 4).
Then the area of the triangle ∆ with these vertices can be found
−→

−→

by taking the area of the parallelogram spanned by AB and AC and dividing by 2. Thus:
−→

−→

1
|AB × AC |
=
Area(∆) =
2
2

i j k
−1 −1 0
1 −1 1

1√
1√
1
1+1+4=
6.
= | −1, 1, 2 | =
2
2
2

Problem 7(a) - Spring 2008
Find the parametric equations of the line passing through the point (2, 4, 1) that is perpendicular to the plane 3x − y + 5z = 77.
Solution:
The vector part of the line L is the normal vector n = 3, −1, 5 to the plane.
The vector equation of L is: r(t) = 2, 4, 1 + tn
= 2, 4, 1 + t 3, −1, 5 = 2 + 3t, 4 − t, 1 + 5t .
The parametric equations are: x = 2 + 3t y =4−t z = 1 + 5t.

Problem 7(b) - Spring 2008
Find the intersection point of the line L(t) = 2 + 3t, 4 − t, 1 + 5t in part (a) and the plane 3x − y + 5z = 77.
Solution:
By part (a), we have L has parametric equations: x = 2 + 3t y =4−t z = 1 + 5t.
Plug these t-values into equation of plane and solve for t:
3(2 + 3t) − (4 − t) + 5(1 + 5t) = 77,
6 + 9t − 4 + t + 5 + 25t = 77,
35t = 70;

=⇒ t = 2.

So L intersects the plane at time t = 2.
At t = 2, the parametric equations give the point:
2 + 3 · 2, 4 − 2, 1 + 5 · 2 = 8, 2, 11 .

Problem 8(a) - Spring 2008
A plane curve is given by the graph of the vector function u(t) = 1 + cos t, sin t , 0 ≤ t ≤ 2π.
Find a single equation for the curve in terms of x and y by eliminating t.
Solution:
Rewriting u, we get: u(t) = 1 + cos t, sin t = 1, 0 + cos t, sin t .
Since cos t, sin t is the parametrization of the circle of radius 1 centered at the origin, then u is a circle of radius r = 1 centered at (1, 0).
So the answer is:
(x − 1)2 + (y − 0)2 = 12 or (x − 1)2 + y 2 = 1.

Problem 8(b) - Spring 2008
Consider the space curve given by the graph of the vector function r(t) = 1 + cos t, sin t, t , 0 ≤ t ≤ 2π.
Sketch the curve and indicate the direction of increasing t in your graph. Solution:
The sketch would be the following one translated 1 unit along the x-axis. Problem 8(c) - Spring 2008
Determine parametric equations for the line T tangent to the graph of the space curve for r(t) = 1 + cos t, sin t, t at t = π/3, and sketch T in the graph obtained in part (b).
Solution:
First find the velocity vector r (t) : r (t) = (1 + cos t) , (sin t) , 1 = − sin t, cos t, 1 . π At t = 3 ,
√
π π π π
3
3 π r( ) = 1 + cos , sin ,
= ,
, ,
3
3
3 3
2 2 3
√
π π π
3 1
, ,1 . r ( ) = − sin , cos , 1 = −
3
3
3
2 2
The vector part of tangent line T is r ( π ) and a point on line is
3
r( π ).
3
The vector equation is: T(t) = r( π ) + tr ( π ).
3
3
The parametric equations are: x= y= z= √
3
3
2 − 2 t
√
3
1
2 + 2t π 3 + t.

Problem 9(a) - Spring 2008
Suppose that r(t) has derivative r (t) = − sin 2t, cos 2t, 0 on the
1
interval 0 ≤ t ≤ 1. Suppose we know that r(0) = 2 , 0, 1 .
Determine r(t) for all t.
Solution:
Find r(t) by integration: r(t) =
=

r (t) dt =

− sin 2t, cos 2t, 0 dt

1
1
cos(2t) + x0 , sin(2t) + y0 , z0 .
2
2
1
2 , 0, 1 :
1
x0 , y0 , z0 ) = ( 2 , 0, 1).

Now solve for the point (x0 , y0 , z0 ) using r(0) =
1
( 1 cos(0) + x0 , 2 sin(0) + y0 , z0 ) = ( 1 +
2
2

So x0 = 0, y0 = 0, z0 = 1.
Thus,
r(t) =

1
1
cos 2t, sin 2t, 1 .
2
2

Problem 9(b) - Spring 2008
Suppose that r(t) has derivative r (t) = − sin 2t, cos 2t, 0 on the
1
interval 0 ≤ t ≤ 1. Suppose we know that r(0) = 2 , 0, 1 . Show that r(t) is orthogonal to r (t) for all t.
Solution:
By part (a),

1
1
cos 2t, sin 2t, 1 ,
2
2
Taking dot products, we get: r(t) =

1
1
r(t) · r (t) = − cos 2t sin 2t + sin 2t cos 2t + 0 = 0.
2
2
Since the dot product is zero, then for each t, r(t) is orthogonal to r (t).

Problem 9(c) - Spring 2008
Suppose that r(t) has derivative r (t) = − sin 2t, cos 2t, 0 on the
1
interval 0 ≤ t ≤ 1. Suppose we know that r(0) = 2 , 0, 1 . Find the arclength L of the graph of the vector function r(t) on the interval 0 ≤ t ≤ 1.
Solution:
Recall that the length of r(t) on the interval [0, 1] is gotten by integrating the speed |r (t)|.
Calculating, we get:
1

1

|r (t)| dt =

L=
0

| − sin 2t, cos 2t, 0 | dt
0
1

1

sin2 2t + cos2 2t dt =

=

|1| dt = t
0

0

Thus
L = 1.

1

= 1.
0

Problem 10(a) - Spring 2008
If r(t) = (2t)i + (t 2 − 6)j − ( 1 t 3 )k represents the position vector
3
of a moving object (where t ≥ 0 is measured in seconds and distance is measured in feet),
Find the speed s(t) and the velocity v(t) of the object at time t.
Solution:
Recall that the velocity v(t) vector of r(t) at time t is r (t) and the speed s(t) is its length |r (t)|.
Calculating with r(t) = 2t, t 2 − 6, − 1 t 3 :
3
v(t) = r (t) = 2, 2t, −t 2 , s(t) = |r (t)| =

22 + (2t)2 + (−t 2 )2 =

4 + 4t 2 + t 4 .

Problem 10(b) - Spring 2008
If r(t) = (2t)i + (t 2 − 6)j − ( 1 t 3 )k represents the position vector of a
3
moving object (where t ≥ 0 is measured in seconds and distance is measured in feet.)
If a second object travels along a path given defined by the graph of the vector function w(s) = 2, 5, 1 + s 2, −1, −5 , show that the paths of the two objects intersect at a common point P.

Solution:
Note that w(s) = 2 + 2s, 5 − s, 1 − 5s and r(t) = 2t, t 2 − 6, − 1 t 3 .
3
Setting the x and y -coordinates of w(s) and r(t) equal, we obtain: x = 2t = 2 + 2s =⇒ t = s + 1 y = t 2 − 6 = 5 − s =⇒ (s + 1)2 − 6 = s 2 + 2s − 5 = 5 − s
=⇒ s 2 + 3s − 10 = 0 =⇒ (s + 5)(s − 2) = 0.
So, (s = 2 and t = 3) or (s = −5 and t = −4).
Since

r(3) = 6, 3, −9 = w(2), the paths intersect at P = (6, 3, −9).

Problem 10(c) - Spring 2008
If r(t) = (2t)i + (t 2 − 6)j − ( 1 t 3 )k represents the position vector
3
of a moving object (where t ≥ 0 is measured in seconds and distance is measured in feet),
If s = t in part (b), (i.e. the position of the second object is w(t) when the first object is at position r(t)), do the two objects ever collide? Solution:
Set t = s in part (b).
Then the x-coordinate of r(t) is 2t and the x-coordinate of w(t) = 2 + 2t, 5 − t, 1 − 5t is 2 + 2t, and 2t = 2 + 2t for all
t.
Since r(t) and w(t) have different x-coordinates for all values of t, then they never collide.

Problem 11(a) - Spring 2007
Find parametric equations for the line L which contains A(7, 6, 4) and B(4, 6, 5).
Solution:
A vector parallel to the line L is:
−→

v = AB = 4 − 7, 6 − 6, 5 − 4, = −3, 0, 1 .
A point on the line is A(7, 6, 4).
Therefore parametric equations for the line L are: x = 7 − 3t y =6 z = 4 + t.

