MATHEMATICS 226, FALL 2014, PROBLEM SET 2
Solutions1
Section 10.3, Question 13: 4 marks
If u + v + w = 0, then
v × w = v × (−u − v) = −v × u − v × v = u × v,
w × u = (−u − v) × u = −u × u − v × u = u × v.
Section 10.3, Question 15: Not graded
The tetrahedron is spanned by the vectors h0, 2, 0i, h1, 2, 2i, h−1, 3, 2i. Using
the formula from Question 14 (you do not need to prove it), we get
0 2 0
1 1 2 1
1 = · 2(2 + 2) = 4
V = 1 2 2 = 2 · −1 2 6
6 6
3
−1 3 2
Section 10.3, Question 27: 4 marks
If we had (−i+2j+3k)×x = i+5j, then i+5j would have to be orthogonal to
both of the vectors on the left. But (−i+2j+3k)·(i+5j) = −1+10 = 9 6= 0,
a contradiction.
Section 10.4, Question 5: Not graded
The plane is parallel to the vectors i − j + 2k and −2i + 3j + k, so the normal
vector can be found using
i
j
k
n = (i − j + 2k) × (−2i + 3j + k) = 1 −1 2 = −7i − 5j + k.
−2 3 1 The plane also passes through (1, 1, 0), so its equation is −7(x − 1) − 5(y −
1) + z = 0, or after simplifying, 7x + 5y − z = 12.
Section 10.4, Question 9: 8 marks
The plane we are looking for is parallel to the normal vector n to the plane
2x + 3y + 4z = 5. From the equation of that plane, we get n = 2i + 3j + 4k.
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Also, we rewrite the equations of the given line in a parametric form: x =
2−y, y = y, z = y−3, so that the direction vector of the line is v = −i+j+k.
This should also be parallel to the plane we are looking for. Therefore a
normal vector to that plane is
i j k
(−i + j + k) × (2i + 3j + 4k) = −1 1 1 = i + 6j − 5k.
2 3 4
Pick a point on the given line, e.g. (2, 0, −3). Then the equation of the
plane we are looking for is (x − 2) + 6y − 5(z + 3) = 0, or after simplifying,
x + 6y − 5z = 17.
Section 10.4, Question 17: 8 marks
The line of intersection lies in both of the given planes, hence is perpendicular
to their normal vectors. These vectors are i + 2j − k and 2i − j + 4k. Hence
the line is parallel to the vector
i j
k
(i + 2j − k) × (2i − j + 4k) = 1 2 −1 = 7i − 6j − 5k.
2 −1 4 The line through 0 parallel to this vector has the vector parametric equation
r = t(7i − 6j − 5k),
scalar parametric equations
x = 7t, y = −6t, z = −5t,
and standard equations
y
z
x
=
=
.
7
−6
−5
Section 10.4, Question 19: 8 marks
Let v = hv1 , v2 , v3 i be the direction vector of the line, then we need
v·i
v·j
v·k
=
=
|v|
|v|
|v|
2
so that v1 = v2 = v3 . We can let v = h1, 1, 1i. Then the line we are looking
for has the vector parametric equation
r = (1 + t)i + (2 + t)j + (−1 + t)k,
scalar parametric equations
x = 1 + t, y = 2 + t, z = −1 + t,
and standard equations
x − 1 = y − 2 = z + 1.
Section 10.5, Question 23: 8 marks
Consider the plane z = bx (tilted in the x-direction), for some b to be determined. This plane intersects the cylinder in an ellipse with the principal
axes in the planes x = 0 and y = 0. In the plane x = 0, we have z = 0
from the equation of the plane, and y 2 = 1, y = ±1, from the equation of
the cylinder. In the plane y = 0, we have 2x2 = 1, x = ± √12 , and z = ± √b2 .
The semi-axis in the plane x = 0 has length 1, and the semi-axis in the plane
y = 0 has length
r
2
1
b
√ , 0, √ = 1 + b
2
2
2
2 The cross-section is a circle if the two semi-axes have equal length, so that
r
1 b2
+
= 1, 1 + b2 = 2, b2 = 1.
2
2
Thus we could take b = ±1 and the plane z = ±x, or x ± z = 0. This
corresponds to a = i + k and a = i − k. (For the correct answer, just one of
these vectors is sufficient.)
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