Math 126 Homework hints
Note: As I’m sure you’ve all seen in section, I make mistakes too! I can’t guarantee that everything
here will be right 100% of the time. If you think there’s an error in something, please e-mail me
and I’ll try to get a correction up. Same goes for any clarification or further questions; if I’ve made
something even more confusing, please e-mail or come to office hours to clear it up!
12.5 Part 2
Problem 4:
Recall our “lines and planes” worksheet. We solved for the angle between two planes by finding
the angle between their normal vectors. What are the normal vectors between these two planes?
They can be read off from the coefficients on the variables in the equation for the planes, so:
n1 = h1, 1, 1i
n1 = h1, 3, 2i
Now refer to the equation from the worksheet:
| cos θ| =
|n1 · n2 |
|n1 ||n2 |
Problem 7: Find equations of the planes that are parallel to the plane x + 2y − 2z = −2 and two
units away from it.
Let’s say that we want to find some plane that is parallel and two unit away from this given plane.
Re-write this first plane in the form given in the problem statement:
x + 2y − 2z + 2 = 0
So, with respect to the equation that is given to us, we have d1 = 2. The equation of the plane
that we’re looking for will look like:
x + 2y − 2z + d2 = 0
All we need to find is some d2 that satisfies the equation:
D=√
|d1 − d2 |
a2 + b2 + c2
Now plug in a = 1, b = 2, c = −2, d1 = 2. There will be two possible values of d2 , giving two
different solutions for planes that are two units away.
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12.5 Part 3
Problem 3: Find the equation of the plane that passes through the point (2, 4, 5) and contains the
line x = 5t, y = 2 + t, z = 4 − t.
In our “lines and planes” worksheet, we found how to find the equation of a plane that passes
through three different points. Note that in doing so, we formed two non-parallel vectors from
these three points, so we need to make sure that our three points don’t all lie on the same line. We
can check and see that (2, 4, 5) doesn’t lie on the line that’s been given, so use that point and any
two others on the line, then apply the process from the worksheet.
Problem 4: Find the point at which the line intersects the given plane
Check out problem 2 from the “lines and planes” worksheet.
Problem 8: (a) Find the point at which the given lines intersect
r = h1, 2, 0i + th2, −2, 2i
r = h3, 0, 2i + sh−2, 2, 0i
Let’s first write these two equations in (x, y, z) form. Note that (unforunately) both lines are
expressed as a variable r. To simplify, I’ll use equations r1 (t) and r2 (s):
r1 (t) = (1 + 2t, 2 − 2t, 2t)
r2 (s) = (3 − 2s, 2s, 2)
So, let’s set these coordinates equal to each other:
1 + 2t = 3 − 2s
2 − 2t = 2s
2t = 2
Now solve for t and s, and plug into either equation to get the point.
(b) Find an equation of the plane that contains these lines
Just like in problem 3, we need to get three points that don’t all lie on the same line. Lucky for
us, we’ve been given equations for two DIFFERENT lines, so as long as we don’t pick all of our
points from the same line, we’re good.
Alternately, all we really need to write the equation for a plane is a vector that we know is perpendicular to the plane and a point that lies in the plane. The two direction vectors for the lines
lie in the plane, so their cross product will be perpendicular. What point could lie in this plane?
Hmmm.
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12.6
Poblem 22:
The equation for a hyperboloid of one sheet whose axis is the z-axis is of the form:
x2 y 2 z 2
+ 2 − 2 =1
a2
b
c
Note that this has the “minimum” cross-section when z = 0. The problem has told us two things
(your numbers may/will be different): the top of the tower (when z = 0) should be the equation of
a circle with diameter (not radius!) 180m. At the base of the tower (when z = −500), this should
be the equation of a circle with diameter 260m. These facts together allow you to solve for a, b,
and c. I’d suggest doing the z = 0 case first.
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