Problem 11(b) - Spring 2007
Find the parametric equations for the line L of intersection of the planes x − 2y + z = 5 and 2x + y − z = 0.
Solution:
A vector v parallel to the line is the cross product of the normal vectors of the planes: i j k v = 1, −2, 1 × 2, 1, −1 = 1 −2 1
2 1 −1
−2 1
1 1
1 −2
=
i− j+ k = 1, 3, 5 .
1 −1
2 −1
2 1
A point on L is any (x0 , y0 , z0 ) that satisfies both of the plane equations. Setting z = 0, we obtain the equations x − 2y = 5 and 2x + y = 0 and find such a point (1, −2, 0).
Therefore parametric equations for L are: x =1+t y = −2 + 3t z = 5t.

Problem 12(a) - Spring 2007
Find an equation of the plane which contains the points
P(−1, 0, 2), Q(1, −2, 1) and R(2, 0, −1).
Solution:
A normal vector to the plane can be found by taking the cross product of any two vectors that lie in the plane. Two vectors
−→

that lie in the plane are PQ = 2, −2, −1 and
−→

PR = 3, 0, −3 .
So the normal vector is i j k 2 −2 −1 =
3 0 −3
−2 −1
2 −1
2 −2 i− j+ k = 6, 3, 6 .
0 −3
3 −3
3 0
A point on the plane is P(−1, 0, 2). Therefore,
6(x − (−1)) + 3(y − 0) + 6(z − 2) = 0, n = 2, −2, −1 × 3, 0, −3 =

or simplified,

6x + 3y + 6z − 6 = 0.

Problem 12(b) - Spring 2007
Find the distance D from the point P1 = (1, 0, −1) to the plane
2x + y − 2z = 1.
Solution:
The normal to the plane is n = 2, 1, −2 and the point
P0 = (0, 1, 0) lies on this plane. Consider the vector from P0 to
P1 = (1, 0, −1) which is b = 1, −1, −1 . The distance D from
(1, 0, −1) to the plane is equal to: n 1
|compn b| = b ·
= | 1, −1, −1 · 2, 1, −2 | = 1.
|n|
3

Problem 12(c) - Spring 2007
Find the point P in the plane 2x + y − 2z = 1 which is closest to the point (1, 0, −1). (Hint: You can use part (b) of this problem to help find
P or first find the equation of the line passing through P and the point
(1, 0, −1) and then solve for P.)
Solution:
First find the parametric equations of the line that goes through the point (1, 0, −1) that is normal to the plane: x = 1 + 2t, y = t, z = −1 − 2t; here n = 2, 1, −2 is a normal to the plane.
The point P in the plane closest to (1, 0, −1) is the intersection of this line and the plane.
Substitute the parametric equations of the line into the plane equation: 2(1 + 2t) + (t) − 2(−1 − 2t) = 1.
Simplifying and solving for t,
1
9t + 4 = 1 =⇒ t = − .
3
Plugging this t-value into the parametric equations, we get the coordinates of the point of intersection: x = 1 + 2(− 1 ) = 1 ,
3
3 y = − 1 , z = −1 − 2(− 1 ) = − 1 .
3
3
3
1
1
So the point on the plane closest to (1, 0, −1) is P = ( 3 , − 1 , − 3 ).
3

Problem 13(a) - Spring 2007
Consider the two space curves r1 (t) = cos(t − 1), t 2 − 1, 2t 4 , r2 (s) = 1 + ln s, s 2 − 2s + 1, 2s 2 , where t and s are two independent real parameters. Find the cosine of the angle θ between the tangent vectors of the two curves at the intersection point (1, 0, 2).
Solution:
The point (1, 0, 2) corresponds to the t-value t = 1 for r1 and s-value s = 1 for r2 . r1 (t) = − sin(t − 1), 2t, 8t 3 is the tangent vector to r1 (t).
At t = 1, r1 (1) = − sin(1 − 1), 2(1), 8(13 ) = 0, 2, 8 . r2 (s) =

1 s , 2s

− 2, 4s is the tangent vector to r2 (s).

At s = 1, r2 (1) =

1
1 , 2(1)

− 2, 4(1) = 1, 0, 4 .

Therefore, cos(θ) =

0, 2, 8 · 1, 0, 4
32
=√ √ .
| 0, 2, 8 || 1, 0, 4 |
68 17

Problem 13(b) - Spring 2007
Find the center and radius of the sphere x 2 + y 2 + 2y + z 2 + 4z = 20.
Solution:
Completing the square in the y and z variables, we get x 2 + (y 2 + 2y + 1) + (z 2 + 4z + 4) = 20 + 1 + 4.
Rewriting, we have x 2 + (y + 1)2 + (z + 2)2 = 25 = 52 .
Hence, the center is C = (0, −1, −2) and the radius is r = 5.

Problem 14(a) - Spring 2007
The velocity vector of a particle moving in space equals v(t) = 2ti − 2tj + tk at any time t ≥ 0.
At the time t = 4, this particle is at the point (0, 5, 4). Find an equation of the tangent line T to the position curve r(t)at the time t = 4.
Solution:
This line goes through the point (0, 5, 4) and has vector part parallel to the tangent vector v(4) = 8, −8, 4 .
The vector equation is: T(t) = 0, 5, 4 + t 8, −8, 4
So the line T has the parametric equations: x = 8t y = 5 − 8t z = 4 + 4t.

Problem 14(b) - Spring 2007
The velocity vector of a particle moving in space equals v(t) = 2ti − 2tj + tk at any time t ≥ 0.
Find the length L of the arc traveled from time t = 2 to time t = 4.
Solution:
Using the arclength formula,
4

4

(2t)2 + (−2t)2 + t 2 dt

|v(t)| dt =

L=
2

2
4√

=
2

=

3 2 t 2

4

9t 2 dt =

3t dt
2

4
2

3
= (16 − 4) = 18.
2

Problem 14(c) - Spring 2007
Find a vector function r(t) which represents the curve of intersection of the cylinder x 2 + y 2 = 1 and the plane x + 2y + z = 4.
Solution:
Since the first equation is the equation of a circular cylinder, parametrize the x and y coordinates by setting x = cos(t) and y = sin(t).
Next use the second equation z = 4 − x − 2y to solve for z in terms of t: z = 4 − x − 2y = 4 − cos(t) − 2 sin(t).
Therefore,
r(t) = cos(t), sin(t), 4 − cos(t) − 2 sin(t) .

Problem 15(a) - Spring 2008
Consider the points A(2, 1, 0), B(1, 0, 2) and C (0, 2, 1). Find the area A of the triangle ABC . (Hint: If you know how to find the area of a parallelogram spanned by 2 vectors, then you should be able to solve this problem.)
Solution:
The area of the parallelogram is i j k
−1 −1 2
−2 1 1
−1 2
−1 2
−1 −1 i− j+ k 1 1
−2 1
−2 1
√
= | −3, −3, −3 | = 27.
−→

−→

|AB × AC | =

=

So the area of the triangle ABC is
√
27
A=
.
2

Problem 15(b) - Spring 2008
Suppose a particle moving in space has velocity v(t) = sin t, cos 2t, e t and initial position r(0) = 1, 2, 0 . Find the position vector function r(t).
Solution:
We find r(t) by integrating r (t) = v(t): t 1 r(t) = v(t) dt + r(0) = − cos t, sin 2t, e t
2
0

+ 1, 2, 0
0

1
1
sin 2t, e t − − cos 0, sin 0, e 0 + 1, 2, 0
2
2
1
= − cos t, sin 2t, e t − −1, 0, 1 + 1, 2, 0
2
1
= − cos t, sin 2t, e t + 2, 2, −1 .
2
1 r(t) = 2 − cos t, 2 + sin 2t, −1 + e t .
2

= − cos t,

So,

t

Problem 16 - Fall 2007
Find the equation of the plane containing the lines x = 4 − 4t, y = 3 − t, z = 1 + 5t and x = 4 − t, y = 3 + 2t, z = 1
Solution:
To find the equation of a plane, we need to find its normal n and a point on it. Setting t = 0, we find the point (4, 3, 1) on the first line.
The part vector v1 of the first line is −4, −1, 5 and the vector part v2 of the second line is −1, 2, 0 .
Since the vector i j k n = v1 × v2 = −4 −1 5 = −10, −5, −9 ,
−1 2 0 is orthogonal to both v1 and v2 , it is the normal to the plane.
The equation of the plane is:
−10, −5, −9 · x − 4, y − 3, z − 1
= −10(x − 4) − 5(y − 3) − 9(z − 1) = 0.

