HW1 Due 6/30
1. Let ~x = (1, −1, 2), ~y = (3, 1, −1). Find:
6~x, ~x + ~y , ~x − ~y , |~x + ~y |, |~x − ~y |, |~x|, |~y |, ~x · ~y , the angle between ~x and ~y .
√
√
6~x = (6, −6,
0, 1), ~x − ~y = (−2, −2, 3), |~x + ~y | = 42 + 1 = 17,
√ 12), ~x +√~y = (4, √
|~x − ~y | = 17, |~x| = 6, |~y | = 11, ~x · ~y = 3 − 1 − 2 = 0, the angle between ~x and ~y is
π/2 since the dot product is 0.
2. Set ~x = (1, 1, 1), ~y = (0, 1, −1), ~z = (2, 0, 1).
(a) find constants a, b, c so that a~x + b~y + c~z = k.
1st way: solve the system of eqns a + 2c = 0, a + b = 0, a − b + c = 1 to get
a = 2/3, b = −2/3, c = −1/3.
2nd way: (Cramer’s rule) dot both sides by ~y × ~z = (1, −2, −2) to get
y ×~
z)
= −2/ − 3 = 2/3. Likewise you can get b, c by dotting both sides by ~x × ~z
a = k·(~
~
x·(~
y ×~
z)
and ~x × ~y .
(b) Set w
~ = (1, 0, 2). Is it possible to find constants α, β, γ so that α~x + β~y + γ w
~ = i?
No, ~x · (~y × w)
~ = 0 so the system is not solvable. Alternately, if you try to directly solve
the system of eqns you get: α + γ = 1, α + β = 0, α − β + 2γ = 0 leads to
2(α + γ) = 0 and α + γ = 1 or 0=1 so has no solutions.
3. The ~x, ~y , ~z from # 2 determine a Parallelipiped.
(a) determine the length of the main diagonal of this Parallelipiped.
√
The main diagonal is ~x + ~y + ~z = (3, 2, 1) has length 14.
(b) Find the equation of the plane passing through the tips of the three vectors.
The plane has the directions ~y − ~x = (−1, 0, −2) and ~z − ~x = (1, −1, 0) and so normal
~n = (−1, 0, 2) × (1, −1, 0) = (−2, −2, 1). It also passes through the point ~x = (1, 1, 1) say.
So the eqn is 2(x − 1) + 2(y − 1) − (z − 1) = 0 or 2x + 2y − z = 3.
(c) What is the distance from the plane containing ~x, ~y to the opposite side?
√
Scalar project ~z onto ~x × ~y = (−2, 1, 1) = ~v to get |~v · ~z|/|~v | = 3/ 6.
(d) What is the area of the triangle determined by ~x, ~y ?
The triangle is half of the parallelogram formed by ~x, ~y , so it’s area is 12 |~x × ~y | =
(e) What is the volume of the Paralleipiped? The surface area?
√
6/2.
√
The volume is base times height which is (from part (c), (d)) √36 6 = 3 which you could
also get from computing a determinant such as ~z · (~x × ~y ).
For surface area compute
√
√ the area of
√ each face with cross product: SA= 2(|~x × ~y | + |~x ×
~z| + |~y × ~z|) = 2( 6 + 6 + 3) = 4 6 + 6.
4. Let ~v = (2, 3). Find a w
~ ∈ R2 perpendicular to ~v with |w|
~ = 5.
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The vectors perpendicular to ~v are of√the form c(−3,√2) for c ∈ R\0. To find one with
length 5 we need:√5 = |c(−3,
√ 2)| = |c| 13 ⇒ c = ±5/ 13 and the two possible solutions
are w
~ = ±(−15/ 13, 10/ 13).
5. Let ~v = (v1 , v2 , v3 ) and α, β, γ denote the angles between ~v and the x, y, z-axes respectively.
Show that cos2 α + cos2 β + cos2 γ = 1.
Since v1 = ~v · i = |v| cos α and v2 = ~v · j = |v| cos β and v3 = ~v · k = |v| cos γ we have |v|2 =
v12 +v22 +v32 = |v|2 (cos2 α+cos2 β +cos2 γ) dividing by |v|2 gives cos2 α+cos2 β +cos2 γ = 1.
6. Describe the line y = 3x − 2 ⊂ R2 in
(a) the form `(t) = p + ~v t for some p~, ~v .
