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Math 251, Spring 2016-copyright Joe Kahlig
1. False: normal vectors are not parallel so the planes are not parallel.
True: not product of the normal vectors is 0 so the planes are perpendicular.
True: Plugging in x = t and y = 5 into the surface equation does give the corresponding z−value.
True.
False: The center is (−3, 5, 0)
True.
2. cone: 4x2 + 4z 2 = y 2
sphere: 2x2 + 2y 2 + 2z 2 = 50
elliptic cylinder: 3x2 + 4y 2 = 12
hyperbolic paraboloid: z = 6x2 − 6y 2
3. b × a = h−5, 5, −2i
4. Set components equal and solve for t and p.
t = 1 + 3p and t2 = 1 + 9p
(1 + 3t)2 = 1 + 9p
1 + 6p + 9p2 = 1 + 9p
9p2 − 3p = 0
3p(3p − 1) = 0
p = 0 or p = 1/3
p = 0 gives t = 1 and p = 1/3 gives t = 2.
check the z component to see if they work,
points: (1, 1, 1) and (2, 4, 8)
5. Answers may vary depending how you solve.
Step 1: find a point in the intersection: I set x = 0 and solve to get z = 6 and y = 11
Step 2 find direction vector: I did the cross product of the two normal vectors. n1 × n2 =
h1, −11, −3i
Answer: x = t, y = 11 − 11t, z = 6 − 3t
6. The direction vector of the line will be a normal vector for the plane.
v = n = h4, 3, −2i be careful of last term.
Answer: 4x + 3y − 2z = 16
*
7. r(t) =
1,
+
t2 ,
√
√
2t3
. r0 (t) = 0, 2t, 2t2 . and |r0 (t)| = 4t2 + 4t4 = 2t 1 + t2
3
Z2
(a) arc length =
p
2t 1 + t2 dt solve with a u−sub.
1
2 51.5 − 21.5
answer:
3
*
+ 2t
2t2
1
t
(b) e T (t) = 0, √
,√
= 0, √
,√
.
4t2 + 4t4
4t2 + 4t4
1 + t2
1 + t2
(c) Easy method is
|r0 × r00 |
1
=
0
3
|r |
2t(1 + t2 )1.5
Check the back of the page for more problems.
Math 251, Spring 2016-copyright Joe Kahlig
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8. normal vector is the cross product of the two direction vectors. v1 × v2 = h−11, 2, 5i
pick a point that is on either line.(easy to find if you set t = 0)
Answer: −11x + 2y + 5z = −9
9. Method 1: If there is a plane, P2 , containing the line, L, where P2 is parallel to the given plane,
P , then the normal vector of P and the normal vector of P2 are parallel. Since the line lies in P2
then the dot product of the direction vector of the line and the normal vector of P2 will be zero.
In similar fashion the dot product of the direction vector of L and plane P will also be zero.
The direction vector of the line is v = h4, 1, −8i and normal vector of plane P is n = h1, 4, 1i.
v · n = 4 + 4 − 8 = 0.
so the answer is yes there is a plane containing line L that will be parallel to plane P .
Method 2: If there is a plane, P2 , containing the line, L, where P2 is parallel to the given plane,
P , then the line will not intersect the plane. So check to see if there is a value of t will the line
will intersect the plane.
x + 4y + z = 2
1 + 4t + 4(1 + t) + 1 − 8t = 2
8t + 6 − 8t = 2
6=2
This is a contradiction so there is not any value of t where the line intersects the plane.
Thus yes there is a plane containing line L that will be parallel to plane P .
Check the back of the page for more problems.