Problem 17 - Fall 2007
Find the distance D from the point P1 = (3, −2, 7) and the plane
4x − 6y − z = 5.
Solution:
1 +by1 +cz1
Recall the distance formula D = |ax√a2 +b2 +c+d| from a point
2
P = (x1 , y1 , z1 ) to a plane ax + by + cz + d = 0.
In order to apply the formula, rewrite the equation of the plane in standard form: 4x − 6y − z − 5 = 0.
So, the distance from (3, −2, 7) to the plane is:
|(4 · 3) + (−6 · −2) + (−1 · 7) − 5|
12
D=
=√ .
2 + (−6)2 + (−1)2
53
4

Problem 18 - Fall 2007
Determine whether the lines L1 and L2 given below are parallel, skew or intersecting. If they intersect, find the point of y −1 z −2 x intersection.
=
L1 : =
1
2
3
x −3 y −2 z −1
L2 :
=
=
−4
−3
2

Solution:
Rewrite these lines as vector equations:
L1 (t) = t, 2t + 1, 3t + 2
L2 (s) = −4s + 3, −3s + 2, 2s + 1
Equating x and y -coordinates: x = t = −4s + 3 y = 2t + 1 = −3s + 2.
Solving gives s = 1 and t = −1.
L1 (−1) = −1, −1, −1 = −1, −1, 3 = L2 (1). So these lines do not intersect.
Since the lines are clearly not parallel (the direction vectors 1, 2, 3 and −4, −3, 2 are not parallel), the lines are skew.

Problem 19(a) - Fall 2007
Suppose a particle moving in space has the velocity v(t) = 3t 2 , 2 sin(2t), e t .
Find the acceleration of the particle. Write down a formula for the speed of the particle (you do not need to simplify the expression algebraically).
Solution:
Recall the acceleration vector a(t) = v (t). Hence, a(t) = 6t, 4 cos(2t), e t .
Recall that the speed(t) is the length of the velocity vector.
Hence,
speed(t) =

9t 4 + 4 sin2 (2t) + e 2t .

Problem 19(b) - Fall 2007
Suppose a particle moving in space has the velocity v(t) = 3t 2 , 2 sin(2t), e t .
If initially the particle has the position r(0) = 0, −1, 2 , what is the position at time t?
Solution:
To find the position r(t), we first integrate the velocity v(t) and second use the initial position value r(0) = 0, −1, 2 to solve for the constants of integration. r(t) =

3t 2 , 2 sin 2t, e t dt = t 3 +x0 , − cos(2t)+y0 , e t +z0 .

Plugging in the position at t = 0, we get:
03 +x0 , − cos(0)+y0 , e 0 +z0 = x0 , −1+y0 , 1+z0 = 0, −1, 2 .
Thus, x0 = 0, y0 = 0 and z0 = 1.
Hence,

r(t) = t 3 , − cos 2t, e t + 1 .

Problem 20(a) - Fall 2007
Three of the four vertices of a parallelogram ∆ are P(0, −1, 1),
Q(0, 1, 0) and R(3, 1, 1). Two of the sides are PQ and PR.
Find the area of the parallelogram.
Solution:
−→

−→

Consider the vectors PQ = 0, 2, −1 and PR = 3, 2, 0 . Then the
−→

−→

area of the parallelogram ∆ spanned by PQ and PR is:
−→

−→

Area(∆) = |PQ × PR| =

= | 2, −3, −6 | =

i j k
0 2 −1
3 2 0

√
4 + 9 + 36 = 7

Problem 20(b) - Fall 2007
Three of the four vertices of a parallelogram are P(0, −1, 1),
Q(0, 1, 0) and R(3, 1, 1). Two of the sides are PQ and PR.
Find the cosine of the angle between the vector PQ and PR.
Solution:
Note that:
−→

−→

PQ = 0, 2, −1

PR = 3, 2, 0 .

By our formula for dot products:
−→

cos θ =

−→

PQ · PR
−→ −→

|PQ||PR|

=

0, 2, −1 · 3, 2, 0
4
√ √
=√ √ .
5 13
5 13

Problem 20(c) - Fall 2007
Three of the four vertices of a parallelogram are P(0, −1, 1),
Q(0, 1, 0) and R(3, 1, 1). Two of the sides are PQ and PR. Find the coordinates of the fourth vertex.
Solution:
Denote the fourth vertex by S. Then
−→

−→

−→

OS = OQ + PR = 0, 1, 0 + 3, 2, 0 = 3, 3, 0 , where O is the origin. That is,
S = (3, 3, 0).

Problem 21(a) - Fall 2007
Let C be the parametric curve x = 2 − t 2 , y = 2t − 1, z = ln t.
Determine the point(s) of intersection of C with the xz-plane.
Solution:
The points of intersection of C with the xz-plane correspond to the points where the y -coordinate of C is 0.
When y = 0, then 0 = 2t − 1 or t = 1 .
2
Hence,
1
1
3
1
= 1 , 0, − ln 2
2 − ( )2 , 2 · − 1, ln
2
2
2
4 is the unique point of the intersection of C with xz-plane.

Problem 21(b) - Fall 2007
Let C be the parametric curve x = 2 − t 2 , y = 2t − 1, z = ln t.
Determine parametric equations of tangent line to C at (1, 1, 0).
Solution:
Using the y -coordinate of C, note that t = 1 when
(1, 1, 0) ∈ C.
The velocity vector to
C(t) = 2 − t 2 , 2t − 1, ln t is: 1
C (t) = −2t, 2, . t Thus,
C (1) = −2, 2, 1 is the vector part of the tangent line to C at (1, 1, 0).
The parametric equations are: x = 1 − 2t y = 1 + 2t z = t.

Problem 21(c) - Fall 2007
Let C be the parametric curve x = 2 − t 2 , y = 2t − 1, z = ln t.
Set up, but not solve, a formula that will determine the length L of
C for 1 ≤ t ≤ 2.
Solution:
The vector equation of C is r(t) = 2 − t 2 , 2t − 1, ln t with velocity vector v(t) = r (t) = −2t, 2,

1
.
t

Since the length of L is the integral of the speed |r (t)|,
2

| −2t, 2,

L=
1

1
| dt = t 2

4t 2 + 4 +
1

1 dt. t2

Problem 22(a) - Fall 2006
Find parametric equations for the line r which contains A(2, 0, 1) and B(−1, 1, −1).
Solution:
−→

Note that AB = −3, 1, −2 and the vector equation is:
−→

r(t) = A + t AB = 2, 0, 1 + t −3, 1, −2 = 2 − 3t, t, 1 − 2t .
The parametric equations are: x = 2 − 3t y =t z = 1 − 2t.

Problem 22(b) - Fall 2006
Determine whether the lines L1 : x = 1 + 2t, y = 3t, z = 2 − t and L2 : x = −1 + s, y = 4 + s, z = 1 + 3s are parallel, skew or intersecting. Solution:
Vector part of line L1 is v1 = 2, 3, −1 and for line L2 is v2 = 1, 1, 3 . Clearly, v1 is not a scalar multiple of v2 and so these lines are not parallel.
If these lines intersect, then for some values of t and s: x = 1 + 2t = −1 + s =⇒ 2t = −2 + s, y = 3t = 4 + s =⇒ 3t = 4 + s.
Solving yields:

t = 6 and s = 14.

Plugging these values into z = 2 − t = 1 + 3s yields the inequality −4 = 43, which means the z-coordinates are never equal and the lines do not intersect.
Thus, the lines are skew.

Problem 23(a) - Fall 2006
Find an equation of the plane which contains the points
P(−1, 2, 1), Q(1, −2, 1) and R(1, 1, −1).
Solution:
−→

−→

Consider the vectors PQ = 2, −4, 0 and PR = 2, −1, −2 which are parallel to the plane.
The normal vector to the plane is:
−→

−→

n = PQ × PR =

i j k
2 −4 0
2 −1 −2

= 8i + 4j + 6k.

Since P(−1, 2, 1) lies on the plane, the equation of the plane is:
8, 4, 6 · x +1, y −2, z −1 = 8(x +1)+4(y −2)+6(z −1) = 0.

