Hi all, I am having a hard time trying to understand the wording of the following questions.
First of all, for part b), I am really confused by what it means by "in the opposite direction of z perpendicular to z". We're talking about only one vector here so how can one vector both in the opposite direction of z and perpendicular to z at the same time?
For c, again I don't really understand the question, especially the part when it says "this pattern is likewise to the vector v and w". It is very vague to me here. Could someone please further elaborate on it?
Question 12 is another whole new level of confusion. If the vector exists in the first quadrant which by definition means that the angle it makes with the positive x-axis is from 0 to 90 degrees, how can it also make a 60 degree with the x-axis?
Thank you so much for all the help!
View attachment 28315
11(b) is expressed poorly... so poorly that I can't be sure what it means. However, there is some value in it. The vector
z that it gives uniquely defines a family of parallel planes, all of which are perpendicular to
z. Any vector
v on any of these planes will have the property that
z . v = 0 and you seek a vector
v with |
v| = 3. There are infinitely many such vectors and this can't be reduced to a single answer without something to specify the plane and the direction of
v on that plane. The opposite direction bit might be an attempt to do that, but it doesn't make sense. The vector in the opposite direction of
z is, of course, -
z, but there can be no
v in that direction (parallel to -
z) as -
z is also perpendicular to all of the planes on which
v may lie.
11(c) is even more poorly worded. From the answer, you are clearly needing to establish that the angle between
u and
z and the angle between
v and
w have the same cosine, and thus the same angle. I notice that
u clearly makes a 45 degree angle to the
x-axis. The bit that "
u is the position vector of
z who arises from the tip of
v" seems to be saying that
z =
u +
v. I am guessing that the "likewise" bit made sense to the writer (who knew the answer) but is just not obvious without a diagram. Based on past exams and experience, I wonder if this is meant to be proving some geometric property. For example, if it means ) is the origin and
A,
B, and
C are in the first quadrant such that
OA =
v (a general vector),
AB =
u (angled at 45 degrees to the horizontal),
OB =
z =
v +
u, and
C positioned so that
OC =
w and
CB =
v... but this gives
z =
w +
v and thus that
w =
u, making
OABC a parallelogram with
OB and
AC as diagonals. We could alternatively place
C on the same side of
OB as
A but in either case you need to place
B such that
OA and
BC are equal and parallel. If
A and
C are placed on the same side of
OB, then the parallelogram is
OACB with
OC and
AB as diagonals. This has not established a pair of equal angles to match our vector result, however, and guess at interpretation of the question could be totally wrong.
HOWEVER, and no matter whether I am correct in this interpretation or not, it is really poorly / confusingly expressed.
12. This is the only reasonable question here, though it is again expressed. Remember that a vector gives a magnitude and direction but not a starting point and can be moved anywhere so long as the magnitude and direction are unchanged.
Consider this question as:
A and
B are points in the first quadrant such that |
AB| = 2.
CD is a vector that crosses he negative x-axis, making an angle of 60 degrees. Find a vector
AB if
AB =
CD.
You only need to find
CD (
i plus sqrt{3}
j), as any vector can exist in any quadrant, depending on where it starts and
AB =
CD. Had the question been "find A and B such that the vector AB ..." then you could say "Taking
A at (1, 1) in the first quadrant, we then have
OA = i +
j, and so it follows that
OB =
OA +
AB = 2
i + (1 + sqrt(3))
j, placing which places
B at (2, 1 + sqrt(3)), and these coordinates for
A and
B give a vector
AB with the required properties AND location.
