Math
B6C – Chapter
10 Quiz
*
Solutions*
1. Find
a unit vector in the direction of
( 3, 1 ). The y–component equals what?
Just divide this vector by it’s length, and you have a unit vector in
the same direction:
The y-component of this vector is thus
2. A
= (–1, 2) and
B = (2, 3). The vector
projection of A onto B has x–component equal to what?
The vector projection of A onto B is given by:
The
x-component of this projection vector is thus
3. A
= (1, 2, 1) and
B = (1, 1, –1). Find a
unit vector in the direction of
B x A.
Its z–component equals
what?
Dividing
this vector by its length yields a unit vector in this direction:
The z-component of this vector is thus
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4. Find
the cosine of the (acute) angle between vectors A and B in problem #3.
Let
denote this angle.
Then
, so
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5. If
A = (2, –3, 5), then
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6. Find a
unit tangent vector to the curve
at the point
.
The absolute value of the x–component of this tangent vector equals
what?
, which at the given point
becomes
.
Thus a tangent vector to the curve is given by
, since it has the same slope as the tangent to the curve at the given point.
Thus a unit tangent is
, and the absolute value of the x-component of this vector is
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7. Find a
unit normal vector to the curve
at the point
.
The absolute value of the x–component of this normal vector equals
what?
We found in problem #6 that
is tangent to the curve at the
given point. Since
is perpendicular to the tangent vector (since their dot products is zero), it
must be a normal vector to the curve at the given point.
Normalizing it we get a unit normal to the curve,
, hence the absolute value of the x-component of this normal vector is
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8. Find
the area of the parallelogram determined by the vectors
A = (–3, 1, 0) and
B = (5, –2, 1).
This is just the length of the cross product of these
vectors:
Thus the area of the parallelogram is
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9. Find
the distance from the point A =
(3, 1, –2) to the line
parametrized by
.
Consider the diagram:
Let A=(3,1,–2), and P=(1,0,3) (corresponding to
taking t=0 in our parametrization).
Let vector W be in the direction of the line, which looking at the
parametrization we see can be chosen to be the vector
(1,2,–1).
The vector V is given by A–P=(2,1,–5).
We seek the distance d, which by basic trigonometry is given by
But a basic property
of the cross product states that
Thus,
and so we first must
compute the cross product.
And so,
Since 
we have
NOTE: This
can also be written as
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10. Find
the distance from the point (3,
1, –2) to the plane 5x – 2y + z = 2.
Consider the diagram:
Let q = (3, 1, –2), and let let P be the point
(1,1,–1) (any other point on the plane would do just as well).
Let
. We
seek the distance d from q to the plane.
But d is just the length of the projection of vector
onto the normal vector n.
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11.
Parametrize the line through point (1, 5, –3)
parallel to vector (–1, 3, 2).
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12. Find
the cosine of the angle between the planes
x + 3y – z = 5 and
2x – y + 5z = 3.
We seek the cosine of the angle between
the normal vectors to these curves.
As in problem #4, we compute:
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13.
Find the distance from the line of intersection of the planes in
problem #12 to the point (1,1,1).
First we’ll find a parametric representation of the
line of intersection of the planes.
The direction of this line of intersection is the direction of the
cross products of the normal vectors to the planes:
We could use this vector for our line, or the simpler
vector
(which we get by factoring –7
out of that vector). All
we need now is a point on the line of intersection of the planes.
Here’s one way to find a point on this line of intersection…
Let z=0, and solve the system of equations for x and y:
x + 3y = 5,
2x – y = 3.
The solution is x=2, y=1.
So the point (2,1,0)
satisfies both plane equations, and is thus on the line of intersection of the
two planes. Thus the line
of intersection is parametrized by
Finally, to find the distance from this line to point
(1,1,1), we proceed as in problem #9,
with
A=(1,1,1), P=(2,1,0),
W=(–2,1,1), and V=A–P=(–1,0,1).
And so,
Since
we have
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14. Convert
(1, 2, 3) from rectangular coordinates to cylindrical coordinates:
so in cylindrical coordinates
this point is given by
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15. Convert
(1, 2, 3) from rectangular coordinates to spherical coordinates.
(approximately)
We have
and
Also,
Thus
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16.
Convert the equation
from spherical coordinates
to cylindrical coordinates.
Multiplying both sides of
the equation by
(to clear the denominator), we
get:
Since
if we multiply both sides of the
equation by
we get
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