Problem 23(b) - Fall 2006
Find the distance D from the point (1, 2, −1) to the plane
2x + y − 2z = 1.
Solution:
The normal to the plane is n = 2, 1, −2 and the point
P0 = (0, 1, 0) lies on this plane. Consider the vector from P0 to
P1 = (1, 2, −1) which is b = 1, 1, −1 . The distance D from
(1, 2, −1) to the plane is equal to: n 1
5
|compn b| = b ·
= | 1, 1, −1 · 2, 1, −2 | = .
|n|
3
3

Problem 24(a) - Fall 2006
Let two space curves r1 (t) = cos(t − 1), t 2 − 1, t 4 , r2 (s) = 1 + ln s, s 2 − 2s + 1, s 2 , be given where t and s are two independent real parameters. Find the cosine of the angle between the tangent vectors of the two curves at the intersection point (1, 0, 1).
Solution:
When r1 (t) = 1, 0, 1 , then t = 1.
When r2 (s) = 1, 0, 1 , then s = 1.
Calculating derivatives, we obtain: r1 (t) = − sin(t − 1), 2t, 4t 3 r1 (1) = 0, 2, 4 r2 (s) = 1 , 2s − 2, 2s s r2 (1) = 1, 0, 2 .
Hence,
r (1) · r2 (1)
0, 2, 4 · 1, 0, 2
√ √ cos θ = 1
=
|r1 (1)||r2 (1)|
20 5
1
8
4
=√
(0 · 1 + 2 · 0 + 4 · 2) =
= .
10
5
100

Problem 24(b) - Fall 2006
Suppose a particle moving in space has velocity v(t) = sin t, cos 2t, e t and initial position r(0) = 1, 2, 0 . Find the position vector function r(t).
Solution:
The position vector function r(t) is the integral of its derivative r (t) = v(t): r(t) = v(t) dt
1
= sin t, cos 2t, e t dt = − cos(t) + x0 , sin(2t) + y0 , e t + z0 .
2
Now use the initial position r(0) = 1, 2, 0 to solve for x0 , y0 , z0 .
− cos(0) + x0 = −1 + x0 = 1 =⇒ x0 = 2.
1
sin(0) + y0 = 0 + y0 = 2 =⇒ y0 = 2.
2
e 0 + z0 = 1 + z0 = 0 =⇒ z0 = −1.
Hence,

r(t) = − cos(t) + 2,

1 sin(2t) + 2, e t − 1.
2

Problem 25(a) - Fall 2006
Let f (x, y ) = e x fxy .

2 −y

+x

4 − y 2 . Find partial derivatives fx , fy and

Problem 25(b) - Fall 2006
Find an equation for the tangent plane of the graph of f (x, y ) = sin(2x + y ) + 1 at the point (0, 0, 1).
Problem 26(a) - Fall 2006
Let g (x, y ) = ye x . Estimate g (0.1, 1.9) using the linear approximation of g (x, y ) at (x, y ) = (0, 2).

Solutions to these problems:

These types of problems might not be on this exam (check web site).

Problem 26(b) - Fall 2006
Find the center and radius of the sphere x 2 + y 2 + z 2 + 6z = 16.
Solution:
Complete the square in order to put the equation in the form:
(x − x0 )2 + (y − y0 ) + (z − z0 )2 = r 2 .
We get: x 2 + y 2 + (z 2 + 6z) = x 2 + y 2 + (z 2 + 6z + 9) − 9 = 16.
This gives the equation
(x − 0)2 + (y − 0)2 + (z + 3)2 = 25 = 52 .
Hence, the center is C = (0, 0, −3) and the radius is r = 5.

Problem 26(c) - Fall 2006
Let f (x, y ) = 16 − x 2 − y 2 . Draw a contour map of level curves f (x, y ) = k with k = 1, 2, 3. Label the level curves by the corresponding values of k.
Solution:
A problem of this type might not be on this exam (check web site). Problem 27
Consider the line L through points A = (2, 1, −1) and
B = (5, 3, −2). Find the intersection of the line L and the plane given by 2x − 3y + 4z = 13.
Solution:
−→

The vector part of L is AB = 3, 2, −1 and the point A is on the line.
The vector equation of L is:
−→

L = A+t AB = 2, 1, −1 +t 3, 2, −1 = 2+3t, 1+2t, −1−t .
Plugging x = 2 + 3t, y = 1 + 2t and z = −1 − t into the equation of the plane gives:
2(2 + 3t) − 3(1 + 2t) + 4(−1 − t) = −4t − 3 = 13
=⇒ −4t = 16 =⇒ t = −4.
So, the point of intersection is:
L(−4) = 2 − 12, 1 − 8, −1 − (−4) = −10, −7, 3 .

Problem 28(a)
Two masses travel through space along space curve described by the two vector functions r1 (t) = t, 1 − t, 3 + t 2 , r2 (s) = 3 − s, s − 2, s 2 where t and s are two independent real parameters.
Show that the two space curves intersect by finding the point of intersection and the parameter values where this occurs.
Solution:
Equate the x and z-coordinates: x =t =3−s z = 3 + t 2 = 3 + (3 − s)2 = 3 + 9 − 6s + s 2 = s 2
Thus, the parameter values are:
12 − 6s = 0 =⇒ (s = 2 and t = 1).
So, r1 (1) = 1, 0, 4 = r2 (2) is the desired intersection point.

Problem 28(b)
Two masses travel through space along space curve described by the two vector functions r1 (t) = t, 1 − t, 3 + t 2 , r2 (s) = 3 − s, s − 2, s 2 where t and s are two independent real parameters.
Find parametric equation for the tangent line to the space curve r1 (t) at the intersection point. (Use the value t = 1 in part (a)).
Solution:
The velocity vector of r1 (t) at the intersection point is r1 (1).
Since
r1 (t) = 1, −1, 2t , r1 (1) = 1, −1, 2 .
The vector equation of the tangent line is:
T(t) = r1 (1)+t 1, −1, 2 = 1, 0, 4 +t 1, −1, 2 = 1+t, −t, 4+2t .

The parametric equations are: x =1+t y = −t
.
z = 4 + 2t

Problem 29
Consider the parallelogram with vertices A, B, C , D such that B and C are adjacent to A. If A = (2, 5, 1), B = (3, 1, 4),
D = (5, 2, −3), find the point C .
Solution:
After drawing a picture, the point C is easily seen to be:
−→

−→

OA + BD = 2, 5, 1 + 2, 1, −7 = 4, 6, −6 , where O is the origin.

Problem 30(a)
Consider the points A = (2, 1, 0), B = (1, 0, 2) and C = (0, 2, 1).
−→

−→

Find the orthogonal projection projAB (AC ) of the vector AC
−→

onto the vector AB.
Solution:
−→

We just plug in the vectors a = AB = −1, −1, 2 and
−→

b = AC = −2, 1, 1 into the formula: proja b =

a·b
a.
a·a

Plugging in, we get:
−→

proj−→ (AC ) =
AB

−1, −1, 2 · −2, 1, 1
−1, −1, 2 · −1, −1, 2

−1, −1, 2 =

1
−1, −1, 2 .
2

Problem 30(b)
Consider the points A = (2, 1, 0), B = (1, 0, 2) and C = (0, 2, 1).
Find the area of triangle ABC .
Solution:
Consider the points A = (2, 1, 0), B = (1, 0, 2) and
C = (0, 2, 1).
Then the area of the triangle ∆ with these vertices can be

−→

found by taking the area of the parallelogram spanned by AB
−→

and AC and dividing by 2.
Thus:
−→

−→

|AB × AC |
1
Area(∆) =
=
2
2

i j k
−1 −1 2
−2 1 1

1√
1√
1
9+9+9=
27.
= | −3, −3, −3 | =
2
2
2

Problem 30(c)
Consider the points A = (2, 1, 0), B = (1, 0, 2) and C = (0, 2, 1).
Find the distance d from the point C to the line L that contains points A and B.
Solution:
From the figure drawn on the blackboard, we see that the distance d from C to L is the absolute value of the scalar
−→
projection of AC in the direction
−→

−→

v = AC − proj−→ AC .
AB

The vector v lies in the plane containing A, B, C and is
−→

perpendicular to AB.
Hence,
−→

d = |v| =

−→

|AC |2 − |proj−→ AC |2 .
AB

Next, you the student, do the algebraic calculation of d.

Problem 31
Find parametric equations for the line L of intersection of the planes x − 2y + z = 1 and 2x + y + z = 1.
Solution:
The vector part v of the line L of intersection is orthogonal to the normal vectors 1, −2, 1 and 2, 1, 1 . Hence v can be taken to be: i j k
1 −2 1 = −3i + j + 5k.
2 1 1
Choose P ∈ L so the z-coordinate of P is zero. Setting z = 0, we get: x − 2y = 1
2x + y = 1.

v = 1, −2, 1 × 2, 1, 1 =

3
Solving, we find that x = 5 and y = − 1 . Hence, P =
5
on the line L.
The parametric equations are: x = 3 − 3t
5
y = −1 + t
5
z = 5t.

