Download Mathematics 202 Examination 1 Answers 1. (15 points) Let v = (4, −1

Mathematics 202
Examination 1
Answers
1. (15 points) Let v = (4, −1, 3) and w = (6, 3, 4). Compute v × w, v · w, and kvk.
√
Answer: We have v × w = (−13, 2, 18), v · w = 33, and kvk = 26.
2. (15 points) Find an equation for the plane that goes through the points (4, 2, 9), (6, 3, 1), and (3, −2, 8).
Answer: We compute (4, 2, 9) − (3, −2, 8) = (1, 4, 1) and (6, 3, 1) − (3, −2, 8) = (3, 5, −7). We then compute
(1, 4, 1)×(3, 5, −7) = (−33, 10, −7). Therefore, the equation for the plane is (−33, 10, −7)·(x−4, y−2, z−9) =
0, or 33x − 10y + 7z = 175.
3. (15 points) For each of the following two functions, compute ∇f :
f (x, y) = ex cos y
f (x, y, z) = tan z + 2xy
Answer: For the first, we have ∇f = (ex cos y, −ex sin y), and for the second, we have (2y, 2x, sec2 z).
4. (10 points) Find the equation of the tangent plane to the graph of z = xe2y at the point (5, 0, 5).
Answer: We compute zx = e2y = 1 and zy = 2xe2y = 10. Therefore, the equation for the tangent line is
z = 5 + 1(x − 5) + 10(y − 0) = x + 10y.
5. (15 points) Find the area of the parallelogram with vertices (10, 13, −13), (13, 11, −5), (9, 10, −11), and
(14, 14, −7).
Answer: We start by subtracting any one of the four points from the other three, to get a parallelogram with
one vertex at the origin, so that we can use the cross-product of two vectors to find the area. Subtracting
the first vertex from each of the other three, we get (13, 11 − 5) − (10, 13, −13) = (3, −2, 8), (9, 10, −11) −
(10, 13, −13) = (−1, −3, 2), and (14, 14, −7) − (10, 13, −13) = (4, 1, 6). Clearly, the first vector is the sum of
the other two, so we look for the area of a parallelogram with sides (−1, −3, 2) and (4, 1, 6). We know√that
this is given by the length of the cross-product (−1, −3, 2) × (4, 1, 6) = (−20, 14, 11), which has length 717.
6. (15 points) Consider the two lines l1 (t) = t(3, 1, 4) + (14, 12, −1) and l2 (t) = t(4, 2, 1) + (12, 12, −8). Do
these lines intersect? If so, find the point of intersection. Be sure to explain your answer fully.
Answer: We need to see if we can find a simultaneous solution to the two sets of equations
x = 3t1 + 14
y = t1 + 12
z = 4t1 − 1
x = 4t2 + 12
y = 2t2 + 12
z = t2 − 8
We try solving the first two equations simultaneously, and see if we also have a solution to the third equation.
We have
3t1 + 14 = 4t2 + 12
t1 + 12 = 2t2 + 12
The second equation gives t1 = 2t2 , and substitution into the first equation gives 6t2 + 14 = 4t2 + 12, or
t2 = −1. Therefore, t1 = −2. Substitution into the third equation gives z = −9 in either case. Therefore,
the two lines intersect at the point (8, 10, −9).
7. (15 points) The two lines ax + by = 0 and ax + by = c are parallel in the xy-plane. Find the distance
between them in terms of a, b, and c.
Answer: We can start by choosing a point arbitrarily on one of the lines. The simplest thing is to use the
point (0, 0) on the line ax + by = 0. Now, we find a line perpendicular to ax + by = 0 going through (0, 0),
and that happens to have a particularly simple form: bx − ay = 0.
Next, we see where the line bx − ay = 0 intersects ax + by = c by solving those two equations simultaneously:
bx − ay = 0
ax + by = c
Multiply the first equation by b, and the second by a, yielding
b2 x − aby = 0
a2 x + aby = ac
Add, and we get (a2 + b2 )x = ac. Therefore, x = a2ac
+b2 .
Return to the two equations
bx − ay = 0
ax + by = c
and multiply the first by a and the second by b to get
abx − a2 y = 0
abx + b2 y = bc
Subtract, and we get (−a2 − b2 )y = −bc, or y = a2bc
+b2 .
bc
Therefore, the coordinates of the other point are ( a2ac
+b2 , a2 +b2 ). So the distance that we need is the
bc
ac
bc
distance between (0, 0) and ( a2ac
+b2 , a2 +b2 ), which is the length of the vector ( a2 +b2 , a2 +b2 ), or
s
ac
a2 + b2
2
+
bc
a2 + b2
s
2
=
a2 c2 + b2 c2
=
(a2 + b2 )2
r
|c|
c2
=√
.
2
a2 + b2
a + b2
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Mean: 71.79
Standard deviation: 16.87
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