One way is `(t) = (0, −2) + t(1, 3) = (t, 3t − 2).
(b) the form ~n · (x, y) = c for some ~n and constant c.
2 = 3x − y = (3, −1) · (x, y).
7. (a) Parametrize the line through (1, 2, 3) and perpendicular to the yz plane.
One way is `(t) = (1, 2, 3) + t(1, 0, 0) = (1 + t, 2, 3).
(b) Write the equation of the plane perpendicular to the line in part (a) and passing
through the point (4,5,6).
The plane has equation x = 4.
8. Let P be the plane containing the points (1, 2, 0), (1, 1, 1), (0, 1, −1). Parametrize the line
that is the intersection of P with the xz-plane.
The plane is given by 2x − y − z = 0 and xz-plane is y = 0 so the intersection is 2x = z
which we can parametrize by `(t) = (t, 0, 2t).
9. Parametrize the line where the two planes x − 5y + z = 3, 2x + y − z = 0 intersect.
One point on this line is (1, 0, 2). The direction of the line is (1, −5, 1) × (2, 1, −1) =
(4, 3, 11) so one parametrization is `(t) = (1, 0, 2) + t(4, 3, 11) = (1 + 4t, 3t, 2 + 11t).
10. True or False: (explain your reasoning)
(a) Do the three points (−1, −13, 17), (3, 1, 5), (5, 8, −1) lie on the same line?
True, since (3, 1, 5) − (−1, −13, 17) = (4, 14, −12) is a multiple of (5, 8, −1) − (3, 1, 5) =
(2, 7, −6).
(b) (~u × ~v ) × w
~ = ~u × (~v × w)
~ for any vectors ~u, ~v , w.
~
False, take for example ~u = i, ~v = i, w
~ =j
(c) Two planes with normal vectors ~n1 , ~n2 (not zero) are parallel when ~n1 × ~n2 = 0.
True since the cross product being 0 means just that ~n1 = c~n2 .
(d) Any three distinct points determine a unique plane.
False because the three points could all be on a line.
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(e) Two lines `1 (t) = p1 + t~v1 , `2 = p2 + t~v2 in R2 with ~v1 6= λ~v2 must intersect.
True, because they have different slopes.
(f) Two lines `1 (t) = p1 + t~v1 , `2 = p2 + t~v2 in R3 with ~v1 6= λ~v2 must intersect.
False since they could be skew lines.
11. Show that |~x + ~y |2 − |~x − ~y |2 = 4~x · ~y . What does this tell us when the parallelogram
spanned by ~x, ~y is a rectangle?
To show it expand out the dot products: |~x + ~y |2 − |~x − ~y |2 = |~x|2 + 2~x · ~y + |~y |2 − |~x|2 +
2~x · ~y − |~y |2 = 4~x · ~y . This tells us that the parallelogram is a rectangle (~x · ~y = 0) exactly
when the diagonals have equal length (|~x + ~y | = |~x − ~y |).
12. Show that the diagonals of a rhombus are perpendicular (rhombus=diamond, a parallelogram whose sides are all equal length).
Let the sides be ~x, ~y then |~x| = |~y | and the diagonals are ~x +~y , ~x −~y . Then (~x +~y )·(~x −~y ) =
|~x|2 − |~y |2 = 0 so they are perpendicular.
13. The force F~ = (1, 2, 3) acts on a particle constrained to the plane x + y + z = 0. Find the
component of F~ that produces the particles resulting motion.
The question is asking for the component of F~ that lies in the plane. Divide F~ = F~⊥ + F~|| ,
then F~⊥ =
~ ·(1,1,1)
F
(1, 1, 1)
3
= (2, 2, 2), hence F~|| = F~ − F~⊥ = (−1, 0, 1).
14. At t = 0 a bullet moving in a straight line passes through the point (1, 2, 5) with velocity
(−1, 1, 4)smoots/sec.
(a) We are at (−3, −6, 15). Are we going to be hit? If not when and where does the bullet
pass closest to us? How close?
The bullet moves on the line `(t) = (1 − t, 2 + t, 5 + 4t) so if we got hit then we would
need 1 − t = −3 ⇒ t = 4 but then the other coords don’t match so we are safe. One way
to find closest time t0 is it is when ((−3, −6, 15) − `(t0 )) · (−1, 1, 4) = 0 which solves out
to 0 = 4 − t − 8 − t + 40 − 16t = 36 − 18t ⇒ t√
= 2 is when and where
is `(2) = (−1, 4, 13).