3
1
5, −5, 0

lies

Problem 32
Let L1 denote the line through the points (1, 0, 1) and (−1, 4, 1) and let L2 denote the line through the points (2, 3, −1) and
(4, 4, −3). Do the lines L1 and L2 intersect? If not, are they skew or parallel?
Solution:
The vector equations of the lines are:
L1 (t) = 1, 0, 1 + t −2, 4, 0 = 1 − 2t, 4t, 1
L2 (s) = 2, 3, −1 + s 2, 1, −2 = 2 + 2s, 3 + s, −1 − 2s
Equating z-coordinates, we find 1 = −1 − 2s =⇒ s = −1.
Equating y -coordinates with s = −1, we find
1
4t = 3 − 1 =⇒ t = 2 .
1
Equating x-coordinates with s = −1 and t = 2 , we find:

1
L1 ( ) = 0, 2, 1 = L2 (−1).
2
Hence, the lines intersect.

Problem 33(a)
Find the volume V of the parallelepiped such that the following four points A = (1, 4, 2), B = (3, 1, −2), C = (4, 3, −3),
D = (1, 0, −1) are vertices and the vertices B, C , D are all adjacent to the vertex A.
Solution:
The volume V is equal to the absolute value of the determinant of
−→

−→

the matrix with rows AB = 2, −3, −4 , AC = 3, −1, −5 ,
−→

AD = 0, −4, −3 .
2 −3 −4
V = 3 −1 −5
0 −4 −3
= |2 · (−17) + −(−3) · (−9) + (−4) · (−12)| = | − 13| = 13.

Problem 33(b)
Find an equation of the plane through
A = (1, 4, 2), B = (3, 1, −2), C = (4, 3, −3).
Solution:
−→

−→

Consider the vectors AB = 2, −3, −4 and AC = 3, −1, −5 which lie parallel to the plane.
The normal vector is:
−→

−→

n = AB × AC =

i j k
2 −3 −4
3 −1 −5

= 11i − 2j + 7k.

Since A = (1, 4, 2), is on the plane, then the equation of the plane is given by:
11(x − 1) − 2(y − 4) + 7(z − 2) = 0.

Problem 33(c)
Find the angle between the plane through
A = (1, 4, 2), B = (3, 1, −2), C = (4, 3 − 3) and the xy -plane.
Solution:
The normal vectors of these planes are n1 = 0, 0, 1 , n2 = 11, −2, 7 .
If θ is the angle between the planes, then: n1 · n2
7
7 cos θ =
=
=√
.
|n1 ||n2 |
174
112 + (−2)2 + 72 θ = cos−1

1
√
174

.

Problem 34(a)
The velocity vector of a particle moving in space equals v(t) = 2ti + 2t 1/2 j + k at any time t ≥ 0. At the time t = 0 this particle is at the point (−1, 5, 4). Find the position vector r(t) of the particle at the time t = 4.
Solution:
To find the position r(t), integrate the velocity vector field r (t) = v(t).
1
r(t) = v(t) dt =
2t, 2t 2 , 1 dt
4 3
= t 2 + x0 , t 2 + y0 , t + z0 .
3
Now use the initial position r(0) = −1, 5, 4 to find x0 = −1; y0 = 5; z0 = 4.
4 3
Thus,
r(t) = t 2 − 1, t 2 + 5, t + 4
3
32 r(4) = 15,
+ 5, 8 .
3

Problem 34(b)
The velocity vector of a particle moving in space equals v(t) = 2ti + 2t 1/2 j + k at any time t ≥ 0.
Find an equation of the tangent line T to the curve at the time t = 4.
Solution:
Vector equation of the tangent line T to r(t) at t = 4 is:
T(s) = r(4) + sr (4) = r(4) + sv(4).
By part (a), r(4) = 15, 32 + 5, 8 .
3
Since then v(4) = 8i + 4j + k = 8, 4, 1 ,
T(s) = 15,

32
+ 5, 8 + s 8, 4, 1 .
3

Problem 34(c)
The velocity vector of a particle moving in space equals v(t) = 2ti + 2t 1/2 j + k at any time t ≥ 0.
Does the particle ever pass through the point P = (80, 41, 13) ?
Solution:
From part (a), we have
4 3 r(t) = t 2 − 1, t 2 + 5, t + 4 .
3
If r(t) = 80, 41, 13 , then t + 4 = 13 =⇒ t = 9.
Hence the point r(9) = 80, 41, 13 is on the curve r(t).

Problem 34(d)
The velocity vector of a particle moving in space equals v(t) = 2ti + 2t 1/2 j + k at any time t ≥ 0.
Find the length of the arc traveled from time t = 1 to time t = 2.
Solution:
2

2

4t 2 + 4t + 1 dt.

|v(t)| dt =

Length =
1

1

Since we are not using calculators on our exam, then this is the final answer.

Problem 35(a)
Consider the surface x 2 + 3y 2 − 2z 2 = 1.
What are the traces in x = k, y = k, z = k? Sketch a few.
Solution:
For x = k = 1, we get the hyperbolas 3y 2 − 2z 2 = k.
For x = 1, we get the 2 lines y = ± 3 z.
2
For z = 0, we get the ellipse x 2 + 3y 2 = 1.
For z = 1, we get the ellipse x 2 + 3y 2 = 3.
I am leaving it to you to do the sketches!

Problem 35(b)
Consider the surface x 2 + 3y 2 − 2z 2 = 1.
Sketch the surface in the space.
Solution:
Sorry, you need to do the sketch.
Problem 36
Find an equation for the tangent plane to the graph of f (x, y ) = y ln x at (1, 4, 0).
Solution:
A problem of this type might not be on this exam (check web site). Problem 37
Find the distance D between the given parallel planes z = 2x + y − 1,
−4x − 2y + 2z = 3.
Solution:
The normal to the first plane is n = 2, 1, −1 and the point
P0 = (0, 0, −1) lies on this plane. The point P1 = 0, 0, 3 lies on
2
the second plane. Consider the vector from P0 to P1 which is
5
b = 0, 0, 2 . The distance D from P1 to the first plane is equal to: n
1
5
5
|compn b| = b ·
= | 0, 0, · √ 2, 1, −1 | = √ .
|n|
2
6
2 6

Problem 38
Identify the surface given by the equation
4x 2 + 4y 2 − 8y − z 2 = 0. Draw the traces and sketch the curve.
Solution:
Sorry, no sketch given.

Problem 39(a)
A projectile is fired from a point 5 m above the ground at an angle of 30 degrees and an initial speed of 100 m/s.
Write an equation for the acceleration vector.
Solution:
Since the force due to gravity acts downward, we have
F = ma = −mg j, where g = |a| ≈ 9.8 m/s2 . Thus a = −g j.

Problem 39(b) and 34(c)
A projectile is fired from a point 5 m above the ground at an angle of 30 degrees and an initial speed of 100 m/s.
(b) Write a vector for initial velocity v(0).
(c) Write a vector for the initial position r(0)
Solution:
Initial velocity is:

√ v(0) = 100(cos 30◦ i + sin 30◦ j) = 50 3i + 50j,

in units of m/s.
The initial position is: r(0) = 5j, in units of meters m.

Problem 39(d)
A projectile is fired from a point 5 m above the ground at an angle of 30 degrees and an initial speed of 100 m/s.
At what time does the projectile hit the ground?
Solution:
We first find the velocity r(t) and position r(t) functions. r (t) = v(t) = −gtj + v(0)
1
r(t) = − gt 2 j + tv(0) + D.
2
1
Since D = r(0) = 5j, then r(t) = − 2 gt 2 j + tv(0) + 5j.
√
Hence,
1
r(t) = 50 3ti + [50t − gt 2 + 5]j.
2
The projectile hits the ground when 50t − 1 gt 2 + 5 = 0.
2
Applying the quadratic formula, we find t= 100 +

1002 + 40g
.
2g

Problem 39(e)
A projectile is fired from a point 5 m above the ground at an angle of 30 degrees and an initial speed of 100 m/s.
How far did it travel, horizontally, before it hit the ground?
Solution:

√
1
Recall r(t) = 50 3ti + [50t − 2 gt 2 + 5]j and the projectile
√
100+ 1002 +40g hits the ground when t =
.
2g
The horizontal distance d traveled is the value of the
√
100+ 1002 +40g x-coordinate of r(t) at t =
:
2g
√
d = 50 3

100 +

1002 + 40g
2g

.

Problem 40
Explain why the limit of f (x, y ) = (3x 2 y 2 )/(2x 4 + y 4 ) does not exist as (x, y ) approaches (0, 0).
Solution:
A problem of this type might not be on this exam (check web site). Problem 41
Find an equation of the plane that passes through the point
P(1, 1, 0) and contains the line given by parametric equations x = 2 + 3t, y = 1 − t, z = 2 + 2t.
Solution:
The direction vector a = 3, −1, 2 of the line is parallel to the plane. For t = 0, the point Q = 2, 1, 2 on the line and the plane.
−→

So b = PQ = 1, 0, 2 is also parallel to the plane.
To find a normal vector to the plane, take cross products: i j k n = a × b = 3 −1 2 = −2, −4, 1 .
1 0 2
Since (1, 1, 0) is on the plane, the equation of the plane is:
−2, −4, 1 · x − 1, y − 1, z = −2(x − 1) − 4(y − 1) + z = 0.

