10.5#17

Hi Dr. Taylor,

I've attempted this problem 8 times and my answers are
still wrong. I used the equation for the angle between planes. Am I doing
something wrong?

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Well, 142 radians is a lot of radians and 2.49 degrees is very few degrees. That should make you suspicious about which one is which.   In fact, 2.49 radians divided by π x 180 degrees = 142 degrees, whereas 142 radians divided by π x 180 degrees amounts to 8136 degrees. So: at the very least you have degrees and radians reversed.

Now let's see, you didn't show me how you computed your angles but you should have had:
cos(θ)=u.v/(|u| |v|)
          =<4,-4,-1>.<4,-4,-3>/(√(4^2+(-4)^2+(-1)^2)√(4^2+(-4)^2+(-3)^2))
          =(16+16+3)/(√(16+16+1)√(16+16+9))
          =39/(√37√45)
          =0.956
which is a number very close to 1, hence we expect that θ is close to zero.  In fact computing the arccos what we arrive at is θ=0.30 radians, which is about 17.1 degrees.

10.5#11

Following the book, I believe that I have correctly solved for the plane's
equation. The normal vector is expressed as <-2,4,1> and following the book
my specific points should lead to -2(x--4)+4(y--3)+1(z-2)=0. Am I correct
in this assumption or am I missing something?

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Nope, not correct. A parametric equation for a line looks like r(t)=u+tv, where the vector u just gives a point on the line and the vector v determines the direction of the line: the direction of the line is read from the coefficients of t.   Instead you've used the u.

practice exams

MAT267 Page

question regarding section 10.5

Professor,

I’m wondering about an equation, equation 4, given on page 569 of our textbook. Do these statements show the right relationship?

    

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The first equation is fine, it describes the line segment starting at the vector (point) r_0  and passing through to the endpoint vector (point) r_1, because r(0)=r_0 and r(1)=r_1.  Note that this equation can be rewritten as 

r(t) = (1-t)r_0 + t r_1 

      = r_0 - t r_0 + t r_1 

      = r_0 + t (r_1 - r_0)   

      = r_0 + t v   

where v= (r_1 - r_0), which is just the form of a line as we described today in lecture, and allows all real values of t, i.e. does not require 0≤t≤1.

I'm not sure what the second equation means, since I'm not sure what you mean by r_a and r_b, but it would give some line segment not necessarily beginning or ending at  r_a and r_b/


(Edit the morning after: after thinking a bit more about your question I think I understand where you were coming from: you wanted to know if 0 and 1 are special or if any numbers would do.  The answer is no, any numbers will not do and 0 and 1 are special.  The whole point is that

(1-t)u+t v 

interpolates between u and v as t goes from 0 to 1, no matter what vectors u and v are.)

10.4#5

I need help on this problem on how to find the orthogonal vectors
of 'a' and 'b'. I've tried it many times! I computed the cross product of
'a' and 'b' and then used the dot product using each vectors. Please
explain to me what I'm doing wrong.

(you can click on the image to enlarge it)

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Well, you're doing a number of things wrong, starting with your interpretation of the problem.
First of all, your supposed to find two unit vectors u^ and v^, each of which is orthogonal to both of the vectors a and b.   One thing you choose to do is to compute the cross product a x b, because this will give you a vector orthogonal to both a and b.  It looks like you made a simple arithmetic error in this computation however, because for example
 a x b = (-3i-5j+4k)x (-2i+4j+4k)
         = -12( i x j)+10( j x i)-12( i x k)-8( k x i)-20( j x k)+16( k x j)
         = -22( i x j) - 4( i x k) - 36( j x k)
         = -22 k + 4 j - 36 i
         = - 36 i + 4 j - 22 k

While this vector is orthogonal to both a and b,  it is not a unit vector.  Since the norm of this vector is ||a x b||=√(36^2+4^2+22^2)=√1796=2√449, the unit vector pointing in the direction of  a x b is
 a x b/(2√449). Of course another unit vector perpendicular to both  a and b is b x a/(2√449), which points in the opposite direction

Question 10.4#8

Hello, Dr. Taylor.

I solved this problem by projecting vector QP onto vector QR, finding the angle measure between vectors QP and QR with the arccos function, and then calculating sqrt(14)sin(theta) to find the
distance d.  However, I am also aware that using cross products rather than projections is the focus of this lesson - how can this problem be solved using cross products? Using the formula ||U x V|| = ||U|| ||V|| sin(theta), I would end up with 14sin(theta) instead of sqrt(14), since both QP and QR
have magnitudes of sqrt(14).

Thanks for your assistance.











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OK, some short answers are 1) you started on one possible right track using projections, but then you got off the trail and started bushwhacking through the weeds, and 2) that nice formula
||U x V|| = ||U|| ||V|| sin(theta)
you use comes in three pieces, it's up to you to figure out how to take the pieces you want.

But when doing these problems I think it's important to take an organized approach, and the beginning should be TO DRAW A PICTURE OF THE PROBLEM from a perspective that captures the essential features.  This three-D plot captures EVERYTHING:

but maybe it's too much information?  This simpler diagram says what we need to know:

See that right triangle? You know the size of the hypotenuse, and the piece || to QR is the projection so you know it's size.  The other piece is the answer you need.  You can get that in several ways. Hint: cos^2(x)+sin^2(x)=1. 

Lecture Notes 8/29/16-9/2/16

Lecture Notes 8/29/16

Lecture Notes 8/31/16

Lecture Notes 9/2/16