Calculus Homework
Assignment 1
1. Let u = 3i + 4j − 10k and v = 2i + j − 3k. Write u as the sum
of a vector parallel to v and a vector orthogonal to v.
[like §12.3 #19]
Sol. The vector projection of u onto v is of the form
u·v
(3i + 4j − 10k) · (2i + j − 3k)
projv u =
v=
(2i + j − 3k)
2
|v|
(2i + j − 3k) · (2i + j − 3k)
6 + 4 + 30
=
(2i + j − 3k)
4+1+9
40
20
60
=
i+ j− k
7
7
7
Obviously, we have that projv u parallel to v. Let
40
20
60 i+ j− k
w = u − projv u = (3i + 4j − 10k) −
7
7
7
40 20 60 = 3−
i+ 4−
j + − 10 +
k
7
7
7
19
8
10
= − i+ j− k
7
7
7
+ 87 + 30
= 0.
Then w is orthogonal to v, since w · v = − 38
7
7
Hence we have written u = projv u+w, which projv u is parallel
to v and w is orthogonal to v.
√
2. Find the angles between the curves y = x2 , y = − 3 x (two
points of intersection).
[like §12.3 #56]
√
Sol. Let
C1 : y = x2 , C2 : y = − 3 x. Solve the equation
√
x2 = − 3 x ⇒ x6 + x = x(x5 − 1) = 0. This implies that the real
solutions for the equation is x = 0 or x = −1. Thus, C1 and C2
intersect at points O(0, 0) and P (−1, 1). Note that the tangent
lines of C1 and C2 at O are y = 0 and x = 0 respectively.
Therefore, the angle of two curves at O is π2 . Next, the tangent
lines of C1 and C2 at P are L1 : y = −2x+1 and L2 : y = − 13 x+1
respectively. The corresponding normal vectors of tangent lines
at are n1 = 2i + j and n2 = i + 3j respectively. We have that
2+3 1 π
n ·n 1
2
= cos−1 √ √
= cos−1 √ =
θ = cos−1
|n1 ||n2 |
4
5 10
2
Thus, the angle of two curves at (−1, 1) are
π
4
and
3π
.
4
3. Let P (2, −2, 0), Q(0, 1, −1), R(−1, 2, 2).
a. Find the area of the triangle determined by the points P, Q
and R.
b. Find a unit vector perpendicular to plane P QR.
[like §12.4 #18]
Sol.
a. Note that
−*
P Q = h−2, 3, −1i = −2i + 3j − k
−*
P R = h−3, 4, 2i = −3i + 4j + 2k
Also, we have that
i j
k −* −* P Q × P R = −2 3 −1 = 10i + 7j + k
−3 4
2 Hence the area of the triangle determined by the points
P, Q are R is
√
1 −* −*
1√
5 6
area ∆P QR = |P Q × P R| =
100 + 49 + 1 =
2
2
2
−* −*
−*
−*
b. Since P Q × P R is perpendicular to both P Q and P R, so
the unit vector perpendicular to plane P QR is of the form
−* −*
PQ × PR
1
n = ± −* −* = ± √ (10i + 7j + k)
5 6
|P Q × P R|
√
√
√
6
7 6
6 = ±
i+
j+
k
3
30
30
4. If u 6= 0 and if u × v = u × w and u · v = u · w, then does
v = w? Given reasons for your answer.
[like §12.4 #34]
Sol. Yes. Since u×v = u×w and u·v = u·w, so we have that
u×(v−w) = 0 and u·(v−w) = 0. Since u×(v−w) = 0, so u
and w − w are parallel. Suppose that v − w 6= 0, so v − w = ku
for some nonzero constant k. Thus
0 = u · (v − w) = u · ku = k|u|2
Since k is nonzero, we have that |u|2 = 0 ⇒ u = 0, which leads
to a contradiction. Therefore, we have that v−w = 0 ⇒ v = w.
5. Find the point of intersection of the lines x = 2t + 1, y = −t +
2, z = 4t + 3, and x = 2s + 2, y = s + 3, z = 4s + 5, and then
find the plane determined by these lines.