Problem 42(a)
Find all of the first order and second order partial derivatives of the function. f (x, y ) = x 3 − xy 2 + y
Solution:
There is no problem of this type on this exam.
Problem 42(b)
Find all of the first order and second order partial derivatives of the function. f (x, y ) = ln(x + x 2 + y 2 )
Solution:
There is no problem of this type on this exam.
Problem 43
Find the linear approximation of the function f (x, y ) = xye x at
(x, y ) = (1, 1), and use it to estimate f (1.1, 0.9).
Solution:
There is no problem of this type on this exam.

Problem 44
Find a vector function r(t) which represents the curve of intersection of the paraboloid z = 2x 2 + y 2 and the parabolic cylinder y = x 2 .
Solution:
Set t = x.
Since y = x 2 = t 2 , we get from the equation of the paraboloid a vector function r(t) which represents the curve of intersection: r(t) = t, t 2 , 2t 2 + (t 2 )2 = t, t 2 , 2t 2 + t 4 .

Problem 1(a) - Spring 2009
Given a = 3, 6, −2 , b = 1, 2, 3 .
Write down the vector projection of b along a. (Hint: Use projections.) Solution:
We have |a| =

√

9 + 36 + 4 =

√

49 = 7.

Then n= So,

a
1
= a = unit vector parallel to a.
|a|
7 proja b = (b · n)n =

b·a a= |a|2

9
1
1, 2, 3 · 3, 6, −2 3, 6, −2 =
3, 6, −2 .
49
49

Problem 45(b) - Spring 2009
Given a = 3, 6, −2 , b = 1, 2, 3 .
Write b as a sum of a vector parallel to a and a vector orthogonal to a. (Hint: Use projections.)
Solution:
We have
9
9
3, 6, −2 +
3, 6, −2
49
49
1
9
=
22, 44, 165 +
3, 6, −2 .
49
49

b = 1, 2, 3 = 1, 2, 3 −

Here and 9
3, 6, −2 parallel to a = 3, 6, −2
49
1
22, 44, 165 orthogonal to a = 3, 6, −2 .
49

Problem 45(b) Continuation - Spring 2009
Given a = 3, 6, −2 , b = 1, 2, 3 .
Write b as a sum of a vector parallel to a and a vector orthogonal to a. (Hint: Use projections.)
Solution:
Why so? All we did was to write b = b − (b · n)n + (b · n)n where n =

a
7,

n2

= 1.

Of course this is the same as b = (b − proja b) + proja b.
That is, we write b as proja b plus “the rest”. But “the rest” is orthogonal to n (and to a), since
(b − (b · n)n) · n = b · n − (b · n)(n · n) = 0, as n · n = 1.

Problem 45(c) - Spring 2009
Given a = 3, 6, −2 , b = 1, 2, 3 .
Let θ be the angle between a and b. Find cos θ.
Solution:
cos(θ) =

a·b
3, 6, −2 · 1, 2, 3
9
9
=
=√ √ = √ .
|a||b|
| 3, 6, −2 || 1, 2, 3 |
49 14
7 14

Problem 46(a) - Spring 2009
Given A = (−1, 7, 5), B = (3, 2, 2) and C = (1, 2, 3).
Let L be the line which passes through the points A = (−1, 7, 5) and B = (3, 2, 2). Find the parametric equations for L.
Solution:
To get parametric equations for L you need a point through which the line passes and a vector parallel to the line. For
−
→ example, take the point to be A and the vector to be AB.
The vector equation of L is
−→

−→

r(t) = OA+t AB = −1, 7, 5 +t 4, −5, −3 = −1 + 4t, 7 − 5t, 5 − 3t ,

where O is the origin.
The parametric equations are: x = −1 + 4t y = 7 − 5t, z = 5 − 3t

t ∈ R.

Problem 46(b) - Spring 2009
Given A = (−1, 7, 5), B = (3, 2, 2) and C = (1, 2, 3).
A, B and C are three of the four vertices of a parallelogram, while
CA and CB are two of the four edges. Find the fourth vertex.
Solution:
Denote the fourth vertex by D. Then
−→

−→

−→

OD = OA + CB = −1, 7, 5 + 2, 0, −1 = 1, 7, 4 , where O is the origin. That is,
D = (1, 7, 4).

Problem 47(a) - Spring 2009
Consider the points P(1, 3, 5), Q(−2, 1, 2), R(1, 1, 1) in R3 .
Find an equation for the plane containing P, Q and R.
Solution:
Since a plane is determined by its normal vector n and a point on it, say the point P, it suffices to find n. Note that:
−→

−→

n = PQ × PR =

i j k
−3 −2 −3
0 −2 −4

= 2, −12, 6 = 2 1, −6, 3 .

So the equation of the plane is:
(x − 1) − 6(y − 3) + 3(z − 5) = 0.

Problem 47(b) - Spring 2009
Consider the points P(1, 3, 5), Q(−2, 1, 2), R(1, 1, 1) in R3 .
Find the area of the triangle with vertices P, Q, R.
Solution:
The area of the triangle ∆ with vertices P, Q, R can be found by
−→

−→

taking the area of the parallelogram spanned by PQ and PR and dividing it by 2. Thus, using a), we have:
−→

−→

1
|PQ × PR|
= |2 1, −6, 3 |
Area(∆) =
2
2
√
√
= 1 + 36 + 9 = 46.

Problem 48 - Spring 2009
Find parametric equations for the line of intersection of the planes x + y + 3z = 1 and x − y + 2z = 0.
Solution:
A vector v parallel to the line is the cross product of the normal vectors of the planes: i j k
1 1 3 = 5, 1, −2 . v = 1, 1, 3 × 1, −1, 2 =
1 −1 2
A point on L is any (x0 , y0 , z0 ) that satisfies the equations of both planes.
Setting z = 0, we obtain the equations x + y = 1 and x − y = 0 and find such a point ( 1 , 1 , 0). Therefore
2 2 parametric equations for L are: x = 1 + 5t
2
y = 1 +t
2
z = −2t.

Problem 49(a) - Spring 2009
Consider the parametrized curve r(t) = t, t 2 , t 3 , t ∈ R.
Set up an integral for the length of the arc between t = 0 and t = 1. Do not attempt to evaluate the integral.
Solution:
The velocity field is: v(t) = r (t) = 1, 2t, 3t 2 .
Then the speed is r (t) =

1 + 4t 2 + 9t 4 .

Therefore, the length of the arc is:
1

1 + 4t 2 + 9t 4 dt.

L=
0

Problem 49(b) - Spring 2009
Consider the parametrized curve r(t) = t, t 2 , t 3 , t ∈ R.
Write down the parametric equations of tangent line to r(t) at
(2, 4, 8).
Solution:
The parametrized curve passes through the point (2, 4, 8) if and only if t = 2, t 2 = 4, t 3 = 8 ⇐⇒ t = 2.
The velocity vector field to the curve is given by r (t) =

1, 2t, 3t 2

hence r (2) = 1, 4, 12 .

The equation of the tangent line in question is: x =2+τ y = 4 + 4τ, z = 8 + 12τ

τ ∈R

Caution: The parameter along the line, τ , has nothing to do with the parameter along the curve, t.

Problem 50(a) - Spring 2009
Consider the sphere S in R3 given by the equation x 2 + y 2 + z 2 − 4x − 6z − 3 = 0.
Find its center C and its radius R.
Solution:
Completing the square we get
(x − 2)2 − 4 + y 2 + (z − 3)2 − 9 − 3 = 0
⇐⇒
(x − 2)2 + y 2 + (z − 3)2 = 16.
This gives:
C = (2, 0, 3)

R=4

Problem 50(b) - Spring 2009
What does the equation x 2 + z 2 = 4 describe in R3 ? Make a sketch. Solution:
This a (straight, circular) cylinder determined by the circle in the xz-plane of radius 2 and center (0, 0) and parallel to the y -axis.

Problem 51(a) - Spring 2009
Jane throws a basketball at an angle of 45o to the horizontal at an initial speed of 12 m/s, where m denotes meters. It leaves her hand 2 m above the ground.
Assume the acceleration of the ball due to gravity is downward with magnitude
10 m/s2 and neglect air friction.
(a) Find the velocity function v(t) and the position function r(t) of the ball.
Use coordinates in the xy -plane to describe what is happening; assume Jane is standing with her feet at the point (0, 0) and y represents the height.