√
And it’s distance from us at this moment is 4 + 100 + 4 = 6 3 smoots.
(b) When will it pass through the plane x + y + z = 48, where will it pass through this
plane?
Use the parametrization from (a) in the plane equation we need to solve 1−t+2+t+5+4t =
48 ⇒ t = 10 is when and it is at `(10) = (−9, 12, 54).
(c) At t = 1 what is it’s distance to the plane x + y + z = 48?
At t = 1 it is at `(1) = (0, 3, 9) = q. A point on the plane is p√= (48, 0, 0). Scalar
p)·~
n|
projecting ~q − p~ onto ~n = (1, 1, 1) gives the distance: |(~q−~
= 36/ 3
|~
n|
15. This exercise will show the Cauchy/Schwarz inequality in Rn . Let ~x, ~y ∈ Rn .
(a) Set a = |~y |2 , b = 2~x · ~y and c = |~x|2 . Now for t ∈ R, expand 0 ≤ |~x + t~y |2 to get a
quadratic equation in t like: at2 + bt + c ≥ 0.
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(b) Use the quadratic formula to conclude that b2 − 4ac ≤ 0.
Since the quadratic equation
in part (a) is always positive it has at most one root, so the
√
−b± b2 −4ac
must have b2 − 4ac ≤ 0.
quadratic formula
2a
(c) Plug back in the values from (a) for a, b, c and deduce |~x·~y | ≤ |~x||~y | (the Cauchy/Schwarz
inequality).
Plugging in gives 4(~x · ~y )2 − 4|~x|2 |~y |2 ≤ 0 ⇒ (~x · ~y )2 ≤ |~x|2 |~y |2 . Now take sqrt to get
|~x · ~y | ≤ |x||y|.
16. Find the distance from the line `1 (t) = t(1, −1, 0) to the line `2 (t) = (4, 1, 1) + t(1, 2, 3).
Scalar project ~v = (4, 1, 1) onto the vector perpendicular
√ to both lines ~n = (1, −1, 0) ×
√
(1, 2, 3) = (−3, −3, 3) to get |~v · ~n|/|~n| = 12/3 3 = 4/ 3.
17. Find a formula for the distance from the sphere x2 + y 2 + z 2 = r2 to a line `(t) = p + t~v .
First find the distance from the line to the origin. Using projections: Let x = |~p|~v·~v| | be the
scalar projection of p~ onto ~v . Then d2 + x2 = |p|2 where
q d is the distance from the line to
2
the origin. So the distance to the sphere is d − r = |~p|2 − |~p|~v·~v|2| − r if this is positive. If
this is negative it means the line passes through the sphere and the distance to the sphere
is 0.
18. If two vectors ~a, ~b are sides of an equilateral triangle express the center of the equlateral
triangle as a combination of ~a, ~b.
The center is given by
~a+~b
.
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(There are a variety of ways you could get this)
19. Sketch/describe the following surfaces (as best you can):
(a) x2 + y 2 − z 2 = 0.
Horizontal slices are circles, vertical slices are lines (this thing is a cone):
(b) x2 + y 2 − z 2 = 1
Horizontal slices are circles, vertical slices are hyperbolas (this thing is called a hyperboloid):
(c) x = yz
The horizontal slices are lines, contour plot looks like:
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and so it is kindof like a helix, it is hard to draw but here is my try:
(d) r = 2 cos θ (cylindrical coordinates)
Multiply both sides by r gives r2 = 2r cos θ or x2 + y 2 = 2x or (x − 1)2 + y 2 = 1 which is
a cylinder with central axis (1, 0, z)
(e) tan φ = 1 (spherical coordinates)
This is the same as φ = π/4 which is the top half of a cone.
(f) ρ = cos 2θ (spherical coordinates).
Squaring both sides and switching to cylindrical this is r2 + z 2 = cos2 2θ.
Horizontal slices give the contour map:
here is my attempt to draw it:
20. Compute the partials for the following functions:
(a) f (x, y) = xex
fx = ex
2 +y 2
2 +y 2
2 +y 2
+ 2x2 ex
, fy = 2xyex
2 +y 2
.