[like §12.5 #28]
Sol. Let L1 : x = 2t + 1, y = −t + 2, z = 4t + 3, −∞ < t < ∞
and L2 : x = 2s + 2, y = s + 3, z = 4s + 5, −∞ < s < ∞. Solve
the system of equations
2t + 1 = 2s + 2
−t + 2 = s + 3
4t + 3 = 4s + 5
we get t = − 14 , s = − 34 and hence x = 12 , y = 94 , z = 2.
So L1 , L2 intersect at the point P ( 21 , 94 , 2). Let E be the plane
determined by L1 and L2 , then we have that P ∈ E. Note that
the parallel vector corresponds to L1 and L2 are l1 = 2i − j + 4k
and l2 = 2i + j + 4k respectively. So the normal vector of E is
of the form
i
j k n = l1 × l2 = 2 −1 4 = −8i + 4k
2
1 4 Hence, the equation of E is of the form −8 x− 12 +4(z −2) = 0.
That is,
E : 2x − z + 1 = 0
6. Find a plane through the points P1 (2, 1, −1), P2 (1, −2, 1) and
perpendicular to the plane 3x − y + 2z = 6.
[like §12.5 #32]
Sol. Let E be such plane. And let E1 : 3x − y + 2z = 6, which
has normal vector as n1 = 3i−j+2k. Since E ⊥ E1 , so we have
that n1 is parallel to E. On the other hand, since P1 , P2 ∈ E,
−*
so P1 P2 = −1i − 3j + 2k ∈ E. Thus the normal vector of E is
of the form
i
j
k
−*
n = n1 × P1 P2 = 3 −1 2 = 4i − 8j − 10k
−1 −3 2 Hence the equation of E is of the form 4(x − 2) − 8(y − 1) −
10(z + 1) = 0. That is,
E : 2x − 4y − 5z − 5 = 0
7.
a. Find the distance from the point (2, 1, −2) to the line x =
2t + 1, y = t, z = 2t.
b. Find the distance from the line x = 3 + t, y = 1 + t, z =
− 12 − 12 t to the plane x + 2y + 6z = 11.
[like §12.5 #36,46]
Sol.
a. Let P (2, 1, −2) and L : x = 2t+1, y = t, z = 2t, −∞ < t <
∞. The parallel vector of L is v = 2i + j + 2k. Note that
−*
the line L passing through the point S(1, 0, 0), so P S =
−i − j + 2k. Then the distance form P to L is
j k i
−1 −1 2 −*
2
1 2 | P S × v|
d(P, L) =
=p
|v|
(2i + j + 2k) · (2i + j + 2k)
√
√
| − 4i + 6j + 1|
16 + 36 + 1
53
=
=
=
3
3
3
1
1
b. Let L : x = 3 + t, y = 1 + t, z = − 2 − 2 t, −∞ < t <
∞ and E : x + 2y + 6z − 11 = 0. Note that L passing
through the point P 3, 1, − 12 and E passing through the
point S(11, 0, 0). Also, the normal vector of E is n =
i + 2j + 6k and the parallel vector of L is l = i + j − 12 k.
Since
n·l=1+2−3=0⇒n⊥l
so we have that L is parallel to E. Thus
−* n d(L, E) = d(P, E) = P S ·
|n| 1
1
= (8i − j + k) · √
(i + 2j + 6k)
2
1 + 4 + 36
√
1
9 41
= √ |8 − 2 + 3| =
41
41
8. Find parametrization for the line in which the planes 3x − 4y −
z = 3 and 2x + 3y − 2z = 2 intersect.
[like §12.5 #58]
Sol. Let E1 : 3x − 4y − z − 3 = 0, E2 : 2x + 3y − 2z − 2 = 0.
Then E1 and E2 have normal vectors n1 = 3i − 4j − k and
n2 = 2i + 3j − 2k respectively. Let L = E1 ∩ E2 be such line.
Then we have that L ⊥ n1 and L ⊥ n2 . Thus, the parallel
vector of L is
i
j
k
l = n1 × n2 = 3 −4 −1 = 11i + 4j + 17k
2
3 −2 Let z = 0 and solve the system of equations
3x − 4y − 3 = 0
2x + 3y − 2 = 0
we get x = 1, y = 0. This shows that the point P (1, 0, 0) ∈
E1 ∩ E2 = L. Hence the parametric equation for the line L is
of the form
L : x = 1 + 11t, y = 4t, z = 17t, −∞ < t < ∞