Solution:
Acceleration due to gravity is a = 0, −g = 0, −10 . Initial
√ √ π π velocity is
= 6 2, 6 2 . v(0) = 12 cos , sin
4
So the velocity function is 4 t √ √ adτ = v(0) + at = 6 2, 6 2 − 10t . v(t) = v(0) +
0

One can recover the position by integrating the velocity: t r(t) =

v(τ )dτ + r(0).
0

Notice the initial position is r(0) = 0, 2 . This integral yields:
√
√ t2 r(t) = r(0) + v(0)t + a = 6 2t, 2 + 6 2t − 5t 2 .
2

Problem 51(b) - Spring 2009
Jane throws a basketball from the ground at an angle of 45o to the horizontal at an initial speed of 12 m/s, where m denotes meters.
It leaves her hand 2 m above the ground. Assume the acceleration of the ball due to gravity is downward with magnitude 10 m/s2 and neglect air friction.
(b) Find the speed of the ball at its highest point.
Solution:
At the highest point, the vertical component of the velocity is zero, so we only need to calculate the horizontal component which is
√
√
6 2. Thus the speed at the highest point is 6 2.

Problem 51(c) - Spring 2009
Jane throws a basketball from the ground at an angle of 45o to the horizontal at an initial speed of 12 m/s, where m denotes meters.
It leaves her hand 2 m above the ground. Assume the acceleration of the ball due to gravity is downward with magnitude 10 m/s2 and neglect air friction.
(c) At what time T does the ball reach its highest point.
Solution:
When the ball reaches its highest point, the vertical component of its velocity is zero. That is,
√
6 2 − 10t = 0, so T =

√
3 2
5 .

Problem 57(a) - Spring 2010
Consider the parallelogram with vertices A, B, C , D such that B and C are adjacent to A where A = (1, 2, −1), B = (3, 5, 1) and
D = (2, −1, 2). Find the area of the parallelogram.
Solution:
Recall that:

−→

−→

area = AB × BD .
−→

−→

Since AB = 2, 3, 2 and BD = −1, −6, 1 ,

area =

i j k
2
3 2
−1 −6 1

= | 15, −4, −9 | =

√

322.

Problem 57(b) - Spring 2010
Consider the parallelogram with vertices A, B, C, D such that B and C are adjacent to A where A = (1, 2, −1), B = (3, 5, 1) and
D = (2, −1, 2). Find the coordinates of the point C.
Solution:
−→

−→

−→

OC = OA + BD = 1, 2, −1 + −1, −6, 1 = 0, −4, 0 , where O is the origin. That is,
C = (0, −4, 0).

Problem 58(a) - Spring 2010
Consider the points A = (0, 3, −3), B = (−1, 3, 2),

−→

C = (−1, 2, −3). Find the orthogonal projection proj−→ (AC ) of
−→

AB

−→

the vector AC onto the vector AB.
Solution:
−→

We just plug in the vectors a = AB = −1, 0, 5 and
−→

b = AC = −1, −1, 0 into the formula: proja b =

a·b
a.
a·a

Plugging in, we get:
−→

proj−→ (AC ) =
AB

−1, 0, 5 · −1, −1, 0
−1, 0, 5 · −1, 0, 5

−1, 0, 5 =

1
−1, 0, 5 .
26

Problem 58(b) - Spring 2010
Consider the points A = (0, 3, −3), B = (−1, 3, 2),
C = (−1, 2, −3). Find the distance d from the point C to the line
L that contains points A and B.
Solution (Method 1):
From the figure drawn on the blackboard, we see that the distance d from C to L is the absolute value of the scalar
−→
projection of AC in the direction
−→

−→

v = AC − proj−→ AC .
AB

The vector v lies in the plane containing A, B, C and is
−→

perpendicular to AB.
Hence,
d = |v|.
Next, you the student, do the algebraic calculation of d.

Problem 58(b) - Spring 2010
Consider the points A = (0, 3, −3), B = (−1, 3, 2),
C = (−1, 2, −3). Find the distance d from the point C to the line
L that contains points A and B.
Solution (Method 1):
From the figure drawn on the blackboard and by the
Pythagorean Theorem, we see that the distance d from C to
L is equal to:
−→

d=

−→

|AC |2 − |proj−→ AC |2 .
AB

Next, you the student, do the algebraic calculation of d.

Problem 58(b) - Spring 2010
Consider the points A = (0, 3, −3), B = (−1, 3, 2),
C = (−1, 2, −3). Find the distance d from the point C to the line
L that contains points A and B.
Solution (Method 2):
Let P be the point on the line L closest to the point C .
From the figure drawn on the blackboard, we see that the distance d from C to L is equal to the length of the side PC of the right triangle with legs AP, PC and hypotenuse AC .
Then, using the rule |a × b| = |a||b| sin(θ),
−→

−→

d = |PC | = |AC | sin(θ) =

−→

−→

|AC × AB|
−→

.

|AB|
Next, you the student, do the algebraic calculation of d.

Problem 59(a) - Spring 2010
Let P1 be the plane x + 3y + z = 0 and P2 be the plane
2x + y − z = 1. Find the cosine of the angle between the planes.
Solution:
Note that the normals of the planes are: n1 = 1, 3, 1

n2 = 2, 1, −1 .

By the formula for dot products of 2 vectors: cos θ =

1, 3, 1 · 2, 1, −1
4
n1 · n2
√ √
=
=√ √ .
|n1 ||n2 |
11 6
11 6

Problem 59(b) - Spring 2010
Let P1 be the plane x + 3y + z = 0 and P2 be the plane 2x + y − z = 1.
Find the parametric equations of the line of intersection between the 2 planes P1 and P2 .
Solution:
The vector part v of the line L of intersection is orthogonal to the normal vectors 1, 3, 1 and 2, 1, −1 . Hence v can be taken to be: i j k v = 1, 3, 1 × 2, 1, −1 = 1 3 1 = −4i + 3j − 5k.
2 1 −1
Choose P ∈ L so the z-coordinate of P is zero. Setting z = 0, we obtain: x + 3y = 0
Solving, we find that x = on the line L.

3
5

2x + y = 1. and y = − 1 . Hence, P =
5

The parametric equations are: x = 3 − 4t
5
y = − 1 + 3t
5
z = 0 − 5t = −5t.

3
1
5, −5, 0

lies

Problem 59(c) - Spring 2010
Let P1 be the plane x + 3y + z = 0 and P2 be the plane
2x + y − z = 1. Find the distance from the plane P2 to the origin.
Solution:
|ax1 + by1 + cz1 + d|
√
from a a2 + b 2 + c 2 point P = (x1 , y1 , z1 ) to a plane ax + by + cz + d = 0.
Recall the distance formula D =

In order to apply the formula, rewrite the equation of the plane in standard form: 2x + y − z − 1 = 0.
So, the distance from the origin to the plane is:
D=

|(2 · 0) + (1 · 0) + (−1 · 0) − 1|
22

+

12

+

(−1)2

| − 1|
1
= √ =√
6
6

Problem 60(a) - Spring 2010
Let r(t) = cos(2t)i + sin(2t)j + tk. What is the length L of the curve starting at t = 0 and ending at t = 5.
Solution:
Recall that the length of r(t) on the interval [0, 5] is gotten by integrating the speed |r (t)|.
Calculating, we get:
5

5

|r (t)| dt =

L=
0

| −2 sin 2t, 2 cos 2t, 1 | dt
0
5√

5

4 sin2 2t + 4 cos2 2t + 1 dt =

=
0

=
Thus

√

5

5t

0

√
= 5 5.

√
L=5 5

0

5 dt

Problem 60(b) - Spring 2010
Let r(t) = cos(2t)i + sin(2t)j + tk. Find an equation for the tangent line to the graph at the point given by t = 0.
Solution:
The parametrized curve passes through the point r(0) = (1, 0, 0)
The velocity vector field to the curve is given by r (t) = −2 sin(2t), 2 cos(2t), 1 hence r (0) = 0, 2, 1 .
The equation of the tangent line in question is: x =1 y = 2τ, τ ∈ R. z =τ

Problem 61 - Spring 2010
Show that the limit lim(x,y )→(0,0)

x 2 −y 2 x 2 +y 2

does not exist.

Solution:
Let f (x, y ) =

x 2 −y 2
.
x 2 +y 2

Along the line t, t , t = 0, f (x, y ) has the value
Along the line 0, t , t = 0, f (x, y ) has the value

0
= 0.
2t 2 t2 = 1. t2 Since f (x, y ) has 2 different limiting values at (0, 0), it does not have a limit at (0, 0).

Problem 62(a) - Spring 2010
Consider the sphere S in R3 given by the equation x 2 + y 2 + z 2 − 2x − 8y − 2 = 0.
Find the coordinates of its center and its radius.
Solution:
Completing the square we get x 2 −2x+y 2 −8y +z 2 = (x 2 −2x+1)−1+(y 2 +8y +16)−16+(z 2 )
= (x − 1)2 − 1 + (y + 2)2 − 16 + z 2 = 2
⇐⇒
(x − 2)2 + (y + 2)2 + z 2 = 19.
This gives:
Center = (1, 4, 0)

Radius =

√

19

Problem 62(b) - Spring 2010
What does the equation x 2 + y 2 = 64 describe in R3 . Make a sketch. Solution:
This equation describes a circular cylinder of radius 8 centered along the z-axis. Here is a sketch of a cylinder of radius 1 centered

along the z-axis.