(b) f (x, y) = x cos xy cos y
2
fx = cos xy cos y − xy sin xy cos y, fy = xy2 sin xy cos y − x cos xy sin y.
√
(c) f (x, y, z) = z y + 2(x2 + y 2 ) log 1 + xz
2
2
2
2
+y
+y
fx = 2x log(1 + xz) + z x1+xz
, fy = z y log z + 2y log(1 + xz), fz = yz y−1 + x x1+xz
( 2
x
y 6= 0
21. Let f (x, y) = y
.
0 y=0
(a) Compute df0,0~v = limt→0
For v2 6= 0: limt→0
For v2 = 0: limt→0
f (t~v )−f (0,0)
t
f (t~v )−f (0,0)
t
f (t~v )−f (0,0)
t
= limt→0
= limt→0
for ~v = (v1 , v2 ).
t2 v12 /tv2
t
0
t
= v12 /v2
= 0.
(b) Do you think this surface has a tangent plane at (0,0)? Why or why not?
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No, if it did this plane would contain the points (0, 0, 0), (v1 , 0, 0), (v1 , v2 , v12 /v2 ) for any
v1 , v2 but these points do not lie in a plane (we could take v1 = 1, v2 = 0 or v1 = 0, v2 = 1
or v1 = v2 = 1 then we need a plane with the points (0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 1).
22. Where does the plane tangent to z = ex−y at (1,1) intersect the z-axis?
The tangent plane is: (x − 1) − (y − 1) − (z − 1) = 0 or x − y − z = −1. The z-axis is
where x = y = 0, so −z = −1 or z = 1. Hence it intersects the z-axis at (0, 0, 1).
23. Use linear approximation with a suitable function to estimate the following:
(a) (0.99e0.02 )8
Take f (x, y) = (xey )8 , then the tangent plane to z = f (x, y) at x = 1, y = 0 is:
8(x − 1) + 8y − (z − 1) = 0 or z = 8(x − 1) + 8y + 1. This plane is close to f (x, y)
near x = 1, y = 0 so instead of evaluating f (0.99, 0.02) we plug in x = 0.99, y = 0.02
108
into the plane and get: z = 8(−.01) + 8(0.02) + 1 = 100
= 1.08. Which is close to
0.02 8
(0.99e ) = 1.08285...
(b) (0.99)3 + (2.01)3 − 6(0.99)(2.01)
Take f (x, y) = x3 + y 3 − 6xy, then tangent plane to z = f (x, y) at x = 1, y = 2 is:
−9(x − 1) + 6(y − 2) − (z + 3) = 0 or z = −9(x − 1) + 6(y − 2) − 3, evaluating at
x = 0.99, y = 2.01 gives: z = 9/100 + 6/100 − 3 = −285/100 = −2.85. The actual value
is −2.8485.
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24. Let f (x, y) = yexy .
(a) Find a normal vector to the graph z = f (x, y) when x = 1, y = 2.
2
2
2
The normal is given by (fx , fy , −1)|1,2 = (y 3 exy , exy + 2xy 2 exy , −1)|1,2 = (8e4 , 9e4 , −1).
(b) Find the equation for the tangent plane to this graph when x = 1, y = 2.
The tangent plane is: 8e4 (x − 1) + 9e4 (y − 1) − (z − 2e4 ) = 0.
(c) At what point on the parabaloid z = x2 + y 2 is the tangent plane parallel to the plane
in (b)?
A point (x, y, x2 + y 2 ) on the parabaloid has tangent plane with normal (2x, 2y, −1) for
this plane to be parallel to the plane in (b) the normals must point in the same direction
so: (2x, 2y, −1) = λ(8e4 , 9e4 , −1). We can solve this with λ = 1, x = 4e4 , y = 29 e4 .
25. Parametrize the line contained in the tangent plane to z = x2 y 3 at (1,2) that passes
through (1,3,20) and above (2,1,0).
The tangent plane is: 16(x − 1) + 12(y − 2) − (z − 8) = 0.
Since the line is in the tangent plane, when the line passes above (2,1,0) it has z-coordinate
satisfying 16(2 − 1) + 12(1 − 2) − (z − 8) = 0 or z = 12. Hence our line passes through
the points (1,3,20) and (2,1,12) so we can parametrize by
`(t) = (1, 3, 20) + t(−1, 2, 8) = (1 − t, 3 + 2t, 20 + 8t).
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