Problem 63(a) - Spring 2012
Given A = (−1, 7, 5), B = (3, 0, 2) and C = (1, 2, 3).
Let L be the line which passes through the points A = (−1, 7, 5) and B = (3, 0, 2). Find the parametric equations for L.
Solution:
To get parametric equations for L you need a point through which the line passes and a vector parallel to the line. For
−
→ example, take the point to be A and the vector to be AB.
The vector equation of L is
−→

−→

r(t) = OA+t AB = −1, 7, 5 +t 4, −7, −3 = −1 + 4t, 7 − 7t, 5 − 3t ,

where O is the origin.
The parametric equations are: x = −1 + 4t y = 7 − 7t, z = 5 − 3t

t ∈ R.

Problem 63(b) - Spring 2012
Given A = (−1, 7, 5), B = (3, 0, 2) and C = (1, 2, 3).
A, B and C are three of the four vertices of a parallelogram, while
AB and BC are two of the four edges. Find the fourth vertex.
Solution:
Denote the fourth vertex by D.
Then
−→

−→

−→

OD = OA + BC = −1, 7, 5 + −2, 2, 1 = −3, 9, 6 , where O is the origin.
That is,
D = (−3, 9, 6).

Problem 64(a) - Spring 2012
Consider the points P(1, 1, 1), Q(−2, 1, 2), R(1, 3, 5) in R3 .
Find an equation for the plane containing P, Q and R.
Solution:
Since a plane is determined by its normal vector n and a point on it, say the point P, it suffices to find n.
Note that:
−→

−→

n = PQ × PR =

i j k
−3 0 1
0 2 4

= −2, 12, −6 .

So the equation of the plane is:
−2(x − 1) + 12(y − 1) − 6(z − 1) = 0.

Problem 64(b) - Spring 2012
Consider the points P(1, 1, 1), Q(−2, 1, 2), R(1, 3, 5) in R3 .
Find the area of the triangle with vertices P, Q, R.
Solution:
The area of the triangle ∆ with vertices P, Q, R can be

−→

found by taking the area of the parallelogram spanned by PQ
−→

and PR and dividing it by 2.
Thus, using part a), we have:
−→

−→

|PQ × PR|
1
Area(∆) =
= | −2, 12, −6 |
2
2
1√
1√
=
4 + 144 + 36 =
184.
2
2

Problem 65 - Spring 2010
Let P1 be the plane x − 2y + 2z = 10 and P2 be the plane
2x + y + 2z = 0. Find the cosine of the angle between the planes.
Solution:
Note that the normals of the planes are: n1 = 1, −2, 2

n2 = 2, 1, 2 .

By the formula for dot products of 2 vectors: cos θ =

n1 · n2
1, −2, 2 · 2, 1, 2
4
√ √
=
= .
|n1 ||n2 |
9
9 9

Problem 66(a) - Spring 2012
Find the distance d from the point Q = (1, 6, −1) to the plane
2x + y − 2z = 19.
Solution:
Method 1
The distance from the point (x1 , y1 , z1 ) to the plane
Ax + By + Cz + D = 0 is:
|Ax1 + By1 + Cz1 + D|
√
d=
.
A2 + B 2 + C 2
In our case this gives
|2 + 6 + 2 − 19|
9
√ d= = √ =3
9
9

Problem 66(a) - Spring 2012
Find the distance d from the point Q = (1, 6, −1) to the plane
2x + y − 2z = 19.
Solution:
Method 2
Note that P = (0, 19, 0) is on the plane.
The distance between Q and the plane equals the length of the
−→

component of b = PQ which is orthogonal to the plane, i.e., which is parallel to the normal vector to the plane.
If the normal vector of the plane is n, then this is distance is the n absolute value of the scalar projection: d = b · |n| .
−→

Here b = PQ = 1, −13, −1 and n = 2, 1, −2 with
|n| = 22 + 1 + (−2)2 = 3.
Then
d=

| 1, −13, −1 · 2, 1, −2 |
|2 − 13 + 2|
√
=
= 3.
3
9

Problem 66(a) - Spring 2012
Find the distance d from the point (1, 6, −1) to the plane
2x + y − 2z = 19.
Solution:
Method 3 Write the equation of the line through P(1, 6, −1) which is orthogonal to the plane. Find the point Q where it intersects the plane. Find the distance d between P and Q from the distance formula. The line in question passes through (1, 6, −1) and is parallel to n = 2, 1, −2 , so it has parametric equations x = 1 + 2t y = 6 + t, t ∈ P. z = −1 − 2t
It intersects the plane 2x + y − 2z = 19 if and only if
2(1 + 2t) + (6 + t) − 2(−1 − 2t) = 19 ⇔ 9t = 9 ⇔ t = 1.
Substituting t = 1 above gives the point Q(3, 7, −3).
So the distance between P and Q is
√
d = (3 − 1)2 + (7 − 6)2 + (−3 + 1)2 = 4 + 1 + 4 = 3.

Problem 66(b) - Spring 2012
Write the parametric equations of the line L containing the point
T (1, 2, 3) and perpendicular to the plane 2x + y − 2z = 19.
Solution:
The line L passes through (1, 2, 3) and is parallel to n = 2, 1, −2 .
So, L has parametric equations: x = 1 + 2t y = 2 + t, z = 3 − 2t

t ∈ R.

Problem 66(c) - Spring 2012
Find the point of intersection of the line L in part b) with the plane 2x + y − 2z = 19.
Solution:
The line L has parametric equations: x = 1 + 2t y = 2 + t, t ∈ R. z = 3 − 2t
So, L intersects the plane 2x + y − 2z = 19 if and only if
7
2(1 + 2t) + (2 + t) − 2(3 − 2t) = 19 ⇐⇒ 9t = 21 ⇐⇒ t = .
3
7
Substituting t = 3 in the parametric equations of L gives
5
the point Q = ( 17 , 13 , − 3 ).
3 3

Problem 67 - Spring 2012
Find the equation of the sphere with center the point (1, 2, 3) and which contains the point (3, 1, 5).
Solution:
The distance from P = (1, 2, 3) to Q = (3, 1, 5) is the length
−→

of the vector PQ which is 3.
So the radius of the sphere is r = 3.
Hence the equation of the sphere is:
(x − 1)2 + (y − 2)2 + (z − 3)2 = 32 = 9

Problem 68 - Spring 2012
Make a sketch of the surface in R3 described by the equation y 2 + z 2 = 36. In your sketch of the surface, include the labeled coordinate axes and the trace curves on the surface for the planes x = 0 and x = 4.
Solution:
This equation describes a circular cylinder of radius 6 centered along the x-axis. Below is a sketch of a cylinder of radius 1 centered along the z-axis. Th red curve should be removed.

Problem 69 - Spring 2012
Find the equation of the plane which contains the points A(1, 2, 3) and
B(1, 0, 4) and which is also perpendicular to the plane 4x − 2y + z = 8.
Solution:
Since a plane is determined by its normal vector n and a point on it, say the point A, it suffices to find n.
Note that the normal 4, −2, 1 to the plane 4x − 2y + z = 8 and
−→

the vector AB = 0, −2, 1 are both perpendicular to the the normal vector n that we want.
It follows that:
−→

n = AB × 4, −2, 1 =

i
0
4

j
−2
−2

k
1
1

= 0, 4, 8 .

So the equation of the plane is:
0(x − 1) + 4(y − 2) + 8(z − 3) = 4(y − 2) + 8(z − 3) = 0.

Problem 70(a) - Spring 2012
Suppose that a = 2, 1, 2 and b = 8, 2, 0 . Find the vector
−→
projection, call it c, of b in the direction of a .
Solution:
We just plug in the vectors a = 2, 1, 2 and b = 8, 2, 0 into the formula: a·b a. proja b = a·a Plugging in, we get: proja b =

2, 1, 2 · 8, 2, 0
2, 1, 2 · 2, 1, 2

2, 1, 2 = 4, 2, 4 .

Problem 70(b) - Spring 2012
Suppose that a = 2, 1, 2 and b = 8, 2, 0 and c = proja b = a·b a. Calculate the vector b − c and then show that a·a it is orthogonal to a.
Solution:
By the previous problem, c = 4, 2, 4 .
Calculating, we get b − c = 8, 2, 0 − 4, 2, 4 = 4, 0, −4 .
Since
4, 0, −4 · 2, 1, 2 = 0, the vector b − c is orthogonal to a.